NCERT Exemplar Problems Class 12 Biology Chapter 16 Environmental Issues

Multiple Choice Questions
Single Correct Answer Type
1. Non-biodegradable pollutants are created by
(a) Nature (b) Excessive use of resources
(c) Humans (d) Natural disasters
Answer. (c) Non-biodegradable pollutants are created by humans.

2. According to the Central Pollution Control Board, particles that are responsible for causing great harm to human health are of diameter
(a) 2.50 micrometers (b) 5.00 micrometers
(c) 10.00 micrometers (d) 7.5 micrometers
Answer. (a) According to CPCB (Central Pollution Control Board) particulate size 2.5 micrometer or less in diameter (PM 2.5) are responsible for causing the greatest harm to human health.

3. The material generally used for sound proofing of rooms like a recording studio and auditorium, etc. is
(a) Cotton (b) Coir (c) Wood (d) Styro foam
Answer. (d) Styro foam is a kind of expanded polystyrene used especially for making food containers used for sound proofing of rooms like a recording studio and auditorium, etc.

4. Compressed Natural Gas (CNG) is
(a) Propane (b) Methane (c) Ethane (d) Butane
Answer. (b) Compressed Natural Gas (CNG) is methane.

5. World’s most problematic aquatic weed is
(a) Azolla (b) Wolffia (c) Eichhomia (d) Trapa
Answer. (c) The world’s most problematic aquatic weed is water hyacinth (Eichhornia crassipes) also called Terror of Bengal. It were introduced into India for
their lovely flowers (mauve coloured) and shape of leaves. They grow abundantly in eutrophic (nutrient rich) water bodies. Eichhornia caused havoc by their excessive growth by causing blocks in our water ways and lead to an imbalance in the ecosystem dynamics of the water body. They grow faster than our ability to remove them.

6. Which of the following causes biomagnification?
(a) SO₂ (b) Mercury
(c) DDT (d) Both (b) and (c)
Answer. (d) Biomagnification refers to increase in concentration of the toxicant at successive trophic level. This phenomenon is well known for DDT and Hg. In an aquatic food chain concentration of DDT is 0.003 ppb (parts per billion) in water but it can ultimately reaches 25 ppm in fish eating birds through biomagnification. Highest DDT deposition shall occur in birds or sea gull. High concentration of DDT disturb calcium metabolism in birds which causes:
1. Thining of egg shell.
2. Their premature breaking of egg shell.
Thus eventually causing decline in bird population.

7. The expanded form of DDT is
(a) Dichloro diphenyl trichloroethane
(b) Dichloro diethyl trichloroethane
(c) Dichlorodipyrydyltrichloroethane
(d) Dichloro diphenyl tetrachloroacetate
Answer. (a) The expanded form of DDT is dichloro diphenyl trichloroethane.

8. Which of the following material takes the longest time for biodegradation?
(a) Cotton (b) Paper (c) Bone (d) Jute
Answer. (c)

9. Choose the incorrect statement.
(a) The Montreal protocol is associated with the control of emission of ozone depleting substances
(b) Methane and carbon dioxide are greenhouse gases
(c) Dobson units are used to measure oxygen content
(d) Use of incinerators is crucial to disposal of hospital wastes
Answer. (c) The thickness of ozone is measured in terms of Dobson Units (DU). In stratosphere ozone gas is continuously-formed by the action of ultra violet rays on molecular oxygen & also degraded into molecular oxygen.

10. Among the following which one causes more indoor chemical pollution?
(a) Burning coal (b) Burning cooking gas
(c) Burning mosquito coil (d) Room spray
Answer. (a) Burning coal causes more indoor chemical pollution.

11. The green scum seen in the fresh water bodies is
(a) Blue green algae (b) Red algae
(c) Green algae (d) Both (a) and (c)
Answer. (d) The green scum seen in the fresh water bodies is due to blue green algae and green algae.

12. The loudness of a sound that a person can withstand without discomfort is about
(a) 150 dB (b) 215 dB (c) 30 dB (d) 80 dB
Answer. (d) Green muffler is related to pollution of noise. Noise pollution is measured in decibels (dB). Intensity level of whispering sound is 20-40 dB. Intensity of sound in normal conservation 30-60 dB. Day time sound level in silent zone is 50 dB. Noise become uncomfortable above 80 dB. Sound become hazardous noise pollution at level above 80 dB.

13. The major source of noise pollution, worldwide is due to
(a) Office equipment
(b) Transport system
(c) Sugar, textile and paper industries
(d) Oil refineries and thermal power plants
Answer. (b) The major source of noise pollution, worldwide is due to transport system.

14. Match correctly the following and choose the correct option:

The correct matches is
(a) i—C, ii—D, iii—A, iv—B (b) i—A, ii—C, iii—B, iv—D
(c) i—D, ii—A, iii—B, iv—C (d) i—C, ii—D, iii—B, iv—A
Answer. (a)

15. Catalytic converters are fitted into automobiles to reduce emission of harmful gases. Catalytic converters change unburnt hydrocarbons into (a) Carbon dioxide and water (b) Carbon monoxide (c) Methane (d) Carbon dioxide and methane.
Answer. (a) Catalytic converters are fitted into automobiles for reducing emission of poisonous gases.’ Catalytic converters have expensive metals namely platinum—palladium and rhodium (Pt—Pd and Rh) as catalysts. As the exaust passes through the catalytic converter :
1. Unburnt hydrocarbons are converted into CO₂ and H₂0.
2. Carbon monoxide is changed to CO₂.
3. Nitric oxide (NO) is changed to N, gas.
Vehicles equipped with catalytic converters should use unleaded petrol because lead in the petrol inactivates the catalyst.

16. Why is it necessary to remove sulphur from petroleum products?
(a) To reduce the emission of sulphur dioxide in exhaust fumes.
(b) To increase efficiency of automobiles engines.
(c) To use sulphur removed from petroleum for commercial purposes.
(d) To increase the life span of engine silencers.
Answer. (a) To reduce the emission of sulphur dioxide in exhaust fumes is necessary to remove sulphur from petroleum products.

17. Which one of the following impurities is easiest to remove from wastewater?
(a) Bacteria (b) Colloids
(c) Dissolved solids (d) Suspended solids
Answer. (d) Suspended solids impurities is easiest to remove from waste water.

18. Which one of the following diseases is not due to contamination of water?
(a) Hepatitis-B (b) Jaundice (c) Cholera (d) Typhoid
Answer. (a) Sewage from our homes as well as from hospitals are likely to contain many undesirable pathogenic micro-organisms and its disposal into a water without proper treatment may cause outbreak of serious diseases such as:
1. Dysentry, 2. Typhoid, 3. Jaundice, 4. Cholera

19. Nuisance growth of aquatic plants and bloom-forming algae in natural waters is generally due to high concentrations of
(a) Carbon (b) Sulphur (c) Calcium (d) Phosphorus
Answer. (d) Nuisance growth of aquatic plants and bloom-forming algae in natural waters is generally due to high concentrations of phosphorus.

20. Algal blooms impart a distinct colour to water.due to
(a) Their pigments
(b) Excretion of coloured substances
(c) Formation of coloured chemicals in water facilitated by physiological degradation of algae.
(d) Absorption of light by algal cell wall. 
Answer. (a) Algal blooms impart a distinct colour to water due to their pigments.

21. Match the items in column 1 and column II and choose the correct option:

The correct match is:
(a) A—ii, B—i, C—iv, D—iii (b) A—iii, B—ii, C—iv, D—i
(c) A—iii, B—iv, C—i, D—ii (d) A—iii, B—i, C—iv, D—i
Answer. (c)

22. In the textbook you came across Three Mile Island and Chernobyl disasters associated with accidental leakage of radioactive wastes. In India we had Bhopal gas tragedy. It is associated with which of the following?
(a) CO₂ (b) Methyl Iso-Cyanate
(c) CFC’s (d) Methyl Cyanate
Answer. (b) In the textbook we came across Three Mile Island and Chernobyl disasters associated with accidental leakage of radioactive wastes. In India we had Bhopal gas tragedy. It is associated with Methyl Iso-Cyanate (MIC).

Very Short Answer Type Questions
1. Use of lead-free petrol or diesel is recommended to reduce the pollutants emitted by automobiles. What role does lead play?
Answer. Anti knocking agent

2. In which year was the Air (Prevention and Control of Pollution) Act amended to include noise as air pollution?
Answer. 1987.

3. Name the city in our country where the entire public road transport runs on CNG.
Answer. Delhi.

4. It is a common practice to undertake desilting of the overhead water tanks. What is the possible source of silt that gets deposited in the water tanks?
Answer. The soil particles carried by water from the source of supply.

5. What is cultural eutrophication?
Answer. Pollutants from man’s activities like effluents from the industries and homes can radically accelerate the aging process. This phenomenon has been called Cultural or Accelerated Eutrophication.

6. List any two adverse effects of particulate matter on human health.
Answer. When particulate matter inhaled deep into the lungs they cause:

1. Breathing and respiratory symptoms
2.  Irritation
3.  Inflammations
4. Damage to the lungs and premature deaths.

7. What is the raw material for polyblcnd?
Answer. Plastic waste.

8. Blends of polyblend and bitumen, when used, help to increase road life by a factor of three. What is the reason?
Answer. Blends of polyblend and bitumen, when used to lay roads, enhanced the bitumen’s water repellant properties, hence helped to increase road life by a factor of three.

9. Mention any two examples of plants used as wind breakers in the agricultural fields.
Answer. Acacia and Neem.

10. Name an industry which can cause both air and thermal pollution and as well as eutrophication.
Answer. Fertiliser factory.

11. What is an algal bloom?
Answer. Presence of large amounts of nutrients in waters causes excessive growth of planktonic (free-floating) algae, called an algal bloom.

12. What do you understand by biomagnification?
Answer. Biomagnification refers to increase in concentration of the toxicant at successive trophic levels.

13. What are the three major kinds of impurities in domestic wastewater?
Answer. 1. Suspended solids, e.g., Sand, silt and clay.
2. Colloidal materials, e.g., Fecal matter, bacteria, cloth and paper fibres.
3. Dissolved materials, e.g., Nutrients (nitrate, ammonia, phosphate, sodium, calcium).

14. What is reforestation? 
Answer. Reforestation is the process of restoring a forest that once existed but was removed at some point of time in the past.

15. What is the best solution for the treatment of electronic wastes?
Answer. Recycling.

Short Answer Type Questions
1. Is it true that carpets and curtains/drapes placed on the floor or wall surfaces can reduce noise level. Explain briefly.
Answer. Yes, carpets and curtains/drapes placed on the floor or wall surfaces can reduce noise level because these are sound absorber substances.

2. What is hybrid vehicle technology? Explain its advantages with a suitable example?
Answer. Vehicles running on dual mode like petrol and CNG are hybrid vehicle. As CNG is a green fuel there is conservation of fossil fuels and reduction in the environmental pollution.

3. Is it true that if the dissolved oxygen level drops to zero, the water will become septic? Give an example which could lower the dissolved oxygen content of an aquatic body.
Answer. Yes, the water become septic if the dissolved oxygen drops to zero. Organic pollution (biodegradable) is an example.

4. Name any one greenhouse gas and its possible source of production on a large scale. What are the harmful effects of it?
Answer. CO₂ and Methane. CO₂ levels are increasing due to burning of fossil fuels, leading to Global Warming.

5. It is a common practice to plant trees and shrubs near the boundary walls of buildings. What purpose do they serve?
Answer. The plants growing near the boundary wall act as barriers for sound pollution and act as dust catchers.

6. Why has the National Forest Commission of India recommended a relatively larger forest cover for hills than for plains?
Answer. The National Forest Commission of India recommended a relatively larger forest cover for hills than for plains to reduce the chances of landslides.

7. How can slash and bum agriculture become environment friendly?
Answer. Slash and bum agriculture, commonly called as Jhum cultivation in the north-eastern states of India, has also contributed to deforestation. In slash and bum agriculture, the farmers cut down the trees of the forest and bum the plant remains. The ash is used as a fertilizer and the land is then used for farming or cattle grazing. After cultivation, the area is left for several years so as to allow its recovery. The farmers then move on to other areas and repeat this process. In earlier days, when Jhum cultivation was in prevalence, enough time-gap was given such that the land recovered from the effect of cultivation. With increasing population, and Treated cultivation, this recovery phase is done away with, resulting in deforestation.

8. What is the main idea behind “Joint Foregt Management Concept” introduced by the Government of India?
Answer. Realising the significance of participation by local communities, the Government of India in 1980s has introduced the concept of Joint Forest Management (JFM) so as to work closely with “the local communities for protecting and managing forests. In return for their services to the forest, the communities get benefit of various forest products (e.g., fruits, gum, rubber, medicine, etc.), and thus the forest can be conserved in a sustainable manner.

9. What do you understand by Snow-blindness?
Answer. In human eye, cornea absorbs UV-B radiation, and a high dose of UV-B causes inflammation of cornea, called snow-blindness.

10. How has DDT caused decline in bird population?
Answer. High concentrations of DDT disturb calcium metabolism in birds, which causes thinning of eggshell and their premature breaking, eventually causing decline in bird populations.

11. Observe the figure A and B given below and answer the following questions

(i) The power generation by the above two methods is non-polluting. True/False
(ii) List any two applications of solar energy.
(iii) What is a photovoltaic cell?
Answer. (i) True.
(ii) (a) Electricity generation (b) Solar light
(iii) A photovoltaic cell is a semiconductor diode that converts visible light into direct current (DC).

Long Answer Type Questions
1. Write a short note on electronic waste. List the various sources of e-wastes and the problems associated with its disposal.
Answer. Discarded unusable electronic gadgets such as computers, mobile phones, circuits, television sets, etc., form electronic waste. These contain harmful toxic substances like heavy metals to which the unskilled manual workers are directly exposed.

2. What is organic farming? Discuss the benefits of organic fanning as a viable practise in the context of developing nations like India.
Answer. The use of Biofertilisers in agriculture is called organic farming. Biofertilisers are organisms that enrich the nutrient quality of the soil. The main sources of biofertilisers are bacteria, fungi and cyanobacteria. The nodules on the roots of leguminous plants formed by the symbiotic association of Rhizobium. These bacteria fix atmospheric nitrogen into organic forms, which is used by the plant as nutrient. Other bacteria can fix atmospheric nitrogen while free- living in the soil (examples Azospirillum and Azotobacter), thus enriching the nitrogen content of the soil.
Currently, in our country, a number of biofertilisers are available commercially in the market and farmers use these regularly in their fields to replenish soil nutrients and to reduce dependence on chemical fertilisers.

3. Water logging and soil salinity are some of the problems that have come in the wake of the Green Revolution. Discuss their causes and adverse effects to the environment.
Answer. Water logging and soil salinity are some of the problems that have come in the wake of the Green Revolution. Irrigation without proper drainage of water leads to water logging in the soil. Besides affecting the crops, water logging draws salt to the surface of the soil. The salt then is deposited as a thin crust on the land surface or starts collecting at the roots of the plants. This increased salt content is inimical to the growth of crops and is extremely damaging to agriculture.

4. What are multipurpose trees? Give the botanical and local names of any two multipurpose trees known to you and list their uses.
Answer. Multipurpose trees are that which are grown for more than one purpose or benefits. They may provide food in the form of fruit, seeds, or roots; and also provides firewood, medicines, fibres, etc.
(i) Neem (Azadirachta indicd)— Used as insect repellent, antibiotic, firewood and windbreaks.
(ii) Coconut palm (Cocos nucifera)— Used for food, purified water (juice from inside the coconut), roof thatching, firewood, shade and rope making.

5. What are the basic characteristics of a modem landfillsite. List any three and also mention the reasons for their use.
Answer. Characteristics of a modem landfill include:

1. Methods to contain leachate such as lining clay or plastic liners.
2. Compaction and covering of the waste to prevent it from being blown by wind.
3. Installation of a landfill gas extraction system to extract the gas for use in generation of power.

6. How does an electrostatic precipitator work?
Answer. Electrostatic precipitator has electrode wires that are maintained at several thousand volts, which produce a corona that releases electrons. These electrons attach to dust particles giving them a net negative charge. The collecting plates are grounded and attract the charged dust particles. The velocity of air between the plates must be low enough to allow the dust to fall.

7. Observe figure and answer the followiag,questions.
(i) What ecological term is used to describe the DDT accumulation at different trophic levels?
(ii) list any one effect of DDT accumulation on birds.
(iii) Will DDT accumulation lead to eutrophication?
Does it affect the BOD?
Name disease caused by accumulation of any heavy metal.
Answer. (i) Biomagnification
(ii) High concentrations of DDT disturb, calcium metabolism in birds, which causes thinning of eggshell and their premature breaking, eventually causing decline in bird populations.
(iii) Yes.
(iv) Yes, it increases the BOD.
(v) Disease caused by eating fish found in water contaminated with industrial waste having mercury is Minamata disease. Cd pollution associated with disease is called “Itai-Itai”. Arsenic (As) poisoning causes “Black foot disease”.

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NCERT Exemplar Problems Class 12 Mathematics Chapter 1 Relations and Functions

Short Answer Type Questions

Long Answer Type Questions

Objective Type Questions

Fill in the Blanks Type Questions

True/False Type Questions

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NCERT Exemplar Problems Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions

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True/False Type Questions

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NCERT Exemplar Problems Class 12 Mathematics Chapter 3 Matrices

Short Answer Type Questions

Long Answer Type Questions

Objective Type Questions

Fill In the Blanks Type Questions

True/False Type Questions
82. A matrix denotes a number.
Sol. False
A matrix is an ordered rectangular array of numbers of functions.
83. Matrices of any order can be added.
Sol. False
Two matrices are added, if they are of the same order.
84. Two matrices are equal if they have same number of rows and same number of columns.
Sol. False
If two matrices have same number of rows and same number of columns, we cannot say two matrices are equal as their corresponding elements may be different.
85. Matrices of different order cannot be subtracted.
Sol. True
Two matrices of same order can be subtracted
86. Matrix addition is associative as well as commutative.
Sol. True
Matrix addition is associative as well as commutative i.e.,
(A + B) + C = A + (B + C) and A + B = B + A, where A, B and C are matrices of same order.
87. Matrix multiplication is commutative.
Sol. False
If AB is defined, it is not necessary that BA is defined.
Also if AB and BA are defined, it not”necessary that they have same order. Further if AB and BA are defined and have same order, it is not necessary their corresponding elements are equal.
So, in general AB^BA
88. A square matrix where every element is unity is called an identity matrix.
Sol. False
Since, in an identity matrix, the diagonal elements are one and rest are all zero.
89. If A and B are two square matrices of the same order, then A + B = B + A.
Sol. True
Since, matrix addition is commutative i.e., A + B = B +A, where A and B are two square matrices.
90. If A and B are two matrices of the same order, then A-B = B-A.
Sol. False
A-B = -(B-A)
Thus A-B≠B-A
However when A – B = B – A
A-B = 0 or A =B

92. Transpose of a column matrix is a column matrix. False
Sol.Transpose of a column matrix is a row matrix.
93. If A and B are two square matrices of the same order, then AB = BA. False
Sol.For two square matrices of same order it is not always true that AB = BA.

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NCERT Exemplar Problems Class 12 Mathematics Chapter 4 Determinants

Short Answer Type Questions
Direction for Exercises 1 to 6: Using the properties of determinants,

Long Answer Type Questions

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Fill In the Blanks Type Questions

True/False Type Questions

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NCERT Exemplar Problems Class 12 Mathematics Chapter 5 Continuity and Differentiability

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True/False Type Questions

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NCERT Exemplar Problems Class 12 Mathematics Chapter 6 Application of Derivatives

Short Answer Type Questions

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Objective Type Questions

Fill In the Blanks Type Questions
Fill in the blanks in each of the following exercises 60 to 64.

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NCERT Exemplar Problems Class 12 Mathematics Chapter 7 Integrals

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NCERT Exemplar Problems Class 12 Mathematics Chapter 8 Applications of Integrals

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Objective Type Questions

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NCERT Exemplar Problems Class 12 Mathematics Chapter 9 Differential Equations

Short Answer Type Questions

2. Find the differential equation of all non-vertical lines in a plane.

13. Form the differential equation having y = (sin^-1 x)²+ A cos^-1 x + B, where A and B are arbitrary constants, as its general solution.

14. Form the differential equation* of all circles which pass through origin and whose centres lie on y-axis.

19. Solve : (x + y)(dx -dy) = dx + dy
[Hint: Substitute x+y = z after separating dx and dy]

21. Solve the differential equation dy = cos x (2 – y cosec x) dx given that y=2 when x = π /2.

22. Form the differential equation by eliminating A and B in Ax² -By² = 1.

23. Solve the differential equation (1 +y²) tan^-1 x dx + 2y (1+x²)dy=0.

24. Find the differential equation of system of concentric circles with centre (1,2).

Long Answer Type Questions





31. Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point (x, y) is equal to the square of the difference of the abscissa and ordinate of the point.

32. Find the equation of a curve passing through the point (1, 1). If the tangent drawn at any point P(x, y) on the curve meets the co-ordinate axes at A and B such that P is the mid-point of AB.

Objective Type Questions

60. Family y = Ax + A3 of curves will correspond to a differential equation of order ,
(a) 3 (b) 2 (c) 1 (d) not defined

62. The curve for which the slope of the tangent at any point is equal to the ratio of the abscissa to the ordinate of the point is
(a) an ellipse (b) parabola
(c) circle (d) rectangular hyperbola

Fill in the Blanks Type Questions

True/False Type Questions

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NCERT Exemplar Problems Class 12 Mathematics Chapter 10 Vector Algebra

Short Answer Type Questions


3. Find a unit vector in the direction of PQ, where P and Q have co-ordinates (5,0, 8) and (3, 3,2), respectively.


5. Using vectors, find the value of k such that the points (k, -10, 3), (1, -1, 3) and (3, 5, 3) are collinear.




13. Using vectors, find the area of the triangle ABC with vertices A(l, 2, 3), 5(2, -1, 4) and C(4,5,-l).

14. Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.

Long Answer Type Questions

Objective Type Questions

Fill in the Blanks Type Questions

True/False Type Questions

42. Position vector of a point P is a vector whose initial point is origin.

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NCERT Exemplar Problems Class 12 Mathematics Chapter 11 Three Dimensional Geometry

Short Answer Type Questions


5. Prove that the line through A(0, -1, -1) and B(4, 5, 1) intersects the line through C(3, 9,4) and D(- 4,4,4).
Sol. We know that, the equation of a line that passes through two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is

7. Find the equation of a plane which bisects perpendicularly the line joining the points A(2, 3,4) and 5(4, 5, 8) at right angles.


8. Find the equation of a plane which is at a distance 3 √3 units from origin and the normal to which is equally inclined to coordinate axis.

9. If the line drawn from the point (-2, -1,-3) meets a plane at right angle at the point (1, -3, 3), find the equation of the plane. 
Sol. Since, the line drawn from the point B(-2, -1,-3) meets a plane a plane at right angle at the point 4(1,-3, 3).
So, the plane passes through the point 4(1, -3, 3)

10. Find the equation of the plane through the points (2, 1, 0), (3, -2, -2) and (3, 1, 7).
Sol. We know that, the equation of a plane passing through three non-collinear points (x1. y1, z1), (x₂, y₂, z₂) and (x₃, y₃, z₃) is



14. O is the origin and A is (a, b, c). Find the direction cosines of the line OA and the equation of plane through A at right angle to OA.

Long Answer Type Questions



19. Find the equations of the line passing through the point (3,0,1) and parallel to the planes x + 2y = 0 and 3y-z = 0.


20. Find the equation of the plane through the points (2,1, -1) and (-1,3,4), and perpendicular to the plane x-2y + 4z = 10.


22. Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y- z + 5 = 0.



27. Show that the straight lines whose direction cosines are given by 2l+2m -n=0 and mn + nl + lm = 0 are at right angles.

Objective Type Questions

30. If the directions cosines of a line are k, k, ‘k, then

33. The reflection of the point (α, β,γ) in the xy-plane is
(a) (α, β,0) (b) (0,0, γ) (c) (-α, -β, γ) (d) (α, β,-γ)
Sol. (d) In xy-plane, the reflection of the point (α, β,γ) is (α, β,-γ)
34. The area of the quadrilateral ABCD, where A(0,4,1), B(2, 3, -1), C(4, 5, 0) and D(2, 6,2) is equal to
(a) 9 sq. units (b) 18 sq. units (c) 27 sq. units (d) 81 sq. units

35. The locus represented by xy + yz = 0 is
(a) A pair of perpendicular lines (b) A pair of parallel lines
(c) A pair of parallel planes (d) A pair of perpendicular planes

Fill in the Blanks Type Questions

True/False Type Questions

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NCERT Exemplar Problems Class 12 Mathematics Chapter 12 Linear Programming

Short Answer Type Questions

4. Minimize Z = 13x- 15y subject to the constraints: x+y≤ 7,2x-3y + 6 ≥ 0, x ≥ 0, y ≥ 0.
Sol. We have to Minimize Z = 13x – 15y subject to the constraints x + y ≤7, 2x – 3y + 6 ≥ 0, x ≥ 0, y ≥ 0. These inequalities are plotted as shown in the following figure.

8. Refer to Exercise 7 above. Find the maximum value of Z.
Sol. Z is maximum at (3,2) and its maximum value is 47.

9. The feasible region for a LPP is shown in the following figure. Evaluate Z = 4x+y at each of the comer points of this region. Find the minimum value of Z, if it exists.

Now, we see that 3 is the smallest value of Z the comer point (0, 3). Note that here we see that, the region is unbounded, therefore 3 may not be the minimum value of Z. To decide this issue, we graph the inequality 4x + y < 3 and check whether the resulting open half plane has no point in common with feasible region otherwise, Z has no minimum value.
From the shown graph above, it is clear that there is no point in common with feasible region and hence Z has minimum value 3 at (0, 3).

10. In the following figure, the feasible region (shaded) for a LPP is shown. Determine the maximum and minimum value of Z = x + 2y

11. A manufacturer of electronic circuits has a stock of 200 resistors, 120 transistors and 150 capacitors and is required to produce two types of circuits A and B. Type A requires 20 resistors, 10 transistors and 10 capacitors. Type B requires 10 resistors, 20 transistors and 30 capacitors. If the profit on type A circuit is Rs 50 and that on type B circuit is Rs 60, formulate this problem as a LPP so that the manufacturer can maximise his profit.
Sol. Let the manufacture produces x units of type A circuits and y units of type B circuits. Form the given information, we have following corresponding constraint table.

12. A firm has to transport 1200 packages using large vans which can carry 200 packages each and small vans which can take 80 packages each. The cost for engaging each large van is Rs 400 and each small van is Rs 200. Not more than Rs 3000 is to be spent on the job and the number of large vans can not exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimise cost.
Sol. Let the firm has x number of large vans and y number of small vans.
From the given information, we have following corresponding constraint table.

13. A company manufactures two types of screws A and B. All the screws have to pass through a threading machine and a slotting machine. A box of Type A screws requires 2 minutes on the threading machine and 3 minutes on the slotting machine. A box of type B screws requires 8 minutes of threading on the threading machine and 2 minutes on the slotting machine. In a week, each machine is available for 60 hours. On selling these screws, the company gets a profit of Rs 100 per box on type A screws and Rs 170 per box on type B screws. Formulate this problem as a LPP given that the objective is to maximise profit.
Sol. Let the company manufactures x boxes of type A screws and y boxes of type B screws.
From the given information, we have following corresponding constraint table.

14. A company manufactures two types of sweaters: type A and type B. It costs Rs 360 to make a type A sweater and Rs 120 to make a type B sweater. The company can make at most 300 sweaters and spend at most Rs 72000 a day. The number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100. The company makes a profit of Rs 200 for each sweater of type A and ?120 for every sweater of type B. Formulate this problem as a LPP to maximise the profit to the company.
Sol. Let the company manufactures x number of type A sweaters and y number of type B.
The company spend at most Rs 72000 a day.
∴ 360x + 120y ≤ 72000
=> 3x+y≤ 600 …(i)
Also, company can make at most 300 sweaters.
∴ x+y≤ 300 …(ii)
Also, the number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100 i.e., y-x≤ 100
The company makes a profit of Rs 200 for each sweater of type A and Rs 120 for every sweater of type B
So, the objective function for maximum profit is Z = 200x + 120y subject to constraints.
3x+y≤ 600
x+y ≤ 300
x-y ≥ -100
x ≥ 0, y ≥ 0

15. A,man rides his motorcycle at the speed of 50 km/hour. He has to spend Rs 2 per km on petrol. If he rides it at a faster speed of 80 km/hour, the petrol cost increases to Rs 3 per km. He has atmost Rs120 to spend on petrol and one hour’s time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.
Sol. Let the man rides to his motorcycle to a distance x km at the speed of 50 km/h and to a distance y km at the speed of 80 km/h.
Therefore, cost on petrol is 2x + 3y.
Since, he has to spend Rs120 atmost on petrol.
∴ 2x + 3y ≤ 120 …(i)
Also, he has at most one hour’s time.

Long Answer Type Questions
16. Refer to Exercise 11. How many of circuits of Type A and of Type B, should be produced by the manufacturer so as to maximize his profit? Determine the maximum profit.

22. A manufacturer produces two Models of bikes’-Model X and Model Y. Model X takes a 6 man-hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hour available per week. Handling and Marketing costs are Rs 2000 and Rs 1000 per unit for Models X and Y respectively. The total funds available for these purposes are Rs 80,000 per week. Profits per unit for Models X and Y are Rs 1000 and Rs 500, respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.

So, for maximum profit manufacture must produces 25 number of models X and 30 number of model Y bikes.

23. In order to supplement daily diet, a person wishes to take some X and some Y tablets. The contents of iron, calcium and vitamins in X and Y (in milligrams per tablet) are given as below:

The person needs at least 18 milligrams of iron, 21 milligrams of calcium and 16 milligram of vitamins. The price of each tablet of X and Y is Rs 2 and Rs 1 respectively. How many tablets of each should the person take in order to satisfy the above requirement at the minimum cost?

From the figure, we can see the feasible region is unbounded region, with comer points as A(%, 0), B(6,1), C(I, 6), and D(0, 9)

Thus, the minimum value of Z is ‘8’ occurring at B( 1, 6). Since the feasible region is unbounded, ‘8’ may not be the minimum value of Z. To decide this, we plot the inequality 2x+y < 8 and check whether the resulting open half has points common with feasible region or not If it has common point, then 8 will not be the minimum value of Z, otherwise 8 will be the minimum value of Z. Thus, from the graph it is clear that, it has no common point.
Therefore, Z= 2x+y-has 8 as minimum value subject to the given constrains. Hence, the person should take 1 unit of X tablet and 6 units of Y tablets to satisfy the given requirements and at the minimum cost of Rs 8.

24. A company makes 3 models of calculators: A, B and C at factory I and factory II. The company has orders for at least 6400 calculators of model A, 4000 calculator of model B and 4800 calculator of model C. At factory I, 50 calculators of model A, 50 of model B and 30 of model C are made every day; at factory II, 40 calculators of model A, 20 of model B and 40 of model C are made everyday. It costs Rs 12000 and Rs 15000 each day to operate factory I and II, respectively. Find the number of days each factory should operate to minimize the operating costs and still meet the demand.

Objective Type Questions
26. The comer points of the feasible region determined by the system of linear constraints are (0,0), (0,40), (20,40), (60,20), (60,0). The objective function is Z=4x + 3y.
Compare the quantity in Column A and Column B
Column A             Column B
Maximum of Z          325
(a) The quantity in column A is greater
(b) The quantity in column B is greater
(c) The two quantities are equal
(d) The relationship can not be determined on the basis of the information supplied .
Sol. (b)

28. Refer to Exercise 27. Maximum of Z occurs at
(a) (5,0) (b) (6,5) (c) (6, 8) (d) (4, 10)
Sol. (a) Refer to solution 27, maximum of Z occurs at (5, 0)
29. Refer to Exercise 27. (Maximum value of Z + Minimum value of Z) is equal to
(a) 13 (b) 1 (c) -13
Sol. (d) Refer to solution 27,
maximum value of Z + minimum value of Z = 15-32 = -17

31. Refer to Exercise 30. Minimum value of F is
(a) 0 (b) -16 . (c) 12 (d) does not exist
Sol. (b) Referring to solution 30, minimum value of F is -16 at (0,4).

32. Comer points of the feasible region for an LPP are (0, 2), (3, 0), (6,0), (6, 8) and (0,5).
Let F = 4x + 6y be the objective function.
The Minimum value of F occurs at
(a) (0,2) only
(b) (3,0) only
(c) the mid point of the line segment joining the points (0,2) and (3,0) only
(d) any point on the line segment joining the points (0,2) and (3, 0).
Sol. (d)

33. Refer to Exercise 32, Maximum of F – Minimum of F=
(a) 60 (b) 48 (c) 42 (d) 18
Sol. (a) Referring to the solution 32, maximum,of F-minimum of F= 72- 12 = 60

34. Comer points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let Z = px + qy, where p,q>0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1,1) is
(a) p = 2q (b )P=q\2 (c)p = 3q (d) p = q
Sol. (b)

Fill in the Blanks Type Questions
35. In a LPP, the linear inequalities or restrictions on the variables are called————-.
Sol. In a LPP, the linear inequalities or restrictions on the variables are called linear constraints.

36. In a LPP, the objective function is always————.
Sol. In a LPP, objective function is always linear.

37. If the feasible region for a LPP is————, then the optimal value of the objective function Z= axH-fiy may or may not exist.
Sol. If the feasible region for a LPP is unbounded, then the optimal value of the objective function Z = ax + by may or may not exist.

38. In a LPP if the objective function Z=ax+ by has the same maximum value on two comer points of the feasible region, then every point on the line segment joining these two points give the same————value.
Sol. In a LPP, if the objective function Z = ax + by has the same maximum value on two comer points of the feasible region, then every point on the line segment joining these two points gives the same maximum value.

39. A feasible region of a system of linear inequalities is said to be————if it can be enclosed within a circle.
Sol. A feasible region of a system of linear inequalities is said to be bounded, if it can be enclosed within a circle.

40. A comer point of a feasible region is a point in the region which is the———— of two boundary lines.
Sol. A comer point of a feasible region is a point in the region which is the intersection of two boundary lines.

41. The feasible region for an LPP is always a————polygon.
Sol. The feasible region for an LPP is always a convex polygon.

True/False Type Questions
42. If the feasible region for a LPP is unbounded, maximum or minimum of the objective function Z = ax+ by may or may not exist.
Sol. True

43. Maximum value of the objective function Z = ax+ by in a LPP always occurs
at only one comer point of the feasible region. ,
Sol. False

44. In a LPP, the minimum value of the objective function Z = ax+ by is always 0 if origin is one of the comer point of the feasible region.
Sol. False

45. In a LPP, the maximum value of the objective function Z = ax+ by is always finite.
Sol. True

The post NCERT Exemplar Problems Class 12 Mathematics Linear Programming appeared first on Learn CBSE.

NCERT Exemplar Problems Class 12 Mathematics Chapter 13 Probability

Short Answer Type Questions
1. For a loaded die, the probabilities of outcomes are given as under:
P(1) = P(2) = 0.2, P(3) = P(5) = P(6) = 0.1 and P(4) = 0.3.
The die is thrown two times. Let A and B be the events, ‘same number each time’, and ‘a total score is 10 or more’, respectively. Determine whether or not/l and B are independent.


2. Refer to Exercise 1 above. If the die were fair, determine whether or not the events A and B are independent.


4. A bag contains 5 red marbles and 3 black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?


5. Two dice are thrown together and the total score is noted. The events E, F and G are ‘a total of 4’, ‘a total of 9 or more’, and ‘a total divisible by 5’, respectively. Calculate P(E), P(F) and P(G) and decide which pairs of events, if any, are independent.

6. Explain why the experiment of tossing a coin three times is said to have binomial distribution.
Sol. We know that, in a Bindmial distribution,
(i) There are 2 outcomes for each trial
(ii) There is a fixed number of trials
(iii) The probability of success must be the same for all the trials.
When coin is tossed, possible outcomes are Head and Tail.
Since coin is tossed three times, we have fixed number of trials.
Also probability of Head and Tail in each trial is 1/2.
Thus given experiment is said to have binomial distribution.





12. If X is the number of tails in three tosses of a coin, determine the standard deviation of X.


13. In a dice game, a player pays a stake of Rs 1 for each throw of a die. She receives Rs 5 if the die shows a 3, Rs 2 if the die shows a 1 or 6, and nothing otherwise. What is the player’s expected profit per throw over a long series of throws?

14. Three dice are thrown at the same time. Find the probability of getting three two’s, if it is known that the sum of the numbers on the dice was six.


15. Suppose 10,000 tickets are sold in a lottery each for Rs 1. First prize is of Rs 3000 and the second prize is of Rs 2000. There are three third prizes of Rs.500 each. If you buy one ticket, what is your expectation?

16. A bag contains 4 white and 5 black balls. Another bag contains 9 white and 7 black balls. A ball is transferred from the first bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.

17. Bag I contains 3 black and 2 white balls, Bag II contains 2 black and 4 white balls. A bag and a ball is selected at random. Determine the probability of selecting a black ball.


18. A box has 5 blue and 4 red balls. One ball is drawn at random and not replaced. Its colour is also not noted. Then another ball is drawn at random. What is the probability of second ball being blue?

19. Four cards are successively drawn without replacement from a deck of 52 playing cards. What is the probability that all the four cards are kings?

20. A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.

21. Ten coins are tossed. What is the probability of getting at least 8 heads?

22. The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?

23. A lot of 100 watches is known to have 10 defective watches. If 8 watches are selected (one by one with replacement) at random, what is the probability that there will be at least one defective watch?






31. A factory produces bulbs. The probability that any one bulb is defective is 1/50 and they are packed in boxes of 10. From a single box, find the probability that
(i) none of the bulbs is defective
(ii) exactly two bulbs are defective
(iii) more than 8 bulbs work properly

32. Suppose you have two coins which appear identical in your pocket. You know that one is fair and one is 2-headed. If you take one out, toss it and get a head, what is the probability that it was a fair coin?

33. Suppose that 6% of the people with blood group O are left handed and 10% of those with other blood groups are left handed. 30% of the people have blood group O. If a left handed person is selected at . random, what is the probability that he/she will have blood group O?

35. Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice . Determine also the mean of the distribution.




38. A and B throw a pair of dice alternately. A wins the game if he gets a total of 6 and B wins if she gets a total of 7. If A starts the game, find the probability of winning the game by A in third throw of the pair of dice.


40. An um contains m white and n black balls. A ball is drawn at random and is put back into the um along with k additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on k.

Long Answer Type Questions


43. A shopkeeper sells three types of flower seeds A₁,A₂ and A₃. They are sold as a mixture where the proportions are 4:4:2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35%. Calculate the probability
(i) of a randomly chosen seed to germinate
(ii) that it will not germinate given that the seed is of type A₃,
(iii) that it is of the type A₂ given that a randomly chosen seed does not germinate.

44. A letter is known to have come either from TATA NAGAR or from CALCUTTA. On the envelope, just two consecutive letter TA are visible. What is the probability that the letter came from TATA NAGAR.
Sol. Let Ex be the event that letter is from TATA NAGAR and E2 be the event that letter is from CALCUTTA.

45. There are two bags, one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls. A die is thrown. If it shows up 1 or 3, a ball is taken from the 1st bag; but it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball.

46. There are three urns containing 2 white and 3 black balls, 3 white and 2 black balls, and 4 white and 1 black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.


47. By examining the chest X ray, the probability that TB is detected when a person is actually suffering is 0.99. The probability of ah healthy person diagnosed to have TB is 0.001. In a certain city, 1 in 1000 people suffers from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB?

48. An item is manufactured by three machines A, B and C. Out of the total number of items manufactured during a specified period, 50% are manufactured on A, 30% on B and 20% on C. 2% of the items produced on A and 2% of items produced on B are defective, and 3% of these produced on C are defective. All the items are stored at one go down. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?
Sol. Let E₁, E₂, E₃ be the event that item is manufactured on A,B and C respectively. Let E be the event that an item is defective.






=> 124n + 62 = 126n + 42
=> 2n = 20=>n=10
53. Two cards are drawn successively without replacement from a well shuffled deck of cards. Find the mean and standard variation of the random variable X where X is the number of aces.

54. A die is tossed twice. A ‘success’ is getting an even number on a toss. Find the variance of the number of successes.


55. There are 5 cards numbered 1 to 5, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on two cards drawn. Find the mean and variance of X.

Objective Type Questions





70. If two events are independent, then
(a) they must be mutually exclusive
(b) the sum of their probabilities must be equal to 1
(c) (A) and (B) both are correct
(d) None of the above is correct
Sol. (d) If two events A and B are independent, then we know that

If A and B are independent, knowledge that A occurred does not change the probabilities that B may have occurred. Whereas if A and B are disjoint, knowledge that A occurred completely changes the probabilities that B may have occurred by collapsing it to 0.




77. Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is

True/False Type Questions

Fill in the Blanks Type Questions

The post NCERT Exemplar Problems Class 12 Mathematics Probability appeared first on Learn CBSE.

NCERT Exemplar Problems Class 12 Chemistry Chapter 1 Solid State

Multiple Choice Questions

Single Correct Answer Type
Question.1. Which of the following condition favours the existence of a substance in the solid state?
(a) High temperature (b) Low temperature
(c) High thermal energy (d) Weak cohesive forces
Solution: (b) At low temperature substance exists in solid state due to decrease in molecular motion which leads to strong cohesive forces i.e., forces which hold the constituent particles together.

Question.2. Which of the following is not a characteristic of a crystalline solid?
(a) Definite and characteristic heat of fusion
(b) Isotropic nature
(c) A regular periodically repeated pattern of arrangement of constituent particles in the entire crystal
(d) A true solid
Solution: (b) Anisotropy: Crystal¬line solids are anisotropic in nature, that is some of their physical properties like electrical resistance or re¬fractive index show different values when measured along different directions in the same crystal. This arises from different arrangement of particles in different di¬rections arrangement of particles along different directions

Isotropy: In case of amorphous substances, properties such as electrical conductivity, refractive index, thermal expansion, etc. are identical in all directions just as in case of gases or liquids. This property is called isotropy and the substances showing this property are called isotropic.

Question.3. Which of the following is an amorphous solid?
(a) Graphite (C) (b) Quartz glass (Si02)
(c) Chrome alum (d) Silicon carbide (SiC)
Solution: (b)

Question.4. Which of the following arrangement shows schematic alignment of magnetic moments of antiferromagnetic substances?

Solution:(d)

Question.5. Which of the following is true about the value of refractive index of quartz glass?
(a) Same in all directions (b) Different in different directions
(c) Cannot be measured (d) Always zero
Solution: (a) Since quartz glass is an amorphous solid having short range order of constitutents. Hence, value of refractive index is same in all directions, can be measured and not be equal to zero always.

Question.6. Which of the following statement is not true about amorphous solids?
(a) On heating they may become crystalline at certain temperature
(b) They may become crystalline on keeping for a long time
(c) Amorphous solids can be moulded by heating
(d) They are anisotropic in nature
Solution:(d) Amorphous solids are isotropic because they show thermal and optical properties, same in all directions.

Question.7. The sharp melting point of crystalline solids is due to
(a) a regular arrangement of constituent particles observed over a short
distance in the crystal lattice .
(b) a regular arrangement of constituent particles observed over a long distance in the crystal lattice
(c) same arrangement of constituent particles in different directions
(d) different arrangement of constituent particles in different directions
Solution:(b) A solid is said to be crystalline if the various constituent structural units (atoms, ions or molecules) of which the solid is made, are arranged in a definite geometrical pattern within the solid.
The type of forces in crystalline solids are of long range order due to which they have sharp melting point. .

Question.8. Iodine molecules are held in the crystals lattice by
(a) London forces (b) dipole-dipole interactions
(c) covalent bonds (d) coulombic forces
Solution:(a) I2 is a molecular solid

Question.9. Which of the following is a network solid?
(a) S02 (solid) (b) I2
(c) Diamond (d) H20 (ice)
Solution: (c) Diamond is a three-dimensional network solid in which each carbon atom is tetrahedrally bonded with four carbon atoms.

Question.10. Which of the following solids is not an electrical conductor?
Solution: Together by London force or dispersion force. This is soft and non-conductor of electricity.
Water is a hydrogen bonded molecular solid in which H and O are held together by polar covalent bond and each water molecular held together by hydrogen bonding. Due to non-ionic nature, they are not electrical conductor.

Question.11. Which of the following is not the characteristic of ionic solids?
(a) Very low value of electrical conductivity in the molten state
(b) Brittle nature
(c) Very strong forces of interactions
(d) Anisotropic nature
Solution: (a)

Question.12. Graphite is a good conductor of electricity due to the presence of
(a) lone pair of electrons (b) free valence electrons
(c) cations (d) anions
Solution: (b) In graphite one carbon atom is attached to three other carbon atoms. One electron of carbon remains free. Due to this free valence electron graphite is an electrical conductor.

Question.13. Which of the following oxide behaves as conductor or insulator depending upon temperature?
(a) TiO (b) Si02 (c) TiO3 (d) MgO
Solution: (c) TiO3 behaves as conductor or insulator depending on temperature because of variation of energy gap between valence band and conduction band with the variation of temperature.

Question.14. Which of the following oxide shows electrical properties like metals?
(a)SO2 (b) MgO
(c)SO2(s) (d) CrO2
Solution:(d)CrO2, TiO and Re03 are some typical metal oxides which show electrical conductivity similar to metal. While SO2, MgO and SO2 are oxides of metal, semimetal and non-metal which do not show electrical properties.

Question.15. The lattice site in a pure crystal cannot be occupied by
(a) molecule (b) ion
(c) electron (d) atom
Solution:(c) Pure crystals have constituents i.e., atoms or molecules or ions as lattice points which are arranged in fixed stoichiometric ratio. Electron can occupy the lattice site only when there is imperfection in solid and not in a pure crystal.
Hence, existence of free electrons are not possible, it is possible on in case of imperfection in solid.

Question.16. Graphite cannot be classified as
(a) conducting solid (b) network solid
(c) covalent solid (d) ionic solid
Solution:(d) Constituent units of graphite are carbon atoms, held together by covalent bonding in 2D network structure. Thus, it is not an ionic solid.

Question.17. Cations are present in the interstitial sites in
(a) Frenkel defect (b) Schottky defect
(c) vacancy defect (d) metal deficiency defect .
Solution: (a)

Question.18. Schottky defect is observed in crystals when
(a) some cations move from their lattice site to interstitial sites
(b) equal number of cations and anions are missing from the lattice
(c) some lattice sites are occupied by electrons
(d) some impurity is present in the lattice
Solution: (b)

Question.19. Which of the following is true about the charge acquired by p-type semicon-ductors?
(a) Positive
(b) Neutral
(c) Negative
(d) Depends on concentration of p impurity
Solution: (b) p-Type semiconductors are neutral but they conduct electricity through positive holes.

Question.20. To get a n-type semiconductor from silicon, it should be doped with a
substance with valency
(a) 2 (b) 1
(c) 3 (d) 5
Solution:(d) Impurity of higher group is doped to get n-type semiconductor. Thus, silicon (valency = 4) should be doped with the element with valency equal to 5.

Question..21. The total number of tetrahedral voids in the face centered unit cell is
(a) 6 (c) 10
(b) 8 (d) 12
Solution:(b) Fee unit cell contains 8 tetrahedral voids at centre of each 8 smaller cube of a unit cell as shown below

Question.22. Which of the following point defects are shown by AgBr (s) crystals?
(A) Schottky defect (B) Frenkel defect
(C) Metal excess defect (D) Metal deficiency defect
(a) A and B (b) C and D
(c) A and C (d) B and D
Solution: (a) AgBr shows both Schottky and Frenkel defects. In AgBr, both Ag+ and Br ions are absent from the lattice causing Schottky defect. However, Ag+ ions are mobile so they have a tendency to move aside the lattice and trapped in interstitial site, hence cause Frenkel defect. ‘

Question.23. In which pair most efficient packing is present?
(a) hep and bcc (b) hep and ccp
(c) bcc and ccp (d) bcc and simple cubic cell
Solution:(b) Packing efficiency: It is the percentage of total filled space by particles

Since, packing efficiency for hep or ccp is calculated to be 74% which is maximum among all type of crystals.

Question.24. The percentage of empty space in a body centered cubic arrangement is
(a) 74 (b) 68 (c) 32 (d) 26
Solution: (c) Packing efficiency for bcc arrangement is 68% which represents total filled space in the unit cell. Hence, empty space in a body centered arrangement is 100 – 68 = 32%.

Question.25. Which of the following statement is not true about the hexagonal close packing?
(a) The coordination number is 12
(b) It has 74% packing efficiency
(c) Tetrahedral voids of the second layer are covered by the spheres of the third layer
(d) In this arrangement, spheres of the fourth layer are exactly aligned with those of the first layer.
Solution: (d) Hexagonal close packing can be arranged by two layers
A and B one over another which can be diagrammatically represented as

Here, we can see easily that 1st layer and 4th layer are not exactly aligned. Thus, statement (d) is not correct while other statements (a), (b) and (c) are true.

Question.26. In which of the following structure coordination number for cations and anions in the packed structure will be same?
(a) CP ions form fee lattice and Na+ ions occupy all octahedral voids of the unit cell.
(b) Ca2+ ions form fee lattice and F ions occupy all the eight tetrahedral voids of the unit cell
(c) O2 ions form fee lattice and Na+ ions occupy all the eight tetrahedral voids of the unit cell
(d) S2 ions form fee lattice and Zn2+ ions go into alternate tetrahedral voids of the unit cell.
Solution: (a) NaCl crystals have rock salt structure having fee lattice in which Cl” ions are present at fee lattice points and face centre and Na+ occupies all the octahedral voids of given unit cell.
Where, coordination number of Na+ = 6
Coordination number of Cl = 6

Question.27. What is the coordination number in a square close packed structure in two dimensions?
(a) 2 (b) 3 (c) 4 (d) 6
Solution: (c) Coordination number in a square closed packed structure in two dimensions is equal to 4 is shown as:

Question.28. Which kind of defect is introduced by doping?
(a) Dislocation defect (b) Schottky defect
(c) Frenkel defect (d) Electronic defect
Solution:(d) When electron rich or electron deficient impurity is added to a perfect crystal, it introduces electronic defect in them.

Question.29. Silicon doped with electron rich impurity forms
(a) p-type semiconductor (b) n-type semiconductor
(c) intrinsic semiconductor (d) insulator
Solution: (b) Silicon has four valence electrons. If it is doped with an electron rich impurity, the extra electron becomes delocalised and increases the conductivity. Since the increase in conductivity is due to negatively charged electron, hence it is called n-type semiconductor.

Question.30. Which of the following statement is not true?
(a) Paramagnetic substances are weakly attracted by magnetic field
(b) Ferromagnetic substances cannot be magnetized permanently
(c) The domains in antiferromagnetic substances are oppositely oriented with respect to each other
(d) Pairing of electrons cancel their magnetic moment in the diamagnetic substances.
Solution: (b) Ferromagnetic species are strongly attracted in the magnetic field and can be permanently magnetised.
Hence, choice (b) is the correct answer while other three choices are correct.

Question.31. Which of the following is not true about the ionic solids?
(a) Bigger ions form the close packed structure
(b) Smaller ions occupy either the tetrahedral or the octahedral voids depending upon their size
(c) Occupation of all the voids is not necessary
(d) The fraction of octahedral or tetrahedral voids occupied depends upon the radii of the ions occupying the voids.
Solution: (d) The fraction of octahedral or tetrahedral voids occupied depends upon the radii of the ions present at the lattice points. As we know the radii of octahedral or tetrahedral void is related to radii of atoms (r) as Radius of octahedral void (R0) = 0.414 r .
Radius of tetrahedral void (R, ) = 0.225 r Where, r = radius of bigger atom involved.

Question.32. A ferromagnetic substance becomes a permanent magnet when it is placed in a magnetic field because
(a) all the domains get oriented in the direction of magnetic field
(b) all the domains get oriented ill the direction opposite to the direction of magnetic field
(c) domains get oriented randomly
(d) domains are not affected by magnetic field.
Solution:(a) Ferromagnetic solids can be permanently magnetised and then all the domains get oriented in the direction of applied magnetic field.

Question.33. The correct order of the packing efficiency in different types of unit cells is…………
(a) fee < bee < simple cubic (b) fee > bee simple cubic
(c) fee < bee > simple cubic (d) bee < fee > simple cubic
Solution:(b) Packing efficiency in different types of unit cells can be tabulated as

Hence, correct order is fee (74%) > bee (68%) > simple cubic (52%).

Question.34. Which of the following defects is also known as dislocation defect?
(a) Frenkel defect (b) Schottky defect
(c) Non-stoichiometric defect (d) Simple interstitial defect
Solution: (a) In Frenkel defect, some cations occupy interstitial site and hence it is also called dislocation defect.

Question.35. In the cubic close packing, the unit cell has
(a) 4 tetrahedral voids each of which is shared by four adjacent unit cells
(b) 4 tetrahedral voids within the unit cell
(c) 8 tetrahedral voids each of which is shared by four adjacent unit cells
(d) 8 tetrahedral voids within the unit cells.
Solution: (d) In the cubic close packing the unit cell has 8 tetrahedral voids within it and are located at each eight smaller cub? of a unit cell.

Question.36. The edge lengths of the unit cells in terms of the radius of spheres constituting fee, bcc and simple cubic unit cells are respectively

Solution:(a) Note: Distance between two atoms is always measured from their centres
(i) If the crystal lattice consists of SCC, the atom which is present at the comers touch each other

(ii) In case of FCC, atom present at the comer and the centre of the face touch each other.

(iii)In case of BCC atom present at the corner and center of the body touch each other

Question.37. Which of the following represents correct order of conductivity in solids?

Solution: (a) Conductivity of metal, insulator and semiconductors can be represented in the term of k (Kappa) which depends upon energy gap between valence band and conduction band.

Question.38. Which of the following is not true about voids formed in three dimensional hexagonal close packed structure?
(a) A tetrahedral void is formed when a sphere of the second layer is present above triangular void in the first layer
(b) All the triangular voids are not covered by the spheres of the second layer
(c) Tetrahedral voids are formed when the triangular voids in the second layer lie above the triangular voids in the first layer and the triangular voids in the first layer and the triangular shapes of these voids do not overlap
(d) Octahedral voids are formed when the triangular voids in the second layer exactly overlap with similar voids in the first layer.
Solution:(c, d) Tetrahedral voids are formed when the triangular void in the second layer lie exactly above the triangular voids in the first layer and the triangular shape of these voids oppositely overlap.

Octahedral voids are formed when triangular void of second layer is not exactly overlap with similar void in first layer.

Question.39. The value of magnetic moment is zero in the case of antiferromagnetic substances because the domains …
(a) get oriented in the direction of the applied magnetic field
(b) get oriented opposite to the direction of the applied magnetic field
(c) are oppositely oriented with respect to each other without the application of magnetic field
(d) cancel out each other’s magnetic moment
Solution. (c, d) In the case of antiferromagnetic substances, the magnetic moment becomes zero because the domains are oppositely oriented with respect to each other without the application of magnetic field which cancel out each other.

Question.40. Which of the following statements are not true?
(a) Vacancy defect results in a decrease in the density of the substance
(b) Interstitial defects results in an increase in the density of the substance
(c) Impurity defect has no effect on the density of the substance
(d) Frenkel defect results in an increase in the density of the substance
Solution:(c, d) Statements (c) and (d) can be correctly written as (c) Impurity defect
changes the density of substance as impurity has different than the ion present on perfect crystal e.g., When SrCl2 is added to the NaCl crystal, it causes impurity defect, (d) Frenkel defect results neither decrease nor increase in density of substance.

Question.41. Which of the following statements are true about metals?
(a) Valence band overlap with conduction band
(b) The gap between valence band and conduction band is negligible
(c) The gap between valence band and conduction band cannot be determined
(d) Valence band may remain partially filled.
Solution: (a, b, d) In metal, valence band overlap with conduction band. The gap between valence band and conduction band is negligible and valence band may remain partially filled.

Question.42. Under the influence of electric field, which of the following statements are true about the movement of electrons and holes in a p-type semiconductor?
(a) Electron will move towards the positively charged plate through electron holes
(b) Holes will appear to be moving towards the negatively charged plate
(c) Both electrons and holes appear to move towards the positively charged plate
(d) Movement of electrons is not related to the movement of holes
Solution: (a, b) In p-type semiconductor, the conductivity is due to existence of hole. When electric field is applied to p-type semiconductor hole starts moving towards negatively charged plate and electron towards positively charged plate.
Flow of holes in p-type semiconductors Hole .

Question.43. Which of the following statements are true about semiconductors?
(a) Silicon doped with an electron rich impurity is a p-type semiconductor
(b) Silicon doped with an electron rich impurity is an n-type semiconductor
(c) Delocalised electrons increase the conductivity of doped silicon
(d) An electron vacancy increases the conductivity of type semiconductor
Solution:(b, c) Silicon (valence electron – 4) doped with electron rich impurity is an
n-type semiconductor due to extra electron and the delocalised electrons increase the conductivity of doped silicon.

Question.44. An excess of potassium ions makes KC1 crystals appear violet or Lilac in
colour since
(a) some of the anionic sites are occupied by an unpaired electron
(b) some of the anionic sites are occupied by a pair of electrons
(c) there are vacancies at some’anionic sites
(d) F-centres are created which impart colour to the crystals
Solution:(a, d) .
When KC1 is heated in vapour of K, some of the Cl” leave their lattice site and create anion vacancies. This chloride ion wants to combine with K vapour to form potassium chloride. For doing so K atom loses electrons form K ions. This released electron diffuses into the crystal to get entrapped in the anion vacancy called F-centre. When visible light falls on the crystal, this entrapped electron gains energy, goes to the higher level when it comes back to the ground state, energy is released in the form of light.

Question.45. The number of tetrahedral voids per unit cell in NaCl crystal is
(c) twice the number of octahedral voids
(d) four times the number of octahedral voids
Solution: (b, c) NaCl has fee arrangement of CF ions. Thus,
Number of CF ions in packing per unit cell = 4
Number of tetrahedral voids = 2 x No. of particles present in close packing
=2×4=8
Number of tetrahedral voids = 2 x No. of octahedral voids

Question.46. Amorphous solids can also be called
(a) pseudo solids (b) true solids
(c) super cooled liquids (d) super cooled solids
Solution: (a, c) Amorphous solid has short range order which has a tendency to flow very slowly. Hence, it is also known as pseudo solids or super cooled liquids. Glass panes fixed to windows or doors of old buildings are invariably observed to be thicker at bottom than at the top. These are examples of amorphous solids.

Question.47. A perfect crystal of silicon (fig) is doped with some elements as given in the options. Which of these options show n-type semiconductors?


Solution: (a, c)
When group 15 elements are doped into a perfect crystal, it leads to the formation of n-type semiconductor.
Here, in (a) as (group 15, period 3) is doped to perfect Si-crystal and in (c) as (group 15, period 2) is doped to perfect Si-crystal.

Question.48. Which of the following statements are correct?
(a) Ferrimagnetic substances lose ferrimagnetism on heating and become paramagnetic
(b) Ferrimagnetic’substances do not lose ferrimagnetism on heating and remain ferrimagnetic
(c) Antiferromagnetic substances have domain structure similar to ferromagnetic substances and their magnetic moments are not cancelled by each other
(d) In ferromagnetic substances, all the domains get oriented in the direction of magnetic field and remain as such even after removing magnetic field.
Solution: (a, d) Ferrimagnetic substances lose ferrimagnetism on heating and become paramagnetic. In ferromagnetic substance, domains are aligned in parallel and antiparallel direction in unequal numbers.
In ferromagnetic substances, all the domains get oriented in the direction of magnetic field and remain as such even after removing magnetic field.

Question.49. Which of the following features are not shown by quartz glass?
(a) This is a crystalline solid
(b) Refractive index is same in all the directions
(c) This has definite heat of fusion
(d) This is also called super cooled liquid
Solution: (a, c) Quartz glass is an amorphous solid so it has not definite heat of fusion. This is due to short range order of molecule while quartz glass is also known as super cooled liquid and isotropic in nature.

Question.50. Which of the following cannot be regarded as molecular solid?
(a) SiC (b) AIN
(c) Diamond (d) I2
Solution: (a, b, c) SiC, AIN and diamond are examples of network solid as they have three dimensional structure while, I2 is a molecular solid, because such solid particles are held together by dipole-dipole interactions. SiC and AIN are interstitial solids. s

Question.51. In which of the following arrangements octahedral voids are formed?
(a) hep (b) bcc (c) simple cubic (d) fee
Solution: (a, d) In hep and fee arrangement, octahedral voids are formed. In fee, the octahedral voids are observed at edge and centre of cube while in bcc and simple cubic, no any octahedral voids are observed. In bcc, cubic voids formed.

Question.52. Frenkel defect is also known as
(a) stoichiometric defect (b) dislocation defect
(c) impurity defect (d) non-stoichiometric defect
Solution:(a, b) In Frenkel defect, dislocation of cations takes place and there is no change in stoichiometry of the crystal.

Question.53. Which of the following defects decrease the density?
(a) Interstitial defect (b) Vacancy defect
(c) Frenkel defect (d) Schottky defect
Solution:(b, d) Vacancy and Schottky defect which lead to decrease the density both are the types of a stoichiometric defect. In case of Frenkel defect and interstitial defect, there is no change in density of substance.

Short Answer Type Questions

Question.54. Why are liquids and gases categorized as fluids?
Solution:The liquids and gases have a property to flow i.e., the molecules of liquids and gases can easily move fast and tumble over one another freely. Because of their tendency to flow, these have been categorized as fluids.

Question.55. Why are solids incompressible?
Solution: The intemuclear distance between the constituent particles (atoms, molecules or ions) in solids are very less. On bringing them further closer, there will be large repulsive force between electron clouds of these particles. Therefore, solids cannot be compressed.

Question.56. In spite of long range order in the arrangement of particles why are the crystals usually not perfect?
Solution:Crystals have long range in the arrangement of particles but usually the crystals are not perfect this is because when crystallisation occurs at a fast rate or moderate rate, the constituent particles may not get sufficient time to arrange themselves in a perfect order.

Question.57. Why does table salt, NaCl sometimes appear yellow in colour?
Solution: The yellow colour of sodium chloride crystals is due to metal excess defect. In this defect, the unpaired electrons get trapped in anion vacancies. These sites are called F-centres. The‘yellow colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals.

Question.58. Why is FeO(s) not formed in stoichiometric composition?
Solution: Iron oxide (FeO) has rock salt structure.
In this case, O2 ions adopt on sites and Fe2+ ions should occupy non- stoichiometric. This is the ideal arrangement.
This oxide is always non-stoichiometric i.e., the composition of Fe2+ and O2 ions is not 1 : 1. It is 0.95 : 1 i.e. Fe0 95O(Wustite)
This composition can be obtained if a small number of Fe2+ ions are replaced by two-thirds of’Fe3+ ions in Oh sites.
Eventually there would be less amount of metal as compared to stoichiometric composition.

Question.59. Why does white ZnO (s) become yellow upon heating?
Solution: When ZnO is heated, it splits up to give Zn2+, electrons and colour because of the following reasons:
The excess Zn ions thus formed get entrapped in the interstitial site and electron in the neighborhood vacant interstitial sites. This electron is responsible for the colour and electrical conductivity in crystals.

Question.60. Why does the electrical conductivity of semiconductors increase with rise in temperature?
Solution:The energy gap between valence band and conduction band is small. At room temperature, they do not conduct electricity but when temperature is raised large number of electron from valence band get sufficient energy to jump to conduction band. This is known as thermodynamic conduction in intrinsic semiconductors. Thus, they become more conducting as the temperature increases.

Question.61. Explain why does conductivity of germanium crystals increase on doping with gallium?
Solution: p-type semiconductor:

1. When Ge is doped with group 13 elements, for example, gallium, the structure of crystal lattice does not change.
2. 3 valence electrons of gallium are used up in the normal covalent bond.
3. For one dopant atom, one hole is created because the place where fourth electron is missing is called vacancy or hole and is responsible for conduction of germanium doped with gallium.
Electron from neighbouring atom comes and fills the hole, thereby creating a hole in its original position.
Under the influence of electric field electrons move towards positively charged plates through these and conduct electricity. The holes appear to move towards negatively charged plates.

Question.62. In a compound, nitrogen atoms (N) make cubic close packed lattice and metal atoms (M) occupy one-third of the tetrahedral voids present. Determine the formula of the compound formed by M and N?
Solution: In ccp, no. of atoms per unit cell = 4
Thus, of tetrahedral voids = 2 x No. of atoms in ccp =2×4=8
Only one-third of tetrahedral voids are occupied by metal M. No. of

Question.63. Under which situations can an amorphous substance change to crystalline form?
Solution: An amorphous solid on heating at some temperature may become crystalline. Slow heating and cooling over a long period makes an amorphous solid acquires some crystalline character.

Matching Column Type Questions

Question.64. Match the defects given in Column I with the statements given in Column II.

Solution:(i) -> (c); (ii) -> (a); (iii) ->(d); (iv) ->(b)

Question.65. Match the type of unit cell given in Column I with the features given in Column II.

Solution: (i) —»(b, c); (ii) —»(c, d); (iii) -o (c, e); (iv) —» (a, d)
(i) For primitive unit cell, a = b = c
Total number of atoms per unit cell = 1/8 x 8 = 1
Here, 1/8 is due to contribution of each atom present at comer.
(ii) For body centered cubic unit cell, a = b = c.
This lattice contains atoms at comer as well as body centre. Contribution due to atoms at comer = 1/8 x 8 = 1 contribution due to atoms at body centre = 8

Question.66. Match the types of defect given in Column I with the statement given in Column II.

Solution: (i) —»(c); (ii) —»(a); (iii) —» (b)
(A) (i) Impurity defects: The defects introduced in the crystal lattice due to presence of the certain impurity are called impurity defects.
Example: Substitution of Na+ ions in NaCl by Sr2+ ions.
Structure with defect:

Impurity defect due to substitution of Na+ ions in NaCl by Sr2+ ions (Cation vacancy) ‘Schottky Defect’
(B) When NaCl is heated in vapour of sodium some of the Cf leave their lattice site and create anion vacancies. This chloride ion wants to combine with sodium vapour to form sodium chloride. For doing so sodium atom loses electrons form Na+ ions. This released electron diffuses into the crystal to get entrapped in the anion vacancy called F-centre.

(C) Metal deficiency is caused due to cation vacancy created by replacement of some lower valent ions by its higher valentions.
Note: Cation vacancies are found in crystals in which metals have different oxidation states.
Example: FeO, FeS, NiO

Question.67. Match the items given in Column I with the items given in Column II.

Solution:(i) -> (d); (ii)-> (c); (iii)->(b); (iv) ->(a)
(i) Mg in solid state show electronic conductivity due to presence of free electrons hence, they are known as electronic conductors.
(ii) MgCl2 in molten state show electrolytic conductivity due to presence of electrolytes in molten state.
(iii) Silicon doped with phosphorus contain one extra electron due to which it shows conductivity under the influence of electric field and known as p-type semiconductor.
(iv) Germanium doped with boron contains one hole due to which it shows conductivity under the influence of electric field and known as n-type semiconductor.

Question.68. Match the type of-packing given in Column I with the items given in Column II.

Solution: (i) —> (c); (ii) —» (a); (iii) —»(d); (iv) —» (b)
(i) Square close packing in two dimensions each sphere have coordination number 4, as shown below.

(ii) Hexagonal close packing in two dimensions each sphere has coordina¬tion number 6 as shown below and creates a triangular void

(iii)Hexagonal close packing in 3 dimensions is a repeated pattern of sphere in alternate layers also known as ABAB pattern

(iv) Cubic close packing in a 3 dimensions is a repeating pattern of sphere in every fourth layer.

Assertion and Reason Type Questions:

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices.
(a) Assertion and Reason both are correct statements and Reason is the correct explanation for Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation for Assertion.
(c) Assertion is correct but Reason is wrong.
(d) Assertion is wrong but Reason is correct.

Question.69. Assertion (A): The total number of atoms present in a simple cubic unit cell is one.
Reason (R): Simple cubic unit cell has atoms at its comers, each of which is shared between eight adjacent unit cells.
Solution: (a) In simple cubic unit cell, only comers are occupied by atoms. Thus, total number of atoms present in the unit cell will be one.

Question.70. Assertion (A): Graphite is a good conductor of electricity, however, diamond belongs to the category of insulators.
Reason (R): Graphite is soft in nature on the other hand diamond is very hard and brittle.
Solution: (b) Diamond is bad conductor of electricity because all valence e of carbon are involved in bonding. In graphite however 3 out of 4 valence electrons are involved in bonding, fourth electron remains free between adjacent layers which makes it a good conductor.
Graphite is soft because parallel layers are held together by week van der Waals force. However, diamond is hard due to compact three-dimensional network of bonding.

Question.71. Assertion (A): Total number of octahedral voids present in unit cell of cubic close packing including the one that is present at the body centre, is four. Reason (R): Besides the body centre, there is one octahedral void present at the centre of each ofthe six faces of the unit cell and each of which is shared between two adjacent unit cells.
Solution:(c) All edge centres and body centre represent octahedral void.
Total number of octahedral voids = 12 x 1/4 +1 = 4

Question.72. Assertion (A): The packing efficiency is maximum for the fee structure. Reason (R): The coordination number is 12 in fee structures.
Solution: (b) In fee unit cell, there is cep arrangement with packing efficiency of 74.01% which is maximum. In cep arrangement, coordination number is 12.

Question.73. Assertion (A): Semiconductors are solids with conductivities in the intermediate range from

Reason (R): Intermediate, conductivity in semiconductor is due to partially filled valence band.
Solution:(c) Conductance of semiconductors lies between metals and insulators, i.e., in the range of

Long Answer Type Questions

Question.74. With the help of a labelled diagram show that there are four octahedral voids per unit cell in a cubic close packed structure.
Solution: In cep, each cube consists of eight cubic components, number of atoms per unit cell in ccp is

Question.75. Show that in a cubic close packed structure, eight tetrahedral voids are present per unit cell.
Solution: In ccp, each cube consists of eight cubic components. Number of atoms per unit cell in ccp is

Position of tetrahedral Voids = At the centre of each cubic component Number of tetrahedral voids per unit cell in cubic close packing = 8×1=8 Number of tetrahedral Voids = 8.

Question.76. How does the doping increase the conductivity of semiconductors?
Solution:The conductivity of semiconductors is increased by adding an appropriate amount of suitable impurity or doping. Doping can be done with an impurity which is electron rich or electron deficient as compared to the intrinsic semiconductor, silicon or germanium. Such impurities introduce electronic defects in them. When silicon is doped with electron rich impurities the extra electron becomes delocalized. These delocalized electrons increase the conductivity of doped silicon due to the negatively charged electron, hence silicon doped with electron-rich impurity is called n-type semiconductor while electron-deficit impurities increase the conductivity through positive holes and this type of semiconductors are called /?-type semiconductors.

Question.77. A sample of ferrous oxide has actual formula Fe0 93 O100. In this sample, what fraction of metal ions are Fe2+ ions? What is the type of non-stoichiometric defect present in this sample? ’
Solution:

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NCERT Exemplar Problems Class 12 Chemistry Chapter 2 Solution

Multiple Choice Questions

Single Correct Answer Type
Question.1. Which of the following unit is useful in relating concentration of solution with its vapour pressure?
(a) Mole fraction (b) Parts per million
(c) Mass percentage (d) Molality
Solution: (a) Mole fraction is useful in’relating vapour pressure with concentration of solution. According to Raoult’s law, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.
A is one component.

Question.2. On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the following case dissolution of sugar will be most rapid?
(a) Sugar crystals in cold water (b) Sugar crystals in hot water
(c) Powdered sugar in cold water (d) Powdered sugar in hot water
Solution: (d) Since the solution is cool to touch, the dissolution is endothermic. Therefore, high temperature will favour dissolution. Further, powdered sugar has large surface area and is favourable for dissolution.

Question.3. At equilibrium the rate of dissolution of a solid solute in a volatile liquid
solvent is
(a) less than the rate of crystallisation
(b) greater than the rate of crystallisation
(c) equal to the rate of crystallisation
(d) zero
Solution: (c) At equilibrium the rate of dissolution of solid solute is equal to rate of crystallisation.

Question.4. A beaker contains a solution of substance ‘A’. Precipitation of substance ‘A’ takes place when small amount of ‘A’ is added to the solution. The solution is
(a) saturated (b) supersaturated
(c) unsaturated (d) concentrated
Solution: (b) When small amount of solute is added to its solution and it does not dissolve and get precipitated then this solution is supersaturated solution.

Question.5. Maximum amount of a solid solute that can be dissolved in a specified
amount of a given liquid solvent does not depend upon
(a) temperature (b) nature of solute
(c) pressure (d) nature of solvent
Solution: (c) Solubility of a solid in liquid does not depend on pressure because solid is practically incompressible.

Question.6. Low concentration of oxygen in the blood and tissues of people living at high altitude is due to –
(a) low temperature
(b) low atmospheric pressure
(c) high atmospheric pressure
(d) both low temperature and high atmospheric pressure
Solution: (b) At high altitudes the atmospheric pressure is less but the body temperature remains same hence concentration of oxygen in the air as well as blood is less.

Question.7. Considering the formation, breaking and strength of hydrogen bond, predict which of the following mixture will show a positive deviation from Raoulf s law?
(a) Methanol and acetone (b) Chloroform and acetone
(c) Nitric acid and water (d) Phenol and aniline
Solution: (a) Mixture of Methanol and acetone shows positive deviation because methanol-methanol and acetone-acetone interactions are more than methanol-acetone. The more number of hydrogen bonds are broken the less number of new H-bonds are formed.

Question.8. Colligative properties depend on
(a) the nature of the solute particles dissolved in solution
(b) the number of solute particles in solution
(c) the physical properties of the solute particles dissolved in solution
(d) the nature of solvent particles
Solution: (b) Colligative properties depend upon the number of solute particles in the solution and independent of its nature.

Question.9. Which of the following aqueous solution should have the highest boiling point?
(a) l.OMNaOH (b) 1.0MNa₂SO4
(c) 1.0MNH4NO3 (d) l.OMKNO3
Solution: (b) 1.0 M Na₂SO4 since it furnishes maximum number of ions (2Na+ + SO4~).

Question.10. The unit of ebullioscopic constant is , – ;

Solution:

The unit of ebullioscopic constant is K kg mol1 or K molality-1.

Question.11. In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl₂ solution is 1
(a) the same (b) about twice
(c) about three times (d) about six times
Solution: (c) 0.01 M solution of glucose does not ionize while 0.01 M MgCl₂ solution furnishes 3 ions (Mg₂+ + 2Cl”) in the solution, hence the value of colligative property for MgCl₂ solution is about 3 times.

Question.12. An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because 
(a) it gains water due to osmosis
(b) it loses water due to reverse osmosis
(c) it gains water due to reverse osmosis
(d) it loses water due to osmosis
Solution: (d) Water starts moving out of mango (lower concentration) to the salt solution (higher concentration) due to osmosis.

Question.13. At a given temperature, osmotic pressure of a concentrated solution of a substance
(a) is higher than that of a dilute solution
(b) is lower than that of a dilute solution
(c) is same as that of a dilute solution
(d) cannot be compared with osmotic pressure of dilute solution
Solution: (a)

where C is concentration of the solution. So, the higher the concentration of solution at given temperature the higher will be the osmotic pressure.

Question.14. Which of the following statement is false?
(a) Two different solutions of sucrose of same molality-prepared in different solvents will have the same depression in freezing point.
(b) The osmotic pressure of a solution is given by the equation 3.14(pi) = CRT (where C is the molarity of the solution).
(c) Decreasing order of osmotic pressure for 0.01 M aqueous solutions of barium chloride, potassium chloride, acetic acid and sucrose is
BaCl₂> KCl > CH₂COOH > sucrose
(d) According to Raoult’s law, the vapour pressure exerted by a volatile component of a solution is directly proportional to its mole fraction in the solution.
Solution: (a) Value of A is different for different solvents. Thus, according to relation,

even for same molal solutions depression in freezing point will be different.

Question.15. The value of van’t Hoff factor.s for KC1, NaCl and K2S04, respectively, are
(a) 2, 2 and 2 (b) 2, 2 and 3 (c) 1, 1, and 2 (d) 1, 1 and 1
Solution: (b) KCl (K+ + Cl ) and NaCl (Na+ + Cl ) ionize to give 2 ions and K₂SO₄(2K+ + SO₄_) ionizes to give 3 ions thus, van’t Hoff factors for KCl, NaCl and K₂SO4 are 2, 2 and 3 respectively.

Question.16. Which of the following statement is false?
(a) Units of atmospheric pressure and osmotic pressure are the same.
(b) In reverse osmosis, solvent molecules move through a semipermeable membrane from a region of lower concentration of solute to a region of higher concentration.
(c) The value of molal depression constant depends on nature of solvent.
(d) Relative lowering of vapour pressure, is a dimensionless quantity.
Solution: (b) In reverse osmosis, solvent molecules move through a semipermeable
membrane from higher concentration of solute to lower concentration.

Question.17. Value of Henry’s constant KH
(a) increases with increase in temperature
(b) decreases with increase in temperature
(c) remains constant
(d) first increases then decreases
Solution: (a) Value of Henry’s constant increases with increase in temperature.

Question.18. The value of Henry’s constant KH is
(a) greater for gases with higher solubility
(b) greater for gases with lower solubility
(c) constant for all gases
(d) not related to the solubility of gases
Solution: (b) The higher the value of KH, the lower is the solubility of gas in the liquid.

Question.19. Consider the figure and mark the correct option.

(a) Water will move from side (A) to side (B) if a pressure lower than osmotic pressure is applied on piston (B).
(b) Water will move from side (B) to side (A) if a pressure greater than osmotic pressure is applied on piston (B).
(c) Water will move from side (B) to side (A) if a pressure equal to osmotic pressure is applied on piston (B).
(d) Water will move from side (A) to side (B) if pressure equal to osmotic pressure is applied on piston (A).
Solution: (b) Water will move from side 3 (concentrated sodium chloride solution) to side A (fresh water) if a pressure greater than osmotic pressure is applied on piston B.

Question.20. We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with
concentrations 0.1 M, 0.01 M and 0.001 M, respectively. The value of van’t Hoff factor for these solutions will be in the order

Solution: (c) The value of van’t Hoff’s factor will be iA=iB =iC due to complete dissociation of strong electrolyte (NaCl) in dilute solutions. On complete dissociation value of i for NaCl is 2.

Question.21. On the basis of information given below mark the correct option. Information:
(I) In bromoethane and chloroethane mixture intermolecular interactions of A-A and B-B type are nearly same as A-B type interactions.
(II) In ethanol and acetone mixture A-A or B-B type intermolecular interactions are stronger than A-B type interactions.
(III) In chloroform and acetone mixture A-A or B-B type intermolecular interactions are weaker than A-B type interactions.
(a) Solution (II) and (III) will follow Raoult’s law.
(b) Solution (I) will follow Raoult’s law.
(c) Solution (II) will show negative deviation from Raoult’s law.
(d) Solution (III) will show positive deviation from Raoult’s law.
Solution: (b) Solution A will follow Raoult’s law due to nearly same interactions between A-A, B-B and A-B. The solution formed will be nearly ideal.

Question.22. Two beakers of capacity 500 mL were taken. One of these beakers, labelled “A”, was filled with 400 mL water whereas’the beaker labelled “B” was filled with 400 mL of 2 M solution of NaCl. At the same temperature both the beakers were placed in closed containers of same material and same capacity as shown in figure.

At a given temperature, which of the following statement is correct about the vapour pressure of pure water and that of NaCl solution.
(a) Vapour pressure in container (A) is more than that in container (B).
(b) Vapour pressure in container (A) is less than that in container (B).
(c) Vapour pressure is equal in both the containers.
(d) Vapour pressure in container (B) is twice the vapour pressure in container (A).
Solution: (a) Since NaCl is a non-volatile solute, it will reduce vapour pressure of the solution. Hence, vapour pressure of pure water in container A is more than in container B.

Question.23. If two liquids A and B form minimum boiling azeotrope at some specific composiiion’hien
(a) A-B interactions are stronger than those between A-A or B-B.
(b) vapour pressure of solution increases because more number of molecules of liquids A and B can escape from the solution.
(c) vapour pressure of solution decreases because less number of molecules of only one of the liquids escape from the solution.
(d) A-B interactions are weaker than those between A-A or B-B.
Solution: (d) If A-B interactions < A-A or B-B the vapour pressure will be more and the result will be positive deviation. The solutions which show positive deviation form minimum boiling azeotropes.

Question.24. 4 L of 0.02 M aqueous solution of NaCl was diluted by adding one litre of water. The molality of the resultant solution is
(a) 0.004 (b) 0.008 (c) 0.012 (d) 0.016
Solution:

Question.25. On the basis of information given below, mark the correct option. Information: On adding acetone to methanol some of the hydrogen bonds between methanol molecules break.
(a) At specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show positive deviation from Raoulf s law.
(b) At specific composition methanol-acetone mixture forms maximum boiling azeotrope and will show positive deviation from Raoulf s law.
(c) At specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show-negative deviation from Raoult’s law.
(d) At specific composition methanol-acetone mixture will form maximum boiling azeotrope and will show negative deviation from Raoulf s law.
Solution: (a) At specific composition methanol-acetone mixture will show positive
deviation from Raoulf s law as it has lesser interactions than methanol- methanol and acetone-acetone interactions. Hence it forms minimum boiling azeotrope.

Question.26. KH value for Ar(g),CO₂(g), HCHO(g) and CH4(g) are 40.39, 1.67, 1.83 x 10.5
and 0.413 respectively. 
Arrange these gases in the order of their increasing solubility.

Solution: (c) The higher the value of KH, the lower is the solubility of the gas in the liquid. Hence the order of increasing solubility of the gases will be Ar <  CO₂<CH4< HCHO

Question.27. Which of the following factor(s) affect the solubility of a gaseous solute in the fixed volume of liquid solvent?
(i) Nature of solute (ii) Temperature
(iii) Pressure
(a) (i) and (iii) at’constant T (b) (i) and (ii) at constant P
(c) (ii) and (iii) only (d) (iii) only
Solution: (a, b) At constant temperature, the solubility of a gaseous solute in liquid depends on nature of solute and pressure. At constant pressure, solubility is dependent upon nature of solute and temperature.

Question.28. Intermolecular forces between two benzene molecules are nearly of same strength as those between two toluene molecules. For a mixture of benzene and toluene, which of the following are not true?

Solution: (c, d) For an ideal solution,

and benzene and toluene will form an ideal solution. Also, ideal solutions do not form minimum boiling azeotropes.

Question.29. Relative lowering of vapour pressure is a colligative property because
(a) it depends on the concentration of a non-electrolyte solute in solution and does not depend on the nature of the solute molecules.
(b) it depends on number of particles of electrolyte solute in solution and does not depend on the nature of the solute particles.
(c) it depends on the concentration of a non-electrolyte solute in solution as well as on the nature of the solute molecules.
(d) it depends on the concentration of an electrolyte or a non-electrolyte solute in solution as well as on the nature of solute molecules
Solution:(a, b) Relative lowering of vapour pressure depends on concentration of non¬electrolyte solute in solution and number of particles of electrolyte solute. It does not depend on nature of solute in both conditions.

Question.30.van’t Hoff factor i is given by the expression

Solution:

Question.31. Isotonic solutions must have the same
(a) solute (b) density
(c) elevation in boiling point (d) depression in freezing point
Solution:

Isotonic solutions must have same osmotic pressure at a given temperature hence must have same volume and number of moles i.e., same molar concentration. Thus, the isotonic solutions have same elevation in boiling point, and depression in freezing point.

Question.32. Which of the following binary mixtures will have same composition in liquid and vapour phase?
(a) Benzene-Toluene (b) Water-Nitric acid
(c) Water-Ethanol (d) 6-Hexane-n-Heptane
Solution: (b, c) Non-ideal solution mixtures with same composition in liquid and vapour phase are azeotropes. Water-nitric acid has maximum boiling azeotrope and water-ethanol has a minimum boiling azeotrope.

Question.33. In isotonic solutions
(a) solute and solvent both are same
(b) osmotic pressure is same
(c) solute and solvent may or may not be same
(d) solute is always same solvent may be different
Solution: (b, c) For isotonic solutions osmotic pressure is same, solute or solvent may not be same.

Question.34. For a binary ideal liquid solution, the variation in total vapour pressure versus composition of solution it given by which of the curves?

Solution:(a,d)for idea solution.

Question.35. Colligative properties are observed when
(a) a non-volatile solid is dissolved in a volatile liquid
(b) a non-volatile liquid is dissolved in an another volatile liquid
(c) a gas is dissolved in a non-volatile liquid
(d) a volatile liquid is dissolved in an another volatile liquid
Solution: (a, b) Colligative properties are observed when a non-volatile solid or liquid are dissolved in a volatile liquid.

Short Answer Type Questions

Question.36. Components of a binary mixture of two liquids A and B were being separated by distillation. After some time separation of components stopped and composition of vapour phase became same as that of liquid phase. Both the components started coming in the distillate. Explain why does this happen?
Solution: Since both the components are coming in the distillate and composition of liquid and vapour phase become same, this shows that liquids have formed azeotropic mixture. Therefore, these components cannot be separated at this stage by distillation.

Question.37. Explain why on addition of 1 mol of NaCl to 1 litre of water, the boiling point of water increases, while addition of 1 mol of methyl alcohol to one litre of water decreases its boiling point.
Solution: NaCl is a non-volatile solute, therefore, addition of NaCl to water lowers the vapour pressure of water. As a result boiling point of water increases. On the other hand, methyl alcohol is more volatile than water, therefore its addition increases the total vapour pressure over the solution. As a result, boiling point of water decreases.

Question.38. Explain the solubility rule “like dissolves like” in terms of intermolecular forces that exist in solutions.
Solution: A substance dissolves in a solvent if the intermolecular interactions are similar in both the components. For example, polar solutes dissolve in polar solvents and non-polar solutes in non-polar solvents. Thus, we can say “like dissolves like”.

Question.39. Concentration terms such as mass percentage, ppm, mole fraction and molality are independent of temperature, however molarity is a function of temperature. Explain.
Solution: Molarity of a solution is defined as the number of moles of solute dissolved per litre of solution. Since volume depends on temperature and changes with change in temperature, therefore, the molarity will also change with change in temperature. On the other hand, mass does not change with change in temperature, and therefore, concentration terms such as mass percentage, mole fraction and molality which do not involve volume are independent of temperature.

Question.40. What is the significance of Henry’s law constant KH?
Solution: According to Henry’s law:

Thus, the higher is the values of Henry’s law constant, the lesser will be the solubility of gas in liquid.

Question.41. Why are aquatic species more comfortable in cold water in comparison to warm water?
Solution: At a given pressure the solubility of oxygen in water increases with decrease in temperature. Therefore, the concentration of oxygen in sea is more in cold water and thus presence of more oxygen at lower temperature makes the aquatic species more comfortable in cold water.

Question.42. (a) Explain the following phenomena with the help of Henry’s law.
(i) Painful condition known as bends.
(ii) Feeling of weakness and discomfort in breathing at high altitude.
(b) Why does soda water bottle kept at room temperature fizz on opening?
Solution: (a)
(i) Deep sea divers depend upon compressed air for breathing at high pressure under water. The compressed air contains  N2 in addition to 02, which are not very soluble in blood at normal pressure. However, at great depths when the diver breathes in compressed air from the supply tank, more N2 dissolves in the blood and other body fluids because the pressure at that depth is far greater than the surface atmospheric pressure. When the diver comes towards the surface, the pressure decreases, N2 comes out of the body quickly forming bubbles in the blood stream. These bubbles restrict blood flow, affect the transmission of nerve impulses. The bubbles can even burst the capillaries or block them and starve the tissues of 02. This condition is called the bends, which are painful and life-threatening.
(ii) At high altitudes the partial pressure of 02 is less than that at the ground level. This results in low concentration of oxygen in the blood and tissues of the people living at high altitudes or climbers. The low blood oxygen causes climbers to become weak and unable to think clearly known as anoxia.
(b) To increase the solubility of C02 in soft drinks, the soda water bottles are sealed under high pressure. When the bottle is opened at room temperature under normal atmosphere conditions, the pressure inside the bottle decreases to atmospheric pressure and excess C02 fizzes out.

Question.43. Why is the vapour pressure of an aqueous solution of glucose lower than that of water?
Solution: In pure liquid water, the entire surface of liquid is occupied by the molecules of water. When a non-volatile solute, such as glucose is dissolved in water some of the surface is covered by non-volatile glucose molecules. Therefore, the fraction of surface covered by the solvent molecules escaping. As a result
number of solvent molecules escaping from the surface also gets reduced and consequently the vapour pressure of aqueous solution of glucose is reduced.

Question.44. How does sprinkling of salt help in clearing the snow covered roads in hilly areas? Explain the phenomenon involved in the process.
Solution: When salt is spread over snow covered roads, it lowers the freezing point of water to such an extent that water does not freeze to form ice. As a result, the snow starts melting from the surface and therefore, it helps in clearing the roads. Hence, common salt acts as de-icing agent.

Question.45. What is a “semipermeable membrane”?
Solution: A membrane that permits the flow of solvent molecules not the solute molecules is called semipermeable membrane. During osmosis and reverse osmosis, only solvent molecules move across the semipermeable membrane.

Question.46. Give an example of a material used for making semipermeable membrane for carrying out reverse osmosis.
Solution: Cellulose acetate, potassium ferrocyanide, etc. are used as semipermeable membrane for carrying out reverse osmosis.

Matching Column Type Questions

Question.47. Match the items given in Column I and Column II.

Solution: (i —> d), (ii —> c); (iii-> a); (iv b), (v —> f); (vi —> e)
(i) Saturated solution: A solution which contains maximum amounts of solute that can be dissolved in a given amounts of solute that can be dissolved in a given amount of solvent at a given temperature.
(ii) Binary solution: A solution with two components is known as binary solution.
(iii) Isotonic solution: A solution having same osmotic pressure at a given temperature as that of given solution is known as isotonic solution.
(iv) Hypotonic solution: A solution whose osmotic pressure is less than another is known as hypotonic solution.
(v) Solid solution: A solution in solid phase is known as solid solution.
(vi) Hypertonic solution: A solution whose osmotic pressure is greater than that of another is known as hypertonic solution.

Question.48. Match the items given in Column 1 with the type of solutions given in Column II.

Solution: (i -» e), (ii -» c), (iii -» d); (iv —> b), (v —> a)
(i) Soda water: A solution of gas in liquid, e.g.,CO₂  in soft drinks.
(ii) Sugar solution: A solution of solid in liquid in which sugar particles (solid) are dissolved in water (liquid).
(iii) German silver: This is an alloy which is a solid solution of solid in solid. It is an alloy of Cu, Zn and Ni.
(iv) Air: A solution of gas in gas..Air is a mixture of various gases.
(v) Hydrogen gas in palladium: This is an example of solution of gas in solid. This is used as a reducing agent.

Question.49. Match the laws given in Column I with expressions given in Column II.

Solution:

Question.50. Match the terms given in Column I with expression given in Column II.

Solution. (i —» d), (ii —» c), (iii —» b), (iv -> e), (v ->a)

Assertion and Reason Type Questions

In the following questions, a statement of Assertion (A) followed by a statement of
Reason (R) is given. Choose the correct answer out of the following choices:
(a) Assertion and reason both are correct statements and reason is the correct explanation for Assertion.
(b) Assertion and reason both are correct statements but reason is not the correct explanation for Assertion.
(c) Assertion is correct statement but reason is wrong.
(d) Assertion and reason both are incorrect.
(e) Assertion is wrong but reason is correct.

Question.51. Assertion (A): Molarity of a solution in liquid state changes with temperature. Reason (R): The volume of a solution changes with change in temperature.
Solution: (a) Molarity changes with temperature because volume changes with temperature.

Question.52. Assertion (A): When methyl alcohol is added to water, boiling point of water increases.
Reason (R): When a volatile solute is added to a volatile solvent, elevation in boiling point is observed.
Solution: (d) When methyl alcohol (volatile) is added to water, boiling point of water decreases because vapour pressure increases when volatile solute is added to volatile solvent;

Question.53. Assertion (A): When NaCl is added to water, a depression in freezing point is observed.
Reason (R): The lowering of vapour pressure of a solution causes depression in the freezing point.
Solution: (a) When a non-volatile solute is added to water, freezing point lowers due to lowering of vapour pressure.

Question.54. Assertion (A): When a solution is separated from the pure solvent by a semipermeable membrane, the solvent by a semipermeable membrahe, the solvent molecules pass through it from pure solvent side to the solution side. Reason (R): Diffusion of solvent occurs from a region of high concentration solution to a region of low concentration solution.
Solution: (c) Solvent molecules pass through the semipermeable membrane from region of low concentration solution to the region of high concentration solution.

Long Answer Type Questions

Question.55. Define the following modes of expressing the concentration of a solution. Which of these modes are independent of temperature and why?
(i) w/w (mass percentage)
(ii) x (mole fraction)
(iii) VIV (volume percentage)
(iv) M (molarity)
(v) wIV (mass by volume percentage)
(vi) m (molality)
(vii) ppm (parts per million)
Solution:

Mass percentage, ppm, mole fraction and molality are independent of temperature since mass does not change with temperature.

Question.56. Using Raoult’s law explain how the total vapour pressure over the solution is related to mole traction of components in the given solutions.
(a) CHCl₂ andCH4Cl_2(l)
(b) NaCl(s) and _H2O(l)
Solution:

Question.57. Explain the terms ideal and non-ideal solutions in the light of forces of interactions operating between molecules in liquid solutions.
Solution: Ideal solutions: The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solution. For an ideal solution,

Question.58. Why is it not possible to obtain pure ethanol by fractional distillation? What general name is given to binary mixtures which show deviation from Raoult’s law and whose components cannot be separated by fractional distillation? How many types of such mixtures are there?
Solution: The solution or mixture having same composition in liquid as well as in vapour phase and boils at a constant temperature is known as azeotropes. Due to constant composition it cannot be separated by fractional distillation. There are two types of azeotropes
(i) Minimum boiling azeotropes: Solution which shows large positive deviation from Raoult’s law form minimum boiling azeotropes at a specific composition, e.g., ethanol-water mixture
(ii) Maximum boiling azeotropes: Solutions which show large negative deviation from Raoult’s law form maximum boiling azeotropes, e.g., solution having composition 68% HN03 and 32% water by mass.

Question.59. When kept in water, raisin swells in size. Name and explain the phenomenon involved with the help of a diagram. Give three applications of the phenomenon.
Solution: Raisins swell in size on keeping in water. This happens due to the phenomenon of osmosis. The outer skin of raisin acts as a semipermeable membrane. Water moves from a place of lower concentration to a place of higher concentration through the semipermeable membrane. Thus, water enters inside the raisins and make them swell.

Applications of the phenomenon
(i) Movement of water from soil into plant roots and subsequently into upper portion of the plant is partly due to osmosis.
(ii) Preservation of meat against bacterial action by adding salt.
(iii) Preservation of fruits against bacterial action by adding sugar. Bacterium in canned fruit loses water through the process of osmosis, shrivels and dies.
(iv) Reverse osmosis is used for desalination of water.

Question.60. Discuss biological and industrial importance of osmosis.
Solution: Some essential importance of osmosis are given below:
(i) In animals, circulation of water to all parts of body takes place due to osmosis.
(ii) Plant roots absorb water from soil due to osmosis. Concentration of cell sap inside the root hair cells is higher than that of water present in the soil. Water enters the root cells due to endosmosis.
(iii) Water absorbed by plant roots is circulated in the entire plant body and reaches to the top of a tall tree due to osmosis.
(iv) Osmosis helps in plant growth and germination of seeds.
(v) Red blood cells burst when placed in water; it is due to endosmosis.
(vi) Various functions of plants are controlled by osmosis, e.g., stretching of leaves and flowers, opening and closing of flowers.
(vii) Use of salt and sugar in‘pickles and jams acts as preservatives. It prevents growth of bacteria and fungi by osmosis.
(viii) Dead bodies swell under water due to endosmosis.
(ix) When dried fruits and vegetables are placed in water, they slowly swell and return to the original form. It is again due to endosmosis of water into the fruits and vegetables.
(x) Edema: Due to excess intake of salt by a person, the tissues become puffy, it is called edema. It is due to retention of water in the tissue owing to osmosis.

Question.61. How can you remove the hard calcium carbonate layer of the egg without damaging its semipermeable membrane? Can this egg be inserted into a bottle with a narrow neck without distorting its shape? Explain the process involved.
Solution: This can be achieved as under:
(i) Place the egg in a mineral acid solution for about 2 hours. The outershell of the egg dissolves. Remove any portion of it is left with your fingers.
(ii) Place the egg in a saturated solution (hypertonic) of sodium chloride for about 3 hours. Size of the egg is reduced as the egg shrivels due to osmosis.
(iii) Insert the egg in a bottle with a narrow neck. Add water to the bottle. Water will act as hypotonic solution, Egg regains shape due to osmosis. This is shown diagrammatically as under:

Question.62. Why is the molar mass determined by measuring a colligative property in case of some solutes abnormal? Discuss it with the help of van’t Hoff factor.
Solution: The compounds which dissociate or associate in the solvent show abnormal molecular masses.
(i) Association: Compounds like benzoic acid or ethanoic acid dimerise in benzene due to hydrogen bonding as a result of which the number of particles in the solution decreases. Since colligative properties depend upon number of particles, such solutes show lower colligative property.
(ii) Dissociation: Electrolytes like NaCl, KC1, etc. dissociate into ions which result in increased number of particles, hence higher value of colligative property.
To account for association or dissociation van’t Hoff introduced a factor T known as van’t Hoff factor. It is defined as Expected molar mass Abnormal molar mass _ Observed colligative property Calculated colligative property
Total number of moles of particles _ after association/dissociation Total number of moles of particles before association/dissociation

The post NCERT Exemplar Problems Class 12 Chemistry Solution appeared first on Learn CBSE.

NCERT Exemplar Problems Class 12 Chemistry Chapter 3 Electrochemistry 

Multiple Choice Questions

Single Correct Answer Type
Question.1. Which cell will measure standard electrode potential of copper electrode?

Solution: (c) When copper electrode is connected to standard hydrogen electrode, it acts as cathode and its standard electrode potential can be measured.

To calculate the standard electrode potential of the given cell it is coupled with the standard hydrogen electrode in which pressure of hydrogen gas is one bar and the cone, of H+ ion in the solution is one molar and also the concentrations of the oxidized and the reduced forms of the species in the right hand half-cell are unity.

Question.2. Electrode potential for Mg electrode varies according to the equation


Solution:

Question.3. Which of the following statements is correct?

Solution: (c) EceU is an intensive property as it does not depend upon mass of species (number of particles) but ArG of the cell reaction is an extensive property because this depends upon mass of species (number of particles).

Question.4. The difference between the electrode potentials of two electrodes when no
current is drawn through the cell is called 
(a) cell potential (b) cell emf
(c) potential difference (d) cell voltage
Solution: (b) EMF is the difference between the electrode potentials of two electrodes • when no current is drawn through the cell.

Question.5. Which of the following statements is not correct about an inert electrode in a cell?
(a) It does not participate in the cell reaction.
(b) It provides surface either for oxidation or for reduction reaction.
(c) It provides surface for conduction of electrons.
(d) It provides surface for redox reaction.
Solution: (d) Inert electrode does not participate in redox reaction and acts only as source or sink for electrons. It provides surface either for oxidation or for reduction reaction.

Question.6. An electrochemical cell can behave like an electrolytic cell when

Solution: (c) If an external opposite potential is applied on the galvanic cell and increased reaction continues to take place till the opposing voltage reaches the value 1.1 V.
At this stage no current flow through the cell and if there is any further increase in the external potential then reaction starts functioning in opposite direction.
Hence, this works as an electrolytic cell.

Question.7. Which of the following statement about solutions of electrolytes is not correct?
(a) Conductivity of solution depends upon size of ions.
(b) Conductivity depends upon viscosity of solution.
(c) Conductivity does not depend upon salvation of ions present in solution.
(d) Conductivity of solution increases with temperature.
Solution: (c) Conductivity depends upon salvation of ions present in the solution. The greater the salvation of ions, the lesser is the conductivity.

Question.8.Using given below find strongest reduction agent.

Solution: (b) A negative value of standard reduction potential for Cr3+ to Cr means that the redox couple is a stronger reducing agent.

Question.9. Use the data given in Q. 8 and find out which of the following is the strongest oxidizing agent.

Solution: (c) The higher the positive value of standard reduction potential of metal ion, the higher will be its oxidizing capacity.
Since,

has value equal to 1.51 V hence it is the strongest oxidizing agent.

Question.10. Using the data given in Q. 8, find out in which option the order of reducing power is correct.

Solution: (b) The lower the reduction potential, the higher is the reducing power.
The order of reducing power is

Question.11. Use the data given in Q. 8 and find out the most stable ion in its reduced form.

Solution: (d) Mn⁺² is most stale in its reduced form due to highest E° value.

Question.12. Use the data given in Q.8 and find out the most stable oxidized species.

Solution:

Question.13. The quantity of charge required to obtain one mole of aluminium from

Al₂O₃ is (a) IF (b) 6F (c) 3F (d) 2F
Solution:

Question.14. The cell constant of a conductivity cell
(a) changes with change of electrolyte
(b) changes with change, of concentration of electrolyte
(c) changes with temperature of electrolyte
(d) remains constant for a cell
Solution: (d) The cell constant of a conductivity cell (a) remains constant for a cell.

Question.15. While charging the lead storage battery
(a)PbSO₄ anode is reduced to Pb
(b)PbSO₄ cathode is reduced to Pb
(c)PbSO₄ cathode is oxidized to Pb
(d)PbSO₄ anode is oxidized to Pb02
Solution: (a) While charging the lead storage battery the reaction occurring on cell is reversed and PbSO₄(s) on anode and cathode is converted into Pb and Pb02 respectively.
Hence, option (a) is the correct choice The electrode reactions are as follows:

Question.16.

Solution:

Question.17. In the electrolysis of aqueous sodium chloride solution, which of the half-cell reaction will occur at anode?

Solution: (d) During electrolysis of aqueous

(ii)The reaction at anode with lower value of E° should be preferred, but oxidation of 02 is kinetically slow process and needs high voltage thus reaction (i) takes place.
More than One Correct Answer Type

Question.18. The positive value of the standard electrode potential of Cu⁺²/Cu indicates
that
(a) this redox couple is a stronger reduction agent than the H^{/H₂ couple}
(b) this redox couple is a stronger oxidizing agent than H⁺/H₂
(c) Cu can displace H₂ from acid
(d) Cu cannot displace H₂  from acid
Solution: (b, d) The Lesser the E° value of redox couple, the higher the reducing power ’.

Since, 2 H⁺/H₂ has lesser SRP than Cu⁺²/Cu redox couple. Therefore,
(i) This redox couple is a stronger oxidizing agent than H⁺/H₂
(ii) Cu cannot displace H₂ from acid.
Hence, (b) and (d) are correct.

Question.19. E°ell for some half-cell reactions are given below. On the basis of these marks the correct answer will be

(a) In dilute sulphuric acid solution, hydrogen will be reduced at cathode.
(b) In concentrated sulphuric acid solution, water will be oxidized at anode.
(c) In dilute sulphuric acid solution, water will be oxidized at anode.
(d) In dilute sulphuric acid solution, SO^2-ion will be oxidized to tetrathionate ion at anode.
Solution: (a, c) In dilute sulphuric acid solution, hydrogen will be reduced at cathode.

while in concentrated solution of sulphuric acid, SO^-2 ions are oxidized to tetrathionate (SO2) ions.

Question.20. E°en = 1.1 V for Daniell cell. Which of the following expressions are correct description of state of equilibrium in this cell?

Solution:

Question.21. Conductivity of an electrolytic solution depends on
(b) concentration of electrolyte
(d) distance between the electrodes
Solution: (a, b) Conductivity of electrolyte solution is due to presence of mobile ions in the solution. This type of conductance is known as ionic conductance. Conductivity of these type of solutions depend upon
(i) the nature of electrolyte added
(ii) size of the ion produced and their solvation
(iii) concentration of electrolyte
(iv) nature of solvent and its viscosity
(v) temperature
While power of source or distance between electrodes has no effect on conductivity of electrolyte solution.
Hence, options (a) and (b) are the correct choices.

Question.22.

Solution:

However, the sum of molar conductivities of constituent ions gives the molar conductivity of water but here _NH4OH is a weak electrolyte of which complete decomposition is not possible.

Question.23. What will happen during the electrolysis of aqueous solution of CuS04 by using platinum electrodes?
(a) Copper will deposit at cathode.
(b) Copper will deposit at anode.
(c) Oxygen will be released at anode.
(d) Copper will dissolve at anode.
Solution:

The reaction with lower value of E° will be preferred at anode, hence 02 is released at anode.

Question.24. What will happen during the electrolysis of aqueous solution of CuS04 in the presence of Cu electrodes?
(a) Copper will deposit at cathode.
(b) Copper will dissolve at anode.
(c) Oxygen will be released at anode.
(d) Copper will deposit at anode.
Solution: (a, b) Electrolysis of CuS04 can be represented by two half-cell reactions these occurring at cathode and anode respectively as

Here, Cu will deposit at cathode while copper will dissolved at anode. Hence, options (a) and (b) are the correct choices.

Question.25. Conductivity K, is equal to

Solution:

Question.26. Molar conductivity of ionic solution depends on .
(a) temperature
(b) distance between electrodes
(c) concentration of electrolytes in solution
(d) surface area of electrodes
Solution: (a, c) Molar conductivity of ionic solution depends on temperature and concentration of electrolytes in solution.

Question.27. For the given cell, Mg|Mg²⁺||Cu²⁺ || Cu
(a) Mg is cathode
(b) Cu is cathode
(c) The cell reaction Mg + Cu²⁺+ —»Mg²⁺ + Cu
(d) Cu is the oxidizing agent
Solution: (b, c) Left side of cell reaction represents oxidation half-cell i.e., oxidation of Mg and right side of cell represents reduction half-cell reactions i.e., reduction of copper.
(ii) Cu is reduced and reduction occurs at cathode.
(iii) Mg is oxidized and oxidation occurs at anode.
(iv) Whole cell reaction can be written as

Hence, options (b) and (c) both are correct choices.

Short Answer Type Questions

Question.28. Can absolute electrode potential of an electrode be measured?
Solution: No, absolute electrode potential of an electrode cannot be measured.

Question.29. Can

for cell reaction ever be equal to zero?
Solution:

Question.30. Under what condition is

Solution: At equilibrium i.e., when the cell is completely discharged,

Question.31. What does the negative sign in the expression

mean? Zn /Zn
Solution:

Question.32. Aqueous copper sulphate solution and aqueous silver nitrate solution are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells. Will the mass of copper and silver deposited on the cathode be same or different? Explain your answer.
Solution:

Mass of silver will be different because the equivalent mass of Ag is different.

Question.33. Depict the galvanic cell in which the cell reaction is Cu + 2Ag⁺ > 2Ag +Cu2
Solution: Cu | Cu²⁺ (aq, 1M) || Ag⁺ (aq, 1M) | Ag

Question.34. Value of standard electrode potential for the oxidation of Cl ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is Cl- oxidized at anode instead of water?
Solution: Under the conditions of electrolysis of aqueous sodium chloride, oxidation of water at anode requires over potential and therefore, Cu is oxidised instead of water.

Question.35. What is electrode potential?
Solution: The electrical potential difference set up between the metal and its solution is called electrode potential.

Question.36. Consider the following diagram in which an electrochemical cell is coupled to an electrolytic cell. What will be the polarity of electrodes ‘A’ and ‘B’ in the electrolytic cell?

Solution: ‘A’ will have -ve polarity and ‘B’ will have +ve polarity.

Question.37. Why is alternating current used for measuring resistance of an electrolytic solution?
Solution: The alternating current is used to prevent electrolysis so that the concentration of ion in the solution remains constant.

Question.38. A galvanic cell has electrical potential of 1.1 V. If an opposing potential of 1.1 V is applied to this cell, what will happen to the cell reaction and current flowing through the cell?
Solution: When the opposing potential becomes equal to electrical potential, the cell reaction stops and no current flows through the cell. Thus, there is no chemical reaction.

Question.39. How will the pH of brine (aq. NaCl solution) be affected when it is electrolysed?
Solution: Aqueous solution of brine contains Na+, CP, H⁺ and H^–. Electrode process are given as follows:

Remaining solution will contain NaOH, which is base, therefore, pH will increase.

Question.40. Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why?
Solution: Electrolyte is not consumed in the cell process of mercury cell hence it will deliver the current at constant potential throughout its life.

Question.41. Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Am of ‘B’ increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer.
Solution: Electrolyte ‘B’ will be strong electrolyte because it is completely ionised and on dilution the molar conductance will increase to the small extent due to increase in speed of ions only.

Question.42. When acidulated water (dil. H2SO₄ solution) is electrolysed, will the pH of the solution be affected? Justify your answer.
Solution: The pH of solution will not be affected because there is no change in the concentration of  H⁺ions.

Question.43. In an aqueous solution, how does specific conductivity of electrolytes change with addition of water?
Solution: On the addition of water, number of ions per unit volume decreases and therefore conductivity decreases.

Question.44. Which reference electrode is used to measure the electrode potential of other electrodes?
Solution: The standard hydrogen electrode is used as a reference electrode whose electrode potential is taken to be. zero. The electrode potential of other electrodes is measured with respect to it.

Question.45. Consider a cell given below:
Cu |Cu²⁺ || Cl |Cl₂, Pt
Write the reactions that occur at anode and cathode.
Solution: The given cell is:

Question.46. Write the Nemst equation for the cell reaction in the Daniell cell. How will the Ecell be affected when concentration of Zn²⁺ ions is increased?
Solution: Daniell cell
Zn(s) | Zn²⁺ || Cu²⁺ | Cu(s)

Above equation shows that the cell potential will decrease with increase in the concentration of Zn2+ ion.

Question.47. What advantage do the fuel cells have over primary and secondary batteries?
Solution: Primary batteries contain a limited amount of reactants and are discharged when the reactants have been consumed. Secondary batteries can be recharged but take a long time to recharge. Fuel cell runs continuously as long as the reactants are supplied to it and products are removed continuously.

Question.48. Write the cell reaction of a lead storage battery when it is discharged. How does the density of the electrolyte change when the battery is discharged?
Solution:

Density of electrolyte decreases because water is formed and sulphuric acid consumed as the product during discharge of the battery.

Question.49. Why on dilution the Am of CH3COOH increases drastically, while that of
CH4COONa increases gradually? .
Solution: In the case of CH3COOH, which is a weak electrolyte, the number of ions increase on dilution due to an increase in degree of dissociation.

In the case of strong electrolyte such as CH3COONa, the number of ions remains the same but the interionic attraction decreases.

Matching Column Type Questions

Question.50. Match the terms given in Column 1 with the units given in Column II.

Solution: (i —» c), (ii —> d), (iii —> a), (iv —> b)

Question.51. Match the terms given in Column I with the items given in Column II.

Solution: (i -> d), (ii -> a), (iii -> b), (iv -> c)

Question.52. Match the items of Column I and Column II.

Solution: (i —> d), (ii -4 c) (iii -4 a), (iv -4 b)

Question.53.Match the items of Column I and Column II.

Solution:(i -> d), (ii -> c), (iii -> b), (iv -> a)

Question.54.Match the items of Column I and Column II.

Solution: (i -> d), (ii -> c), (iii —> a, e), (iv -> b)

Question.55. Match the items of Column I and Column II on the basis of data given below:

Solution:(i->c),(ii->a),(iii->g),(iv->e),(v->d),(vi->b),(vii->f).

Assertion and Reason Type Questions

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
(a) Both Assertion and Reason are true and the Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true and the Reason is not correct explanation for Assertion
(c) Assertion is true but the Reason is false.
(d) Both Assertion and Reason are false.
(e) Assertion is false but Reason is true.

Question.56. Assertion (A): Cu is less reactive than hydrogen.
Reason (R): E°u2+/Cu is negative.
Solution: (c) Cu is less reactive than hydrogen because E°u2+ Cu is positive.

Question.57. Assertion (A): Ecell should have a positive value for the cell to function. Reason (R). Ecadlode Eanode
Solution: (c) Ecell = Ecathode – Eanode. To have positive value of Ecell, Ecathode > Eanode.

Question.58. Assertion (A): Conductivity of all electrolytes decreases on dilution.
Reason (R): On dilution number of ions per unit volume decreases.
Solution: (a) Conductivity depends on number of ions per unit volume which decreases on dilution of electrolytes.

Question.59. Assertion (A): Lm for weak electrolytes shows a sharp increase when the electrolytic solution on dilution of solution.
Electrockemistrij 63
Reason (R): For weak electrolytes dissociate partially in concentrated solution. On dilution, their degree of dissociation increases hence their Am increases sharply.
Solution: (a) Weak electrolytes dissociate partially in concentrated solution. On dilution, their degree of dissociation increases hence their Am increases sharply.

Question.60. Assertion (A): Mercury cell does not give steady potential.
Reason (R): In the cell reaction, ions are not involved in solution.
Solution: (e) Mercury cell gives a steady potential because in the cell reaction ions are not involved in the solution.

Question.61. Assertion (A): Electrolysis of NaCl solution gives chlorine at anode instead of 02.
Reason (R): Formation of oxygen at anode requires overvoltage.
Solution: (a) Formation of oxygen has lower value of E° than formation of chlorine even then it is not formed because it requires overvoltage.

Question.62. Assertion (A): For measuring resistance of an ionic solution an AC source is used.
Reason (R): Concentration of ionic solution will change if DC source is used.
Solution: (a) Alternating current is used in the measurement of resistance of electrolyte solution because concentration changes with DC current due to electrolysis.

Question.63. Assertion (A): Current stops flowing when Ece,| = 0.
Reason (R): Equilibrium of the cell reaction is attained.
Solution: (a) At equilibrium Eccn = 0 and no current flows. ‘

Question.64. Assertion (A): E , increases with increase in concentration of Ag* ions.
Ag /Ag ^
Reason (R): E + has a positive value.
Ag /Ag
Solution:

On increasing [Ag*], EA + will increase and it has a positive value.

Question.65. Assertion (A): Copper sulphate can be stored in zinc vessel.
Reason (R): Zinc is less reactive than copper.
Solution: (d) Zinc will get dissolved in CuS04 solution since zinc is more reactive than copper.

Long Answer Type Questions

Question.66. Consider the following figure and answer the following questions.

(i) Cell ‘A’ has Ecell = 2 V and cell ‘B’ has Ecen = 1.1 V which of the two cells ‘A’ or ‘B’ will act as an electrolytic cell. Which electrode reactions will occur in this cell?
(ii) If cell ‘A’ has Ecen = 0.5 V and cell ‘B’ has Ecen = 1.1 V, what will be the reactions at anode and cathode?
Solution: (i) Cell ‘B’ will act as electrolytic cell because potential of ‘B’ is less than
that of ‘A’. Electrode process in the cell ‘B’ may be given as
<

Question.67. Consider the figure given below and answer the questions (i) to (vi) that follow.

(i) Redraw the diagram to show the direction of electron flow.
(ii) Is silver plate anode or cathode?
(iii)What will happen if salt bridge is removed?
(iv)When will the cell stop functioning?
(v)How will concentration of Zn2+ cell functions?
(vi)How will the concentration of Zn2+ ions and Ag+ ions be affected after the cell becomes ‘dead’?
Solution:(i)The cell is:
Zn(s) | Zn2+ || Ag+ | Ag
(ii)Electron will flow from zinc anode to silver cathode in external circuit. Silver will act as cathode, since its standard reduction potential is greater than that of zinc.
(iii)Potential will drop to zero if salt bridge is suddenly removed.
(iv)Cell will stop functioning when it is discharged i.e., when cell potential is zero.
(v)Nemst equation for the cell is: 0.059,

Cell potential will decrease with increase in concentration of [Zn2+] while it will increase with the concentration of [Ag+].
(vi) When cell is dead or discharged, E will be zero and the cell will be at
equilibrium. Then, concentration of Zn2+ and Ag+ will not change.

Question.68. What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell? When will the maximum work be obtained from a galvanic cell?
Solution: The required relations are:

The post NCERT Exemplar Problems Class 12 Chemistry Electrochemistry appeared first on Learn CBSE.

NCERT Exemplar Problems Class 12 Chemistry Chapter 4 Chemical Kinetics

Multiple Choice Questions

Single Correct Answer Type
Question.1. The role of a catalyst is to change
(a) Gibbs energy of reaction (b) enthalpy of reaction
(c) activation energy of reaction (d) equilibrium constant
Solution: (c) A catalyst lowers the activation energy of a reaction.

Question.2. In the presence of a catalyst, heat evolved or absorbed during reaction
(a) increases (b) decreases
(c) remains unchanged .(d) may increase or decrease
Solution: (c) There is no effect on heat evolved or absorbed during the reaction in the presence of a catalyst since it does not participate in the reaction.

Question.3. Activation energy of a chemical reaction can be determined by
(a) determining the rate constant at standard temperature
(b) determining the rate constants at two temperatures
(c) determining probability of collision
(d) using catalyst
(b) Activation energy of a chemical reaction is related to rate constant of a reaction at two different temperatures i.e.,K₁ and K₂  respectively.
Solution:

Question.4.Consider the following figure and mark the correct option.

(a) Activation energy of forward reaction is E₁ + E₂  and product is less stable than reactant.
(b) Activation energy of forward reaction is E₁ + E₂ and product is more stable than reactant.
(c) Activation energy of both forward and backward reactions is E₁ + E₂ and reactant is more stable than product.
(d) Activation energy of backward reaction is E, and product is more stable than reactant.
Solution: (a) Ea(forward) = E₁ + E₂
Since energy of reactants is less than products and the product is less stable than the reactant.

Question.5. Consider a first order gas phase decomposition reaction given below:

Solution: (b) Consider a first order gas phase decomposition reaction:

Question.6. According to Arrhenius equation rate constant k is equal to Ae^-Ea/RT. Which of the following options represents the graph of In k versus ?

Solution:

Question.7. Consider the Arrhenius equation given below and mark the correct option. k =Ae^-Ea/RT
(a) Rate constant increases exponentially with increasing activation energy _ and decreasing temperature.
(b) Rate constant decreases exponentially with increasing activation energy and decreasing temperature.
(c) Rate constant increases exponentially with decreasing activation energy and decreasing temperature.
(d) Rate constant increases exponentially with decreasing activation energy and increasing temperature.
Solution: (d) K= Ae^-Ea/RT from the equation it is clear that value of rate constant k increases exponentially with decrease in activation energy Ea and increase in temperature.

Question.8. A graph of volume of hydrogen released vs time for the reaction between zinc and dil. HC1 is shown below. On the basis of this, mark the correct option.

Solution: (c) Average rate of reaction up to 40 seconds on the basis of the graph is

Question.9. Which of the following statement is not correct about order of a reaction?
(a) The order of a reaction can be a fractional number.
(b) Order of a reaction is experimentally determined quantity.
(c) The order of a reaction is always equal to the sum of the stoichiometric coefficients of reactants in the balanced chemical equation for a reaction.
(d) The order of a reaction is the sum of the powers of molar concentration of the reactants in the rate law expression.
Solution: (c) Out of the given four statements, option (c) is not correct.
Order of reaction is equal to the sum of powers of concentration of the reactants in rate law expression.
For any chemical reaction

Order of reaction can be a fraction also. Order of reaction is not always equal to sum of the stoichiometric coefficients of reactants in the balanced chemical equation. For a reaction it may or may not be equal to sum of stoichiometric coefficient of reactants.

Question.10. Consider the graph given in Q. 8. Which of the following option does not show instantaneous rate of reaction at 40th second?

Solution:(b) Does not show instantaneous rate of reaction at 40th second.

Question.11. Which of the following statement is correct?
(a) The rate of a reaction decreases with passage of time as the concentration of reactants decreases.
(b) The rate of a reaction is same at any time during the reaction.
(c) The rate of a reaction is independent of temperature change.
(d) The rate of a reaction decreases with increase in concentration of reactant(s).
Solution: (a) The rate of a reaction depends upon the concentration of reactants.

Question.12. Which of the following expressions is correct for the rate of reaction given below?

Solution:
Rate law expression for the above equation can be written as

Question.13. Which of the following graph represents exothermic reaction?

Solution:For a exotheramic reaction:

Question.14. Rate law for the reaction ‘B’ is doubled, keeping the concentration of ‘A’
constant, the value of rate constant will be
(a) the same (b) doubled (c) quadrupled (d) halved
Solution: (a) Rate constant of a reaction does not depend upon concentrations of the reactants.

Question.15. Which of the following statement is incorrect about the collision theory of chemical reaction?
(a) It considers reacting molecules or atoms to be hard spheres and ignores their structural features.
(b) Number of effective collisions determines the rate of a reaction.
(c) Collision of atoms or molecules possessing sufficient threshold energy results into the product formation.
(d) Molecules should collide with sufficient threshold energy and proper orientation for the collision to be effective.
Solution: (c) According the postulates of collision theory there are the following necessary conditions for any reaction to be occur:
(i) Molecule should collide with sufficient threshold energy.
(ii) Their orientation must be proper.
(iii) The collision must be effective.

Question.16. A first order reaction is 50% completed in 1.26×1014 s. How much time would it take for 100% completion?
(a) 1 26 x 1015 s (b) 2.52×1014 s 
(c) 2.52 x 1028 s (d) infinite
Solution: (d) Reaction would be 100% complete only after infinite time which cannot be calculated.

Question.17. Compounds ‘A’ and ‘B’ react according to the following chemical equation.

Concentration of either ‘A’ or ‘B’ were changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. The following results were obtained. Choose the correct option for the rate equations for this reaction.

Solution: (b)
Rate=K[A]^x[B]^y  
When concentration of B is doubled keeping the concentration of A constant, the rate of formation of C increases by a factor of four. This indicates that the rate of reactions depends upon the square of concentration of B. When concentration of A is doubled, the rate of formation of C also doubles from the initial value. This shows that the rate depends on first power of concentration of A. Hence Rate= K[A]^x[B]^y.

Question.18. Which of the following statement is not correct for the catalyst?
(a) It catalyses the forward and backward reactions to the same extent.
(b) It alters AG of the reaction.
(c) It is a substance that does not change the equilibrium constant of a reaction.
(d) It provides an alternate mechanism by reducing activation energy between reactants and products.
Solution: (b) Characteristics of catalyst
(a) It catalyses the forward and’backward reactions to the same extent as it decreases energy of activation hence, increases the rate of both the reactions.
(b) Since reaction quotient is ‘the relation between concentration of reactants and products. Hence, catalyst does not alter Gibbs free energy as it is related reaction quotient. Thus, Gibbs free energy does not change during the reaction when catalyst is added to it.

(c) It does not alter equilibrium of reaction as equilibrium constant is also concentration dependent term.
(d) It provides an alternate mechanism by reducing activation energy between reactants and product.

Question.19. The value of rate constant of a pseudo first order reaction
(a) depends on the concentration of reactants present in small amount
(b) depends on the concentration of reactants present in excess
(c) is independent of the concentration of reactants
(d) depends only on temperature
Solution: (b) The value of rate constant of a pseudo first order reaction depends on the concentration of reactants present in excess.

Question.20. Consider the reaction A —> B. The concentration of both the reactants and the products varies exponentially with time. Which of the following figures correctly describes the change in concentration of reactants and products with time?

Solution: A->B

The concentration of reactants decreases with time and concentration of products increases with time.

More than One Correct Answer Type

Question.21. Rate law cannot be determined from balanced chemical equation if
(a) reverse reaction is involved
(b) it is an elementary reaction
(c) it is a sequence of elementary reactions
(d) any of the reactants is in excess
Solution: (a, c, d) Rate law can be determined from balanced chemical equation if it is an elementary reaction.

Question.22. Which of the following statements are applicable to a balanced chemical equation of an elementary reaction?
(a) Order is same as molecularity
(b) Order is less than the molecularity
(c) Order is greater than the molecularity
(d) Molecularity can never be zero
Solution: (a, d) For an elementary reaction, the order and molecularity are same. Molecularity of a reaction can never be zero.

Question.23. In any uni molecular reaction,
(a) only one reacting species is involved in the rate determining step
(b) the order and the molecularity of slowest step are equal to one
(c) the molecularity of the reaction is one and order is zero
(d) both molecularity and order of the reaction
Solution: (a, b) For any unimolecular reaction, only one reactant is involved in rate determining step and the order and molecularity of rate determining or slowest step are one.

Question.24. For a complex reaction
(a) order of overall reaction is same as molecularity of the slowest step
(b) order of overall reaction is less than the molecularity of the slowest step
(c) order of overall reaction is greater than molecularity of the slowest step
(d) molecularity of the slowest step is never zero or non-integer ‘
Solution: (a, d)
(a) For a complex reaction, order of overall reaction = molecularity of slowest step.
As rate of overall reaction depends upon total number of molecules involved in slowest step of the reaction. Hence, molecularity of the slowest step is equal to order of overall reaction.
(b) Since the completion of any chemical reaction is not possible in the absence of reactants. Hence, slowest step of any chemical reaction must
contain at least one reactant. Thus, molecularity’of the slowest step is never zero or non-integer.

Question.25. At high pressure the following reaction is zero order.

Which of the following options are correct for this reaction?
(a) Rate of reaction = Rate constant
(b) Rate of the reaction depends on concentration of ammonia.
(c) Rate of decomposition of ammonia will remain constant until ammonia disappears completely
(d) Further increase in pressure will change the rate of reaction.
Solution: (a, c, d) For a zero order reaction, the rate of reaction is independent of concentration of ammonia.

Question.26. During decomposition of an activated complex
(a) energy is always released (b) energy is always absorbed
(c) energy does not change (d) reactions may be formed
Solution: (a, d) Activated complex is formed at the highest level of energy of the system. It is unstable and decomposes to give products with release of energy. Reactants may be formed if the reaction is reversible.

Question.27. According to Maxwell Boltzmann distribution of energy
(a) the fraction of molecules with most probable kinetic energy decreases at higher temperatures
(b) the fraction of molecules with most probable kinetic energy increases at higher temperatures
(c) most probable kinetic energy increases at higher temperatures
(d) most probable kinetic energy decreases at higher temperatures
Solution: (a, c) Distribution of kinetic energy may be described by plotting a graph of fraction of molecules versus kinetic energy.

Kinetic energy of maximum fraction of molecule is known as most probable kinetic energy. It is important to note that with increase of temperature, peak shifts forward but downward. This means that with increase of temperature
(i) most probable kinetic energy increases.
(ii) the fractions of molecules possessing most probable kinetic energy decreases.

Question.28. In the graph showing Maxwell Boltzmann distribution of energy,

(a) area under the curve must not change with increase in temperature
(b) area under the curve increases with increase in temperature
(c) area under the curve decreases with increase in temperature
(d) with increase in temperature curve broadens and shifts to the right hand side
Solution: (a, d) Area under the curve remains same with increase in temperature and with increase in temperature curve broadens and shifts to the right-hand side.

Question.29. Which of the following statements are in accordance with the Arrhenius equation?
(a) The rate of a reaction increases with increase in temperature.
(b) The rate of reaction increases with decrease in activation energy.
(c) The rate constant decreases exponentially with increase in temperature.
(d) The rate of reaction decreases with decrease in activation energy.
Solution: (a, b) The rate of reaction increases with temperature. It becomes double with every 10° change. It increases with decrease in activation energy.

Question.30. Mark the incorrect statements.
(a) Catalyst provides an alternative pathway to reaction mechanism.
(b) Catalyst raises the activation energy.
(c) Catalyst lower the activation energy.
(d) Catalyst alters enthalpy change of the reaction.
Solution: (b, d) .
(i) As the catalyst is added to the reaction medium rate of reaction increases by decreasing activation energy of molecule. Hence, it follows an alternative pathway.
(ii) Catalyst does not change the enthalpy change of reaction. Energy of ‘ reactant and product remain same in both catalysed and .uncatalysed
reaction.Hence, (a) and (c) are incorrect statements.

Question.31. Which of the following graphs are correct for a zero order reaction?

Solution: (a, d) If we plot [R] against t we get a straight line with slope = -k and intercept equal to[R]⁰ from the equation

Question.32. Which of the following graphs are correct for a first order reaction?


Solution: (a, d) For a first order reaction

Short Answer Type Questions

Question.33. State a condition under which a biomolecular reaction is kinetically first order reaction.
Solution: Let us consider a biomolecular reaction:
A + B –>Product
Rate = k[A] [B] …………. (i)
When concentration of [B] is taken in large excess, rate law will become Rate = k’ [A]
where, k’ = k[B] ,
The order of reaction will be equal to one.

Question.34. Write the rate equation for the reaction 2A + B–>C if the order of the
reaction is zero.
Solution: 2A+B –>C
Rate = k[A]° [B]°

Question.35. How can you determine the rate law of the following reaction?

Solution: The rate law cqn be determined by measuring the rate of this reaction as a function of initial concentration by keeping the concentration of one of the reactants constant and changing the concentration of other reactant or by changing the concentration of both the reactants. From the concentration dependence of rate, rate law can be determined.

Question.36. For which type of reactions, order and molecularity have the same value?
Solution: Elementary reactions have same value of order and molecularity.

Question.37. In a reaction, if the concentration of reactant A is tripled, the rate of reaction becomes twenty-seven times. What is the order of the reaction?
Solution:

Question.38. Derive an expression to calculate time required for completion of zero order reaction. ‘
Solution:

Question.39. For a reaction A + B —> Products, the rate law is — Rate = k[A][B]^3/2. Can – reaction be an elementary reaction? Explain.
Solution:

Question.40. For a certain reaction, large fraction of molecules has energy more than the threshold energy, yet the rate of reaction is very slow. Why?
Solution: Though the reacting molecules may be having energy more than threshold energy, yet they may not be effective due to lack of proper orientation.

Question.41. For a zero order reaction will the molecularity be equal to zero? Explain.
Solution: Molecularity of a reaction can never be equal to zero.

Question.42. For a general reaction A > B, plot of concentration of A versus time is

given in figure. Answer the following questions on the basis of this graph.
(i) What is the order of the reaction?
(ii) What is the slope of the curve?
(iii) What are the units of rate constant?
Solution:

Question.43. The reaction between H₂(g) and O₂(g) is highly feasible yet allowing the gases to stand at room temperature in the same vessel does not lead to the formation of water. Explain.
Solution: 2H₂(g) +O₂(g) >2H₂0(l)
This reaction does not take place at room because the activation energy of the reaction is very high.

Question.44. Why does the rate of a reaction increase with rise in temperature?
Solution: At higher temperatures, larger fraction of colliding particles can cross the energy barrier (i.e. the activation energy), which leads to faster rate.

Question.45. Oxygen is available in plenty in air yet fuels do not bum by themselves at room temperature. Explain.
Solution: The activation energy for combustion reactions of fuels is very high at room temperature, therefore, they do not bum by themselves.

Question.46. Why is the probability of reaction with molecularity higher than three very rare?
Solution: The probability of more than three molecules colliding simultaneously is very small. Therefore, the possibility of molecularity being three is very low.

Question.47. Why does the rate of any reaction generally decreases during the course of the reaction?
Solution: The rate of a reaction depends on the concentration of reactants. As the reaction progresses, the concentration of reactants decreases because the reactants start getting converted to products. Hence the rate decreases.

Question.48. Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.
Solution: Thermodynamics feasibility of a reaction depends on Gibbs free energy i.e., AG must be negative for spontaneous process. Kinetic feasibility depends on the activation energy of reaction, the lesser is the activation energy, the greater is the feasibility of reaction, i.e.,

This process is thermodynamically feasible but it is very slow due to its high activation energy.

Question.49. Why, in the redox titration of KMnO4versus oxalic acid we heat oxalic acid solution before starting the titration.
Solution: The reaction between KMnO4 and oxalic acid is very slow. By raising the temperature we can increase the rate of reaction.

Question.50. Why cannot molecularity be applicable only for elementary reactions and order applicable for elementary as well as complex reaction?
Solution: Molecularity is the number of molecules involved in the elementary process. It is at least one. Thus, zero molecularity is not possible.

Question.51. Why is molecularity applicable only for elementary reactions and order applicable for elementary as well as complex reaction?
Solution: A complex reaction occurs through several elementary reactions. Numbers of molecules involved in each elementary reaction may be different i.e., the molecularity of each step pi ay be different. Therefore, the molecularity
of overall complex reaction is meaningless. On the other hand, order of a complex reaction is experimentally determined by the slowest step in its mechanism and is therefore, applicable even in the case of complex reactions.

Question.52. Why can we not determine the order of reaction by taking into consideration the balanced chemical equation?
Solution: Balanced chemical equation often leads to incorrect order or rate law. For example, the following reaction appears to be a tenth order reaction.

However, this is actually a second order reaction. Actually, the reaction is complex and occurs in several steps. The order of such reaction is determined by the slowest step in the reaction mechanism. Order is determined experi¬mentally and gives the actual dependence of observed rate of reaction on the co’ncentration of reactants.

Matching Column Type Questions

Question.53. Match the graph given in Column I with the order of reaction given in Column II. More than one item in Column I may link to the same item of Column II.

Solution:

Question.54. Match the statements given in Column I and Column II.

Solution: (i —» c), (ii -> a), (iii -> d), (iv -> f), (v -» b), (vi-> e)
(i) Catalyst alters the rate of reaction by lowering activation energy.
(ii) Molecularity cannot be fraction or zero. If molecularity is zero, reaction is not possible.
(iii) Second half-life of first order reaction is same as first because half-life
time is temperature independent. ‘
(iv) e-Ea/RT refers to the fraction of molecules with kinetic energy equal to greater than activation energy.
(v) Energetically favourble reactions are sometimes slow due to improper orientation of molecule cause some ineffective collision of molecules.
(vi) Area under the Maxwell-Boltzmann curve is constant because total probability of molecule taking part in a chemical reaction is equal to one.

Question.55. Match the items of Column I and Column II.

Solution: (i ->b), (ii-> a), (iii-> c)
(i) Diamond cannot be converted into graphite under ordinary condition.
(ii) Instantaneous rate of reaction complete at very short span of time.
(iii) Average rate of reaction occurs to a long duration of time.

Question.56. Match the items of Column I and Column II.

Solution: (i -> b), (ii -> a), (iii —> d), (iv-> c)
(i) Mathematical expression for rate of reaction is known as rate law.
(ii) Rate of reaction for zero order reaction is equal to rate constant
r = k[A] .
(iii) Unit of rate of reaction is same as that or rate of reaction.
(iv) Order of complex reaction is determined by rate of a reaction, which is slowest.

Assertion and Reason Type Questions

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
(a) Both Assertion and Reason are correct and the reason is the correct explanation of Assertion.
(b) Both assertion and reason are correct but Reason is not the correct explanation of Assertion
(c) Assertion is correct but Reason is incorrect.
(d) Both Assertion and Reason are incorrect.
(e) Assertion is incorrect but Reason is correct.

Question.57. Assertion (A): Order of the reaction can be zero or fractional.
Reason (R): We cannot determine order from balanced chemical equation.
Solution: (b) Order of a reaction may be zero or fractional. Order can be determined by rate law expression.

Question.58. Assertion (A): Order and molecularity are same.
Reason (R): Order is determined experimentally and molecularity is the sum of the stoichiometric coefficient of rate determining elementary step.
Solution: (e) Order and molecularity may not be necessarily same. Order is determined experimentally but molecularity is calculated using balanced stoichiometric equation.

Question.59. Assertion (A): The enthalpy of reaction remains constant in the presence of a catalyst.
Reason (R): A catalyst participating in the reaction, forms different activated complex and lowers down the activation energy but the difference in energy of reactant and product remains the same.
Solution: (a) AH = Activation Energy of forward reaction – Activation Energy of reverse reaction.
Catalyst does not alter heat of reaction because it affects activation energy of forward and reverse reactions equally.

Question.60. Assertion (A): All collision of reactant molecules lead to product formation. Reason (R): Only those collisions in which molecules have correct orientation and sufficient kinetic energy lead to compound formation.
Solution: (e) Every collision among reactant molecules does not lead to the formation of product. Only effective collision brings out the formation of product.

Question.61. Assertion (A): Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules.
Reason (R): Reactant molecules undergo chemical change irrespective of their orientation during collision.
Solution: (c) According to Arrhenius equation (k =Ae^-Ea/RT); it is found almost accurate for single as well as complex reaction.
However, orientation is essential for the reactant molecules participating in the reaction.

Long Answer Type Questions

Question.62. All energetically effective collisions do not result in a chemical change. Explain with the help of an example.
Solution: Only effective collision leads to the formation of products. It means that collisions in which molecules collide with sufficient kinetic energy (called threshold energy = activation energy + energy possessed by reacting species). And proper orientation lead to a chemical change because it facilitates the breaking of old bonds between (reactant) molecules and formation of the new ones, i.e., in products.
e.g., formation of methanol from bromomethane depends upon the orientation of the reactant molecules.

The proper orientation of reactant molecules lead to bond formation whereas improper orientation makes them simply back and no products are formed. To account to effective collisions, another factor P (probability or steric factor) is introduced K = PZABe_Ea/RT .

Question.63. What happens to most probable kinetic energy and the energy of activation with increase in temperature?
Solution:

When the temperature is raised’ the maxima of the curve moves to the higher energy value and the curve broadens out so that there is a greater proportion of the molecules with much higher energies. Most probable kinetic energy increases with increase in temperature. As the temperature increases, activation energy decreases which will result in the increase in the rate of reaction.

Question.64. Describe how does the enthalpy of reaction remain unchanged when a catalyst is used in the reaction?
Solution: A catalyst is a substance which increases the speed of a reaction without itself undergoing any chemical change.
According to “intermediate complex formation theory” reactants first combine with the catalyst to form an intermediate complex which is short-lived and decomposes to form the products and regenerating the catalyst.
The intermediate formed has much lower potential energy than the intermediate complex formed between the reactants in the absence of the catalyst.
Thus, the presence of catalyst lower the potential energy barrier and the reaction follows a new alternate pathway which require less activation energy. We know that, the lower the activation energy, the faster is the reaction because more reactant molecules can cross the energy barrier and change into products.
Enthalpy, AH is a state function. Enthalpy of reaction, i.e., difference in energy between reactants and product is constant, which is clear from potential energy diagram.
Potential energy diagram of catalysed reaction is given as

Question.65. Explain the difference between instantaneous rate of a reaction and average rate of a reaction.
Solution: Average rate of a reaction is the change in concentration of reactants or products and the time taken for that change to occur.

It occurs for a long interval of time. It can be determined for multistep as well as elementary reactions.

Question.66. With the help of an example explain what is meant by pseudo first order reaction?
Solution: A reaction in which one reactant is present in large amount and its concentration does not get altered during the course of the reaction, behaves as first order reaction. Such reaction is called pseudo first order reaction.
For example, (i) hydrolysis of ethyl acetate

The post NCERT Exemplar Problems Class 12 Chemistry Chemical Kinetics appeared first on Learn CBSE.

NCERT Exemplar Problems Class 12 Chemistry Chapter 5 Surface Chemistry 

Multiple Choice Questions

Single Correct Answer Type
Question.1. Which of the following process does not occur at the interface of phases?
(a) Crystallisation (b) Heterogeneous catalysis
(c) Homogeneous catalysis ’ (d) Corrosion –
Solution: (c) There is no homogeneous catalysis.

Question.2. At the equilibrium position in the process of adsorption

Solution:

Question.3. Which of the following interface cannot be obtained?
(a) Liquid-liquid (b) Solid-liquid
(c) Liquid-gas (d) Gas-gas
Solution: (d) Gas-gas interface cannot be obtained as they are completely miscible in nature.
For example, air is a mixture of various gases such as, 02, N2, C02, etc.

Question.4. The term ‘sorption’ stands for
(a) absorption (b) adsorption
(c) both absorption and adsorption (d) desorption
Solution: (c) Adsorption and absorption when take place together, it is known as sorption.

Question.5. Extent of physisorption of a gas increases with
(a) increase in temperature
(b) decrease in temperature
(c) decrease in surface area of adsorbent
(d) decrease in strength of van der Waals forces
Solution: (b) Physical adsorption of a gas increases with decrease in temperature because, at higher temperature weak van der Waals forces between gas and the surface become difficult to exist.

Question.6. Extent of adsorption of adsorbate from solution phase increases with
(a) increase in amount of adsorbate in solution
(b) decrease in surface area of adsorbent
(c) increase in temperature of solution
(d) decrease in amount of adsorbate in solution
Solution: (a) Extent of adsorption of adsorbate from solution phase increases with increase in the amount of adsorbate in the solution. As amount of adsorbate in the solution increases interaction of adsorbent increases which leads to increase in extent of adsorption.

Question.7. Which one of the following is not applicable to the phenomenon of adsorption?

Solution: (a) Adsorption is an exothermic process. So,

Question.8. Which of the following is not a favourable condition for physical adsorption?
(a) High pressure
(b) Negative AH
(c) Higher critical temperature of adsorbate
(d) High temperature
Solution: (d) Physical adsorption is favoured at low temperature.

Question.9. Physical adsorption of a gaseous species may change to chemical adsorption with
(a) decrease in temperature
(b) increase in temperature
(c) increase in surface area of adsorbent
(d) decrease in surface area of adsorbent
Solution: (b) On increasing temperature physisorption changes to chemisorption. As the temperature increases, energy of activation of adsorbate particles increases which leads to formation of chemical bond between adsorbate and adsorbent. Hence, physisorption transform into chemisorption.

Question.10. In physisorption, adsorbent does not show specificity for any particular gas because
(a) involved van der Waals forces are universal
(b) gases involved behave like ideal gases.
(c) enthalpy of adsorption is low
(d) it is a reversible process ,
Solution: (a) Physisorption is not specific to any gas since it involves van der Waals forces and no specific bonds are formed.

Question.11. Which of the following is an example of absorption?
(a) Water on silica gel
(b) Water on calcium chloride
(c) Hydrogen on finely divided nickel
(d) Oxygen on metal surface
Solution: (b) Calcium chloride absorbs water. Other examples show adsorption.

Question.12. On the basis of data given below predict which of the following gases shows least adsorption on a definite amount of charcoal?

Solution: (d) The lesser the value of critical temperature of gases the lesser will be the extent of adsorption. Here H2 has lowest value of critical temperature, i.e., 33.
Hence, hydrogen gas shows least adsorption on a definite amount of charcoal.

Question.13. In which of the following reactions heterogeneous catalysis involved?
(i) 2S02(g) + 02(g)–>2S03(g)
(ii) 2S02(g) 2S03(g)
(iii) N2(g) + 3H2(g) –>2NH3(g)
(iv) CH3COOCH3(/) + H20(/) HCI —> CH3COOH(aq) + CH3OH(aq) 
(a) (ii), (iii) (b) (ii), (iii), (iv)
(c) (i), (ii), (iii) (d) (iv)
Solution: (a) In reaction (ii) and reaction (iii), catalysts are in solid state, and reactants and products are gases.

Question.14. At high concentration of soap in water, soap behaves as
(a) molecular colloid (b) associated colloid
(c) macromolecular colloid (d) lyophilic colloid .
Solution: (b) At low concentration soap behaves as strong electrolyte but at higher concentration it behaves as colloid due to formation of aggregates called micelles. These are known as associated colloids.

Question.15. Which of the following will show Tyndall effect?
(a) Aqueous solution of soap below critical micelle concentration.
(b) Aqueous solution of soap above critical micelle concentration.
(c) Aqueous solution of sodium chloride.
(d) Aqueous solution of sugar. 
Solution: (b) Aqueous solution of soap above critical micelle concentration leads to the formation of colloidal solution. Tyndall effect is a characteristic of colloidal solution in which colloidal particles show a coloured appearance when sunlight is passes through it and seen perpendicularly.

Question.16. Method by which lyophobic sol can be protected?
(a) By addition of oppositely charged sol
(b) By addition of an electrolyte
(c) By addition of lyophilic sol
(d) By boiling
Solution: (c) Lyophobic sol can be protected by adding lyophilic sol which is known as protective colloid.

Question.17. Freshly prepared precipitate sometimes gets converted to colloidal solution by •
(a) coagulation (b) electrolysis
(c) diffusion . (d) peptisation
Solution: (d) Peptisation is the process of converting freshly prepared precipitate into colloid.

Question.18. Which of the following electrolytes will have maximum coagulating value

Solution: (b) According to Hardy-Schulze law, the greater the charge on anion, the greater will be its coagulating power.

Question.19. A colloidal system having a solid substance as a dispersed phase and a liquid
as a dispersion medium is classified as
(a) solid sol (b) gel (c) emulsion (d) sol
Solution: (d) It is called sol.

Question.20. The values of colligative properties of colloidal solution are of small order in
comparison to those shown by true solutions of same concentration because colloidal particles
(a) exhibit enormous surface area
(b) remain suspended in the dispersion medium
(c) form .lyophilic colloids
(d) are comparatively less in number
Solution: (d) Colloidal particles are large in size and less in number.

Question.21. Arrange the following diagrams in correct sequence of steps involved in the mechanism of catalysis, in accordance with modem adsorption theory.

Solution: (b) Correct sequence is i —» iii -» ii —»iv -> v
Each transformation denotes a meaningful process as follows
i —»adsorption of A and B on surface
iii —»ii interaction between A and B to form intermediate
ii —»iv starting desorption of A-B
iv —» v complete desorption

Question.22. Which of the following process is responsible for the formation of delta at a place where rivers meet the sea?
(a) Emulsification (b) Colloid formation
(c) Coagulation (d) Peptisatioti
Solution: (c) A delta is formed at a place where rivers meet the sea due to the process of setting down of colloidal particles. The ions which are present in sea water are responsible for coagulation.

Question.23. Which of the following curves is in accordance with Freundlich adsorption isotherm?

Solution:

Question.24. Which of the following process is not responsible for the presence of electric charge on the sol particles?
(a) Electron capture by sol particles
(b) Adsorption of ionic species from solution
(c) Formation of Helmholtz electrical double layer
(d) Absorption of ionic species from solution
Solution: (d) Absorption of ionic species from solution is not responsible for the presence of electric charge on the sol particles. Charge on the sol particles is due to
(i) electrons capture by sol particles during electro dispersion of metal.
(ii) preferential adsorption of ionic species from solution.
(iii) formation of Helmholtz electrical double layer.

Question.25. Which of the following phenomenon is applicable to the process shown in the figure?

Solution: (b) Colour is adsorbed on animal charcoal, hence sugar solution which comes out from the column is colourless.
More than One Correct Answer Type

Question.26. Which of the following options are correct?
(a) Micelle formation by soap in aqueous solution is possible at all temperatures.
(b) Micelle formation by soap in aqueous solution occurs above a particular concentration.
(c) On dilution of soap solution micelles may revert to individual ions.
(d) Soap solution behaves as a normal strong electrolyte at all concentrations.
Solution: (b, c) Micelle formation is possible only above critical micelle concentration
and on dilution the micelles may form electrolytes.

Question.27. Which of the following statements are correct about solid catalyst?
(a) Same reactants may give different product by using different catalysts.
(b) Catalyst does not change AH of reaction.
(c) Catalyst is required in large quantities to catalyse reactions.
(d) Catalytic activity of a solid catalyst does not depend upon the strength of chemisorption.
Solution: (a, b)
(a) Same reactants may give different products by using different catalysts as different catalysis have different specific functions to mould the reaction towards specific product, e.g., starting with H2 and CO, and using different catalysts, we get different products.

(b) Catalyst does not change AH of reaction as AH of reaction is difference between enthalpy of reactants and products. So, it does not change during catalysed reaction.

Question.28. Freundlich adsorption isotherm is given by the expression  x/m=kpVn. Which of the following conclusions can be drawn from this expression?

Solution:

Question.29. H2 gas is adsorbed on activated charcoal to a very little extent in comparison
to easily liquefiable gases due to
(a) very strong van der Waals interaction
(b) very weak van der Waals forces
(c) very low critical temperature
(d) very high critical temperature.
Solution: (b, c) Due to weak van der Waals forces and very low critical temperature, H2 is adsorbed to a limited extent.

Question.30. Which of the following statements are correct?
(a) Mixing two oppositely charged sols neutralizes their charges and stabilizes the colloid.
(b) Presence of equal and similar charges on colloidal particles provides stability to the colloids.
(c) Any amount of dispersed liquid can be added to emulsion without destabilizing it.
(d) Brownian movement stabilizes sols.
Solution: (b, d) Presence of equal and similar charges on colloidal particles provides stability to colloids as repulsive forces between charge particles having same charge prevent them from colliding when they come closer to each other.

Question.31. An emulsion cannot be broken by and
(a) heating
(b) adding more amount of dispersion medium
(c) freezing
(d) adding emulsifying agent
Solution: (b, d) An emulsion cannot be broken by adding more amount of dispersion medium and emulsifying agent. It can be broken by heating, freezing, etc.

Question.32. Which of the following substances will precipitate the negatively charged emulsions?
(a) KC1 (b) Glucose
(c) Urea (d) NaCl
Solution: (a, d) KC1 and NaCl contain K+ and Na+ which will precipitate negatively charged emulsions.

Question.33. Which of the following colloids cannot be coagulated easily?
(a) Lyophobic colloids (b) Irreversible colloids
(c) Reversible colloids (d) Lyophilic colloids
Solution: (c, d) Lyophilic colloids (liquid loving colloids) which are also known as reversible colloid cannot be coagulated easily. The stability of these colloids are due to
(i) Charge on colloidal particles and
(ii) Solvation of colloidal particles.

Question.34. What happens when a lyophilic sol is added to a lyophobic sol?
(a) Lyophobic sol is protected.
(b) Lyophilic sol is protected.
(c) Film of lyophilic sol is formed over lyophobic sol.
(d) Film of lyophobic sol is formed over lyophilic sol.
Solution: (a, c) Lyophobic sols are protected due to formation of film of lyophobic sol.

Question.35. Which phenomenon occurs when an electric field is applied to a-colloidal solution and electrophoresis is prevented?
(a) Reverse osmosis takes place.
(b) Electroosmosis takes place.
(c) Dispersion medium begins to move.
(d) Dispersion medium becomes stationary.
Solution: (b, c) The dispersion medium begins to move in an electric field when electrophoresis is prevented. This is known as electroosmosis.

Question.36. In a reaction, catalyst changes ‘
(a) physically (b) qualitatively
(c) chemically (d) quantitatively
Solution: (a, b) In a reaction catalyst changes physically and qualitatively as it is unaltered during the reaction and remain quantitatively intact after completion of reaction, and does not change chemically.

Question.37. Which of the following phenomenon occurs when a chalk stick is dipped in ink?
(a) Adsorption of coloured substance
(b) Adsorption of solvent
(c) Absorption and adsorption both of solvent
(d) Absorption of solvent.
Solution: (a, d) Both adsorption and absorption take place. Coloured substance is adsorbed while solvent is absorbed.

Short Answer Type Questions

Question.38. Why is it important to have clean surface in surface studies?
Solution: Clean surface of adsorbent is suitable for adsorption because if surface is covered with gases of air, it will not be available for adsorption.

Question.39. Why is chemisorption referred to as activated adsorption?
Solution: Chemisorption involves formation of bonds for which activation energy is required. That is why chemisorption is referred to as activated adsorption.

Question.40. What type of solutions are formed on dissolving different concentrations of soap in water?
Solution: At lower concentration, soap forms a normal electrolytic solution with water. After a certain concentration called critical micelle concentration, colloidal solution is obtained.

Question.41. What happens when gelatin is mixed with gold sol?
Solution: Gelatin stabilises the colloidal solution of gold. Gold sol is lyophobic sol or solvent repelling sol.

Question.42. How does it become possible to cause artificial rain by spraying silver iodide on the clouds?
Solution: Clouds are colloidal in nature and carry a charge. On spraying silver iodide which is an electrolyte, the charge on the colloidal particles is neutralised. Clouds coagulate to form rain.

Question.43. Gelatin which is a peptide added in ice-creams. What can be its role?
Solution: Ice-cream is an emulsion. Gelatin is emulsifying agent which is used to stabilize the emulsion so it is added to ice-creams.

Question.44. What is collodion?
Solution: 4% solution of cellulose nitrate in the mixture of alcohol and ether is called collodion. It is used to filter, colloidal solution.

Question.45. Why do we add alum to purify water?
Solution: River water carries negatively charged colloidal particles of clay and sand; alum (K2S04.A12(S04)3. 24H20) is added to river water to coagulate the colloidal particles of clay and sand. Thus, alum helps to purify water.

Question.46. What happens when electric field is applied to colloidal solution?
Solution: When electric field is applied to colloidal solution then charged colloidal particles move towards oppositely charged electrodes. This phenomenon is called electrophoresis.

Question.47. What causes Brownian motion in colloidal dispersion?
Solution: Brownian movement is the continuous zigzag movement of colloidal particles in a colloidal sol. Brownian motion takes place due to collision between the molecules of dispersion medium and colloidal particles. This movement is responsible for stabilisation of colloidal solution.

Question.48. A colloid is formed by adding FeCl3 in excess of hot water. What will happen if excess sodium chloride is added to this colloid?
Solution: A positively charged colloidal solution of hydrated ferric oxide is formed. On adding sodium chloride, negatively charged chloride ions neutralise the positive charge of the colloidal solution. Coagulation of the sol takes place.

Question.49. How do emulsifying agents stabilize the emulsion?
Solution:The emulsifying agent forms an interfacial layer between suspended particles and the dispersion medium thereby stabilising the emulsion.

Question.50. Why are some medicines more effective in the colloidal form?
Solution: Some medicines are effective in colloidal state due to large surface area and easy assimilation.

Question.51. Why does leather get hardened after tanning?
Solution: Animal skin is colloidal, in nature and carries positive charge. Tannin is a negatively charged colloidal solution. When leather is soaked in tannin, mutual coagulation of the two takes place and leather gets hardened.

Question.52. How does the precipitation of colloidal smoke take place in Cottrell precipitator?
Solution: Charged smoke particles are passed through a chamber containing plates with charge opposite to that of smoke particles, in the Cottrell precipitator. On contact with the plates, the particles lose their charge and get precipitated.

Question.53. How will you distinguish between dispersed phase and dispersion medium in an emulsion?
Solution: Emulsions can be diluted to any extent on adding dispersion medium. The dispersed phase forms a separate layer if added in excess.

Question.54. On the basis of Hardy-Schulze rule explain why the coagulating power of phosphate is higher than chloride?
Solution: Minimum quantity of an electrolyte required to cause precipitation of a sol is called its coagulating value. The greater the charge and smaller the amount of the electrolyte required for precipitation, the higher is the coagulating power of the electrolyte.

Question.55. Why does bleeding stop by rubbing moist alum?
Solution: Blood is a colloidal solution when we rub moist alum on that part it causes coagulation which stops bleeding.

Question.56. Why is Fe(OH)3 colloid positively charged, when prepared by adding FeCl3 to hot water?
Solution: The colloidal solution of hydrated ferric oxide adsorbs positively charged Fe3+ ion and therefore the colloidal solution becomes positively charged.

Question.57. Why do physisorption and chemisorption behave differently with rise in temperature?
Solution: Physisorption involves weak van der Waals forces which become weaker with rise of temperature. The chemisorption involves formation of chemical bonds involving activation energy. With increase of temperature, we reach the activated complex faster. Therefore, chemisorption process becomes faster.

Question.58. What happens when dialysis is prolonged?
Solution: On prolonged dialysis, traces of electrolytes which stabilise the colloids, are completely removed. It makes the colloid unstable causing coagulation.

Question.59. Why does the white precipitate of silver halide become coloured in the presence of dye eosin.
Solution: White precipitate of silver chloride becomes coloured due to adsorption of pigments of eosin dye.

Question.60. What is the role of activated charcoal in gas mask used in coal mines?
Solution: Role of activated charcoal can be explained on the basis of adsorption. In gas mask, charcoal adsorbs the pollutant gases like S02, N02, CH4, C02, etc. present in atmosphere due to heavy traffic.

Question.61. How does a delta form at the meeting place of sea and river water?
Solution: River water is a colloidal solution of clay and sea water contains a lot of electrolytes. Coagulation takes place at the meeting place of sea and river water. The coagulated clay forms delta.

Question.62. Give an example where physisorption changes to chemisorption with rise in temperature. Explain the reason for change.
Solution: Physisorption of hydrogen on finely divided nickel involves weak van der Waals forces. With rise of temperature, hydrogen molecules dissociate into hydrogen atoms which held on to the surface by chemical bonds i.e., chemisorption takes place.

Question.63. Why is desorption important for a substance to act as a good catalyst?
Solution: After the reaction is over between the adsorbed reactants, the process of desorption must take place to remove the product molecules and create space for other reactant molecules to adsorb on the catalyst surface.

Question.64. What is the role of diffusion in heterogeneous catalysis?
Solution: The gas molecules diffuse onto the surface of the catalyst and get adsorbed. After the chemical change, the products formed diffuse away from the surface of the catalyst setting the surface free for other reactant molecules to adsorb on the surface and give the product.

Question.65. How does a solid catalyst enhance the rate of combination of gaseous molecules?
Solution: When gaseous molecules come in contact with the surface of a .solid catalyst, a weak chemical combination between the surface of the catalyst and gaseous molecules takes place. This increases the concentration of the reactants on the surface. According to the law of mass action, rate of a reaction is proportional to the concentration of the reactants. With increased concentration of the reactants, the reaction takes place faster. Also adsorption process is exothermic it releases energy which helps in further increasing the rate of reaction. .

Question.66. Do the vital functions of the body such as digestion get affected during fever? Explain your answer.
Solution: The rate of an enzyme reaction is maximum at a particular temperature range, called optimum temperature. On either side of the optimum temperature, the enzyme activity decreases. The optimum temperature range for enzymatic activity is 298-310K. Normal human body temperature being 310K is suited for enzyme-catalysed reactions. If a person is suffering from fever, the temperature will be over 310 K. This will adversely affect the enzymatic reactions.

Matching Column Type Questions

Question.67. Method of formation of solution is given in Column I. Match it with the type of solution given in Column II.

Solution: (i -»b), (ii —> c), (iii —» d), (iv —> a)
(i) When sulphur vapour passed through cold water, it leads to formation of molecular colloids.
(ii) When soap is mixed with water above critical micelle concentration, it leads to formation of associated colloids.
(iii) White part of egg whipped with water is an example of macro-molecular colloids in which high molecular mass proteneous molecule acts as a colloidal particle.
(iv) Soap mixed with water below critical micelle concentration is known as normal electrolyte solution.

Question.68. Match the statement given in Column I with the phenomenon given in Column II.

Solution: (i —> c), (ii —» d), (iii-> b), (iv —» a)
(i) Dispersion medium moves in an electric field is known as electroosmosis.
(ii) Solvent molecules pass through semipermeable membrane towards
solvent side is known as reverse osmosis.
(iii) Movement of charged colloidal particles under the influence of applied electric potential towards oppositely charge electrodes is known as electrophoresis.
(iv) Solvent molecules pass through semipermeable membranes towards solution side is known as osmosis.

Question.69. Match the items given in Column I and Column II.

Solution: (i -> b), (ii -» c), (iii -» d), (iv-> a)
(i) Lyophobic colloid (solvent hating colloid) are readily protected by small amount of electrolyte. These colloids are also stabilized by addition of lyophilic colloids which makes a protective layer around lyophobic sol. Hence, lyophilic sol are known as protective colloid.
(ii) Liquid-liquid colloid is also known as emulsion if they are partially miscible or immiscible liquids.
(iii) When FeCl3 is added to hot water, it leads to the formation of positively charged colloid.
(iv) When NaOH is added to FeCl3, it leads to the formation of negatively charged colloid.

Question.70. Match the types of colloidal systems given in Column I with the name given in Column II.

Solution: (i -> b), (ii —»c), (iii —» d), (iv —> a)
Colloids are classified on the basis of types of dispersed phase and dispersion medium.

Question.71. Match the items of Column I and Column II.

Solution: (i —> d), (ii -» c), (iii —> a), (iv —> b)
(i) Purification of colloid can be done by dialysis in which ions/particles are removed from solution through semipermeabie membrane.

(ii) Peptisation is a process in which when small quantity of electrolyte (peptizing agent) is added to precipitate. It leads to formation of colloidal solution.
(iii) The process of removing of oily or greasy dirt from the cloth is done by emulsification.
(iv) Process of setting of colloidal particle is called coagulation. Electrophoresis is a process in which on applying electric potential to the electrodes dipped in sol, the oppositely charged particles of colloidal solution move towards oppositely charged electrodes, get discharged and precipitated.

Question.72. Match the items of Column I ahd Column II.

Solution: (i —> d), (ii —> c), (iii —> a), (iv —> b)
(i) Butter is an example of dispersion of liquid in solid.
(ii) Pumice stone is an example of dispersion of gas in solid in which gas bubbles are pierced within solid particles.
(iii) Milk is a dispersion of liquid in which fats and protein are dissolved in milk.
(iv) Paint is an example of solid in liquid.

Assertion and Reason Type Questions

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
(a) Both Assertion and Reason are correct and the Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason arc correct but Reason is not the correct explanation for Assertion.
(c) Assertion is correct but Reason is incorrect.
(d) Both Assertion and Reason are incorrect.
(e) Assertion is incorrect but Reason is correct.

Question.73. Assertion (A): An ordinary filter paper impregnated with collodion solution stops the flow of colloidal particles.
Reason (R): Pore size of the filter paper becomes more than size of colloidal particle.
Solution: (c) Pore size of the filter paper becomes less than the size of colloidal particles hence colloidal particles do not flow through it.

Question.74. Assertion (A): Colloidal solutions show colligative properties.
Reason (R): Colloidal particles are large in size.
Solution: (b) Colloidal particles are large in size and hence the number of particle is lesser than the true solution. The lesser number of particles causes lower colligative properties.

Question.75. Assertion (A): Colloidal solutions do not show Brownian motion.
Reason (R): Brownian motion is responsible for stability of sols.
Solution: (e) Colloidal particles show Brownian movement and it is responsible for the stability of colloidal solution.

Question.76. Assertion (A): Coagulation power of Al3+ is more than NaT
Reason (R): Greater the Valency of the flocculating ion added, greater is its power to cause precipitation (Hardy-Schulze rule).
Solution: (a) According to Hardy-Schulze law, the greater is the valency of the coagulating ion, the more is the power to coagulate the colloidal solution. Thus, coagulation power of Al3+ is greater than that of Na+.

Question.77. Assertion (A): Detergents with low CMC are more economical to use. Reason (R): Cleansing action of detergents involves the formation of micelles. These are formed when the concentration of detergents becomes equal to CMC.
Solution: (a) Cleansing of clothes takes place by micellisation, it starts at CMC. The lesser is the CMC, the better and more economical is the detergent.

Long Answer Type Questions

Question.78. What is the role of adsorption in heterogeneous catalysis?
Solution: The role of heterogeneous catalysts can be explained in terms of adsorption of reactants on the surface of the catalyst. The adsorption helps the reaction in the following ways:
(i) Adsorption increases the concentration of reactants on the surface of the catalyst. Due to increased concentration of the reactants, the reactions proceed rapidly.
(ii) Adsorbed molecules get dissociated to form active species like free radicals which react faster than molecules.
(iii) The adsorbed molecules are not free to move about and, therefore, they collide with other molecules on the surface.
(iv) The heat of adsorption evolved acts as energy of activation for the reaction (chemisorption).
Example:
Adsorption of molecules of the reactant at the active site .
Occurrence of chemical reaction on the surface of catalyst
Desorption of product molecules

Question.79. What are the applications of adsorption in chemical analysis?
Solution: Adsorption has the following applications in chemical analysis:
(i) In qualitative analysis: Certain qualitative tests such as the lake test for the confirmation of Al3+ ions are based upon adsorption, i.e., Al(OH)3 has the capacity to adsorb the colour of blue litmus from the solution.
(ii) In adsorption indicators: Many adsorption indicators are being used in volumetric analysis e.g., dyes such as eosin and fluorescein are used as adsorption indicators. In these methods, the dyes are adsorbed on the surfaces of certain precipitates (such as silver halides) which give characteristic colour at the end point.
(iii) In ion-exchange resins: The organic polymers containing groups like -COOH, -S03H and -NH2, etc. possess the property of selective adsorption of ions from solutions. These are quite useful in the softening of water.
(iv) Separation of inert gases: Due to the difference in degree of adsorption of gases by charcoal, a mixture of noble gases can be separated by adsorption on coconut charcoal at different temperatures.
(v) In froth floatation process: A low grade sulphide ore is concentrated by separating it from silica and other earthly matter by adsorption using pine oil and frothing agent.
(vi) Chromatographic analysis: The chromatographic technique for purification used in analytical and industrial fields is based on the phenomenon of adsorption.

Question.80. What is the role of adsorption in froth floatation process used especially for concentration of sulphide ores?
Solution: Sulphide ores are concentrated by froth floatation process. The finely divided ore is added to pine oil and water. Compressed air is bubbled through the mixture. The foam formed rises to the surface on which ore particles wetted with oil are adsorbed while earthy matter settles down at the bottom.

Question.81. What do you understand by shape selective catalysis? Why are zeolites good shape selective catalysts?
Solution: The catalytic reaction which depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape selective catalysis. Zeolites are good shape selective catalysts because of their honey comb like structures.
(i) Zeolites are microporous aluminosilicates of the general formula Mx/n [(A102)x(Si02)y]mH20.
(ii) Zeolites have an enormous surface area which is largely on the inside of the solid. The zeolites can permit the entry and exit of molecules of a certain size into the active regions within the holes.
(iii) They are used in petrochemical industries for cracking of hydrocarbons and isomerization.
(iv) The reactions in zeolites depend upon the size of the cavities (cages) and pores (tunnels) present in them. The pore size in zeolites generally varies between 260 pm and 740 pm.
(v) Zeolite catalyst known as ZSM-5 converts alcohols to gasoline. The alcohol is dehydrated in the cavities and the hydrocarbons are formed.

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NCERT Exemplar Problems Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

Multiple Choice Questions

Single Correct Answer Type
Question 1.In the extraction of chlorine by electrolysis of brine,

Solution: (a) Extraction of chlorine from brine is based on the oxidation reaction of

Question 2.When copper ore is mixed with silica in a reverberatory furnace, copper matte is produced. The copper matte contains
(a) sulphides of copper (II) and iron (II)
(b) sulphides of copper (II) and iron (III)
(c) sulphides of copper (I) and iron (II)
(d) sulphides of copper (I) and iron (III)
Solution: (c) When copper ore is mixed with silica, iron oxide slags off as iron silicate and copper is produced in the form of copper matte which contains Cu₂S (I) and FeS (II).

Question 3.Which of the following reaction is an example of autoreduction?


Solution:

This reaction includes reduction of copper (I) oxide by copper (I) sulphide. In this process, copper is reduced by itself hence this process is known as autoreduction and the solidified copper so, obtained is known as blister copper.

Question 4. A number of elements are available in the earth’s crust but most abundant
elements are
(a) Al and Fe (b) Al and Cu (c) Fe and Cu (d) Cu and Ag
Solution: (a) Al and Fe are most abundant elements in the earth’s crust.

Question 5. Zone refining is based on the principle that .
(a) impurities of low boiling metals can be separated by distillation
(b) impurities are more soluble in molten metal than in solid metal
(c) different components of a mixture are differently adsorbed on an adsorbent
(d) vapour of volatile compound can be decomposed in pure metal
Solution: (b) Zone refining method is based on the principle that impurities are more soluble in molten metal than in the solid state of the metal.

Question 6. In the extraction of copper from its sulphide ore, the metal is formed by the reduction of Cu₂O with
(a) FeS (b) CO (c) Cu₂S (d) SO₂
Solution: (c) In the extraction of copper from its sulphide ore, the metal is formed by the reduction of Cu2O with Cu₂S. This reaction completes with the process of autoreduction.
Chemical reaction occurring in this reaction is as follows

In this process, copper appears as blister copper.

Question 7. Brine is electrolysed by using inert electrodes. The reaction at anode is

Solution: (a) For oxidation at anode, two possible reactions are oxidation of chlorine and of oxygen. Out of these two, oxidation of chlorine ion is preferred because oxidation of oxygen requires overvoltage.
Chlorine is obtained by electrolysis giving out hydrogen and aqueous NaOH as byproducts.

Question 8. In the metallurgy of aluminium
(a) Al3+ is oxidized to Al(s)
(b) graphite anode is oxidized to carbon monoxide and carbon dioxide
(c) oxidation state of oxygen changes in the reaction at anode
(d) oxidation state of oxygen changes in the overall reaction involved in the process.
Solution: (b) In metallurgy of aluminium, graphite anode is oxidized to carbon monoxide and carbon dioxide hence it is to be replaced after it is used up.

Question 9. Electrolytic refining is used to purify which of the following metals?
(a) Cu and Zn (b) Ge and Si (c) Zr and Ti (d) Zn and Hg
Solution: (a) Copper and zinc are two metals which are generally purified by using electrolyte refining. In this process, impure metal is used as anode and pure metal is used as a cathode. Impurities from the blister copper or impure zinc deposit as anode mud. .

Question 10. Extraction of gold and silver involves leaching the metal with CN⁺ ion. The metal is recovered by
(a) displacement of metal by some other metal from the complex ion
(b) roasting of metal complex
(c) calcinations followed by roasting
(d) thermal decomposition of metal complex
Solution: (a) The metal in cyanide process is recovered by displacement of metal by some more reactive metal from the complex.

Question 11. Choose the correct option of temperature at which carbon reduces FeO to iron and produces CO.
(a) Below temperature at point A
(b) Approximately at the temperature corresponding to point A
(c) Above temperature at point A but below temperature at point D
(d) Above temperature at point A
Solution:

Question 12. Below point ‘A’ FeO can
(a) be reduced by carbon monoxide only
(b) be reduced by both carbon monoxide and carbon
(c) be reduced by carbon only
(d) Cannot be reduced by both carbon and carbon monoxide
Solution: (a) Below point A Gibbs free energy change for the formation of CO2 from CO  has lower value (more negative value) than Gibbs free energy change for the formation of FeO . Hence, FeO will be reduced by CO only below point A.

Question 13. For the reduction of FeO at the temperature corresponding to point D, which of the following statement is correct?

Solution: (a) At point D both the curves of oxidation of iron and oxidation of CO meet each other.

Question 14. At the temperature corresponding to which of the points in figure of Q. 11-13, FeO will be reduced to Fe by coupling the reaction 2FeO —> 2Fe + 02 with all of the following reactions?

Solution: (b, d) The reduction reaction ranges depend upon the points of corresponding intersections in AG° versus T plots.
At point B and E, FeO will be reduced to Fe because at these points AG° value of all three oxidation reactions is lesser than that of reduction reaction.

Question 15. Which of the following options are correct?
(a) Cast iron is obtained by remelting pig iron with scrap iron and coke using hot air blast.
(b) In extraction of silver, silver is extracted as cationic complex.
(c) Nickel is purified by zone refining.
(d) Zr and Ti are purified by van Arkel method.
Solution: (a, d) Iron obtained from blast is called pig iron. It contains about 4% carbon and many impurities such as Mn, P, Si, etc. The pig iron is remelted with scrap iron. This is known as cast iron. It has slightly less carbon content (3%) and is extremely hard and brittle.

Question 16. In the extraction of aluminium by Hall-Heroult process, purified A1203 is mixed with CaF2 to
(a) lower the melting point of A1203.
(b) increase the conductivity of molten mixture.
(c) reduce Al⁺³ into Al(s).
(d) acts as catalyst.
Solution: (a, b) CaF2 mixed with purified Al₂O₃   helps in lowering the melting point of Al₂O₃ and increases the conductivity of the molten mixture.

Question 17. Which of the following statements are correct about the role of substances added in the froth floatation process?
(a) Collectors enhance the non-wettability of the mineral particles.
(b) Collectors enhance the wettability of gangue particles.
(c) By using depressants in the process two sulphide ores can be separated.
(d) Froth stabilizers decrease wettability of gangue.
Solution: (a, c) Collectors like sodium ethyl xanthate are added to enhance the non-wettability of sulphide ore particles.
In the concentration of galena ore, sodium cyanide is used as a depressant for zinc sulphide. NaCN forms a complex with NaCN and deposits on the surface of ZnS which prevents it for the formation of Froth while PbS forms a froth.
Thus, by using depressant, two sulphides ores can be separated.

Question 18. In the froth floatation process, zinc sulphide and lead sulphide can be separated by
(a) using collectors
(b) adjusting the proportion of oil to water
(c) using depressant
(d) using froth stabilisers
Solution: (b, c) Froth floatation method is used to extract metal from sulphide ore. ZnS and PbS can be separated by using depressant and adjusting the proportion of oil to water. Depressant used for this purposes is NaCN. It selectively prevents ZnS from coming to the froth.
Hence, (b) and (c) are correct choices.

Question 19. Common impurities present in bauxite are .
(a) CuO (b) ZnO (c) Fe₂O₃   (d) SiO2
Solution: (c, d) Fe₂O₃ and SiO2 are the common impurities present in bauxite.

Question 20. Which of the following ores are concentrated by froth floatation?
(a) Haematite (b) Galena
(c) Copper pyrites (d) Magnetite
Solution: (b, c) Sulphide ores like galena (PbS) and copper pyrites (CuFeS2) are concentrated by froth floatation.

Question 21. Which of the following reactions occur during calcination?

Solution: (a, c) Calcination involves heating of the ore below is melting point in the absence of air or in limited supply of air. Oxygen containing ores like oxide, hydroxides and carbonates are calcined. Thus, the following reactions occur during calcination.

Question 22. For the metallurgical process of which of the ores, calcined ore can be reduced by carbon?
(a) Haematite (b) Calamine (c) Iron pyrites (d) Sphalerite
Solution: (a, b) Oxides can be reduced by using carbon. Haematite (Fe₂O₃  ) and calamine (ZnCO₃ —> ZnO + CO2) can be reduced by carbon.

Question 23. The main reactions occurring in blast furnace during extraction of iron from
haematite are

Solution: (a, d) In blast furnace during extraction of iron, the following two are main reactions.

Question 24. In which of the following method of purification, metal is converted to its volatile compound which is decomposed to give pure metal?
(a) Heating with stream of carbon monoxide
(b) Heating with iodine
(c) Liquation
(d) Distillation
Solution: (a, b) Vapour phase refining method includes

Question 25. Which of the following statements are correct?
(a) A depressant prevents certain type of particle to come to the froth.
(b) Copper matte contains Cu₂S and ZnS.
(c) The solidified copper obtained from reverberatory furnace has blistered appearance due to evolution of SO₂ during the extraction.
(d) Zinc can be extracted by self-reduction.
Solution: (a, c) Depressants can prevent frothing of a particular sulphide. The solidified copper obtained in reverberatory furnace has blistered appearance hence it is called blister copper.

Question 26. In the extraction of chlorine from brine .

Solution:

Short Answer Type Questions

Question 27.Why is an external emf of the more than 2.2 V required for the extraction of Cl₂ from brine?
Solution: For the reaction of extraction of Cl₂ from brine solution:

Therefore, extraction of Cl₂ from brine solution will require an external emf more than 2.20 V for electrolysis process.

Question 28. At temperatures above 1073 K coke can be used to reduce FeO to Fe. How can you justify this reduction with Ellingham diagram?
Solution:

Question 29. Wrought iron is the purest form of iron. Write a reaction used for the preparation of wrought iron from cast iron. How can the impurities of sulphur, silicon and phosphorus be removed from cast iron?
Solution: The reaction is:

Limestone is added as flux and the impurities of sulphur, silicon and phosphorus change to their oxides and pass into slag.

Question 30. How is copper extracted from low grade copper ores?
Solution: Copper is extracted by hydrometallurgy from low grade ores. It is leached out using acid or bacteria. The solution containing copper ions (Cu⁺²) is treated with scrap iron, zinc or H₂ as:

In this way, copper, is obtained.

Question 31. Write two basic requirements for refining of a metal by Mond’s process and by van Arkel method.Why?
Solution: The basic requirements for refining a metal by Mond’s process and by van Arkel method are:
(i) The metal should form a volatile compound with an available reagent.
(ii) The volatile compound should be easily decomposable, so that metal can be easily recovered.

Question 32. Although carbon and hydrogen are better reducing agents but they are not used to reduce metallic oxides at high temperatures. Why?
Solution: This is because carbon and hydrogen react with metals to form carbides and hydrides respectively at high temperature.
Question 33.How do we separate two sulphide ores by froth floatation method? Explain with an example.
Solution: The separation of two sulphide ores can be achieved by adjusting proportion of oil to water or by using depressants. In the case of an ore containing ZnS and PbS, the depressant used is NaCN. It forms complex with ZnS and prevents it from coming with froth but PbS remains with froth.

Question 34. The purest form of iron is prepared by oxidizing impurities from cast iron in a reverberatory furnace. Which iron ore is used to line the furnace? Explain by giving reaction.
Solution: Haematite (Fe203) is used. It supplies the oxygen and oxidises carbon, silicon, manganese and phosphorus present in the cast iron to CO, Si02, MnO and P205 respectively.

Question 35. The mixture of compounds A and B is passed through a column of Al₂O₃ by using alcohol as eluent. Compound A is eluted in preference to compound B. Which of the compounds A or B is more readily adsorbed on the column?
Solution: Since compound A is eluted in preference to compound B. Compound B is more readily adsorbed on the column.

Question 36. Why is sulphide ore of copper heated in a furnace after mixing with silica? Iron oxide present as impurity in sulphide ore of copper forms a slag of iron silicate and copper is produced in the form of copper matte.
Solution: FeO + SiO₂> FeSiO₃ .

Question 37.Why are sulphide ores converted to oxide before reduction?
Solution: Sulphides are not easily reduced while oxides are easily reduced. Due to this, sulphides are first oxidised and then subjected to reduction.

Question 38.Which method is used for refining Zr and Ti? Explain with equation?
Solution: van Arkel method is used for refining Zr and Ti. In this method crude metal is heated with iodine to form volatile unstable compound. The compound is then decomposed to get pure metal.

Question 39. What should be the considerations during the extraction of metals by electrochemical method?
Solution: The following factors should be kept in mind to ensure proper precautions:
(i) Reactivity of the metal.
(ii) Suitability of the electrode.

Question 40. What is the role of flux in metallurgical processes?
Solution: The ores, even after concentration, contain some gangue. To remove gangue, certain substances are mixed with concentrated ore which combine with the gangue to form a fusible material which is not soluble in molten metal. The substances used are called fluxes.
Fluxes are classified as (i) acidic flux and (ii) basic flux. Acidic flux removes basic impurities while basic flux removes acidic impurities.

Question 41. How are metals used as semiconductors refined? What is the principle of the method used?
Solution: Semiconductor is produced by zone refining method which is based on the principle that the impurities are more soluble in melt than in the solid state of metals.

Question 42. Write down the reactions taking place in blast furnace related to the metallurgy of iron at the temperature range 500-800 K.
Solution: The following reactions take place in blast furnace during the metallurgy of iron at the temperature range 500-800 K.

Question 43. Give two requirements for vapour phase refining.
Solution: (i) The metal should form a volatile compound with available reagent.
(ii) The volatile compound should be unstable and easily decomposable so that the recovery is easy.

Question 44. Write the chemical reactions involved in the extraction of gold by cyanide process. Also give the role of zinc in the extraction.
Solution: Gold particles present in the ore are treated with a dilute solution of NaCN and air is circulated. Gold particles dissolve by forming soluble complex.

Au is recovered from solution by the addition of electropositive metal like zinc.

Extraction of gold by cyanide process involves both oxidation and reduction. Formation of cyanide complex involves oxidation of gold while its recovery involves reduction. Zn acts as a reducing agent.
Matching Column Type Questions

Question 45. Match the items of Column I with items of Column II and assign the correct code:

Solution:
(b) (A ->2), (B —>4), (C —> 5), (D —> 3)
(a) Pendulum is made up of nickel steel.
(b) Molecular formula of malachite is Cu CO₃  .Cu (OH)2

(c) Molecular formula of calamine is ZnCO₃.
(d) Molecular formula of cryolite is Na₃AlF6.

Question 46. Match the items of Column I with the items of Column II and assign the correct code:

Solution:
(b) (A -> 4), (B —> 3), (C -> 1), (D -> 2)
(a) Coloured bands are observed in chromatography.
(b) Impure metal is converted to volatile complex by using Mond’s process.
(c) Ge and Si are purified by zone refining method.
(d) Purification of mercury is done by fractional distillation.

Question 47. Match the items of Column I with the items of Column II and assign the correct code:

Solution: (a) (A –> 4), (B –>2), (C–> 3), (D —> 1)
(a) Cyanide process is used for extraction of Au through formation of anionic complex [Au(N)₂] ^–.
(b) Froth floatation process is used for dressing of ZnS.
(c) Electrolytic reduction method is used for extraction of aluminium. Graphite electrode is used for this purpose.
(d) Zone refining is used for purification of Ge.

Question 48. Match the items of Column I with the items of Column II and assign the correct code:

Solution: (a) (A -> 3), (B -»4), (C -> 2), (D ->1)
(a) Sapphire is a gemstone which contains Co.
(b) Molecular formula of sphalerite is ZnS.
(c) NaCN is used as a depressant in froth floatation method.
(d) Molecular formula of corundum is Al₂O₃.

Question 49. Match the items of Column I with items of Column II and assign the correct code:

Solution: (a) (A -> 2), (B ->3), (C -> 4), (D —> 1)
(a) Blistered Cu can be prepared by means of the following chemical reaction:

Assertion and Reason Type Questions

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices.
(a) Both Assertion and Reason are true and the Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true and the Reason is not the correct explanation of Assertion.
(c) Assertion is true but the Reason is false.
(d) Assertion is false but Reason is true.
(e) Assertion and Reason both are wrong.

Question 50. Assertion (A): Nickel can be purified by Mond’s process.
Reason (R): Ni(CO)₄ is a volatile compound which decomposes at 460 K to give pure Ni.
Solution: (a) Nickel treated with carbon monoxide forms nickel tetracarbonyl Ni(CO)₄  while impurities are left behind. When the vapour of Ni(CO)₄ is heated at 460 K, it is decomposed to give pure nickel while carbon monoxide is removed as gas.

Question 51. Assertion (A): Zirconium can be purified by van Arkel method.
Reason (R): ZrI₄ is volatile and decomposes at 1800 K.
Solution: (a) Zirconium is also purified by vapour phase refining in which it is treated with iodine to form ZrI₄ which on heating decomposes to give pure zirconium.

Question 52. Assertion (A): Sulphide ores are concentrated by froth floatation method. Reason (R): Cresols stabilize the froth in froth floatation method.
Solution: (b) Sulphide ores are wetted by oil and becomes lighter and rises to the surface along with froths while impurities wetted by water become heavier and settle down.

Question 53. Assertion (A): Zone refining method is very useful for producing semiconductors.
Reason (R): Semiconductors are of high purity.
Solution: (b) In zone refining method, the impurities of semiconductors are more soluble in molten zone and the ultrapure semiconductor crystallizes.

Question 54. Assertion (A): Hydrometallurgy involves dissolving the ore in a suitable reagent followed by precipitation by a more electropositive metal.
Reason (R): Copper is extracted by hydrometallurgy.
Solution: (b) Copper is extracted by hydrometallurgical process. In this process, salts of metal are dissolved in suitable solvent like water and then reduced by more electropositive element.

Long Answer Type Questions

Question 55. Explain the following:
(a)CO₂ is a better reducing agent below 710 K whereas CO is a better reducing agent above 710 K.
(b) Generally sulphide ores are converted into oxides before reduction.
(c) Silica is added to the sulphide ore of copper in the reverberatory furnac
(d) Carbon and hydrogen are not used as reducing agents at high temperatures.
(e) Vapour phase refining method is used for the purification of Ti.
Solution: (a) According to Ellingham diagram, at temperature below 710 K,

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