To Study the Effect of Change in Temperature on the Rate of Reaction Between Sodium Thiosulphate and Hydrochloric Acid

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

THEORY
The rate of a chemical reaction depends to a great extent upon temperature. The rate of reaction increases with increase in temperature. Increase in temperature increases kinetic energy of the molecules. Therefore, the fraction of molecules having energy greater than its threshold energy increases which results in the increase in number of effective collisions per second. It has been observed that in most of the cases for every 10°C rise in temperature, the rate of the reaction becomes almost double. The rate of reaction between sodium thiosulphate and hydrochloric acid also increases with increase in temperature.

APPARATUS
Conical flask (250 ml), measuring cylinders (50 ml and 5 ml), stop-watch, thermometer, tripod stand, wire-gauze and burner.

MATERIALS REQUIRED
0.1 M Na₂S₂0₃ solution, 1 M HCl, distilled water and cone. HN0₃.

PROCEDURE

1. Take 50 ml of 0.1 M Na₂S₂0₃ solution in a 100 ml conical flask and note its temperature with the help of a thermometer.
2. Add 10 ml of 1 M HCl to it and start the stop-watch immediately when half of the hydrochloric acid solution has been added.
3. Shake the contents of the flask gently and place it on the tile with a cross-mark as shown in Fig.
4. Observe the cross-mark from the top and note the time taken for the mark to become just invisible.
5. Empty the flask and clean it thoroughly with cone. HN0₃ and then with water.
6. Take again 50 ml of 0.1 M Na₂S₂0₃ in conical flask and heat it so that the temperature of the solution becomes (T + 10°)C.
7. Remove the flask from the tripod-stand and add 10 ml of 1 M HCl to it and start the stop-watch.
8. Shake the contents gently and place it on the tile having a cross-mark.
9. Note the time taken for the mark to become just invisible.
10. Repeat the experiment at (T + 20)°C, (T + 30)°C and (T + 40)°C temperatures and record the observations as given below.

OBSERVATIONS
Volume of 0.1 M Na₂S₂0₃ solution taken each time = 50 ml
Volume of 1 M HCl added each time = 10 ml.

PLOTTING OF GRAPH
Plot a graph by taking 1/t along the ordinate (vertical axis) and temperature along the abscissa (horizontal axis).

RESULT
Rate of reaction between sodium thiosulphate and hydrochloric acid increases with the increase in temperature.

PRECAUTIONS

1. The apparatus must be thoroughly clean. If the same conical flask is to be used again and again, it should be thoroughly washed with cone. HN0₃ and then with water.
2. Measure the volumes of sodium thiosulphate solution, hydrochloric acid and distilled water very accurately.
3. Use the same tile with the same cross-mark for all observation,
4. Complete the experiment at one time only so that there is not much temperature variation.
5. Start the stop-watch immediately when half of the hydrochloric acid solution has been added to sodium thiosulphate solution.
6. View the cross-mark through the reaction mixture from top to bottom from same height for all observations.

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Chemical Kinetics Viva Questions with Answers

Question.1. What is chemical kinetics ?
Answer. Chemical kinetics is that branch of chemistry which deals with the study of the rates of reactions and their mechanisms.

Question.2. What do you understand by the rate of reaction ?
Answer. The rate of a reaction is defined as the change in the molar concentration of any one of the reactant or the product per unit time.

Question.3. What are the units of the rate of reaction ?
Answer. Mol L^-1 s^-1 (Moles per litre per second).

Question.4. What are the factors on which the rate of reaction depends ?
Answer. The rate of reaction depends upon : (i) Nature of the reactants ; (ii) Concentration of the reactants ; (iii) Temperature ; (iv) Presence of catalyst; and (v) Presence of radiations.

Question.5. What is the law of mass action ?
Answer. Law of mass action states that the rate of a reaction is directly proportioned to the product of molar concentrations of the reactants.

Question.6. What is temperature coefficient of a reaction ?
Answer. Temperature coefficient of a reaction is the ratio of rate constants at two temperatures differing by 10°. Its value is generally equal to 2.

Question.7. What are the units of rate constant for zero order reactions ?
Answer. Same as rate of the reaction, i.e. moles/litre/sec.

Question.8. What are the units of rate constant for first order reactions ?
Answer. Sec^-1.

Question.9. What is the effect of temperature on rate constant of a reaction ?
Answer. It increases with increase in temperature.

Question.10. Why certain reactions are very fast ?
Answer. Because they have very low activation energy.

Question.11. What is threshold energy ?
Answer. It is the minimum energy which the colliding molecules must possess so as to have effective collision.

Question.12. Why reactions with molecularity more than three are rare ?
Answer. Because simultaneous collision between more than three particles is rare on the basis of probability considerations.

Question.13. “For an exothermic reaction activation energy for the forward reaction is less than that for the backward reaction.” Is this statement true or false ?
Answer. True.

Question.14. What is rate determining step ?
Answer. In complex reactions, the slowest step determines the over all rate of the reaction. This step is known as rate determining step.

Question.15. What is the effect of catalyst on the activation energy and heat of the reaction ?
Answer. A catalyst decreases the activation energy of the reaction. It has no effect on the heat of the reaction.

Question.16. On increasing the concentration of reactants the rate of the reaction does not change. What can you say about the order of the reaction ?
Answer. It is a zero order reaction.

Question.17. Can order of a reaction be fractional ?
Answer. Yes. For example, for the reaction
CH₃CHO ——> CH₄  + CO
the order is equal to 3/2.

Question.18. What is a complex reaction ?
Answer. A reaction involving more than one step is called a complex reaction.

Question.19. What do you understand by ‘4 volume’ H₂O₂ solution ?
Answer. It is a way of expressing the cone, of H202 solution. 1 litre of ‘4 volume’ H₂O₂ solution gives 4 litres of oxygen at N.T.P. on decomposition.

Question.20. Express the cone, of 1 M H₂O₂ solution in terms of volume strength.
Answer. 2H₂O₂ ——> 2H₂0 + O₂
2 mol                           22.4 L
1 litre of 1 M H₂O₂ contains 1 mole of H₂O₂ and hence would give 11.2 L on complete decomposition. Hence, 1 M H₂O₂ solution is ‘11.2 volume’.

Question.21. What is the equivalent mass of H₂O₂ ?
Answer. 17.

Question.22. What is the normality of 1 M H₂O₂ solution ?
Answer. 2 N.

Question.23. What is the oxidation number of oxygen in H₂O₂ ?
Answer. -1.

Question.24. What is the colour of starch-iodine complex ?
Answer. Blue.

Question.25. What is the effect of increase in cone, of iodide ions on the following reaction : 2H₃0+ + 2I^– +H₂O₂  —–> 4H₂0 + I₂
Answer. Rate of the reaction increases.

Question.26. The reaction under examination is as follows :
S₂O₃^2- (aq) + 2H⁺(aq) —-> H₂O(l) + SO₂(g) + S(g)
Write the conditions under which the rate law expression for this reaction can be written in the following manner.
Rate of precipitation of sulphur = k [S₂O₃^2-] [H⁺]².
Answer. None of the reactants should be used in excess and the reaction should be elementary.

Question.27. Suppose the above rate law expression for the precipitation of sulphur holds good, then on doubling the concentration of S₂O₃^2- ion and H+ ion, by how many times will the rate of the reaction increase ?
Answer. By eight times.

Question.28. How does the rate constant of a reaction vary with temperature ?
Answer. The rate constant of a reaction increases with increase in temperature.

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

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Focal Length of Spherical Lenses

SPHERICAL LENS
(а) Definition: A piece of a transparent medium bounded by atleast one spherical surface, is called a spherical lens.
(b) Types: There are two types of spherical lenses.

1. Convex or Converging Lenses: These are thick in the middle and thin at the edges.
2. Concave or Diverging Lenses: These are thin in the middle and thick at the edges.

(c) Different types of convex lenses: The three types of convex lenses are

1. double convex
2. plano-convex
3. concavo-convex

(d) Different types of concave lenses: The three types of concave lenses are

1. double concave
2. plano-concave
3. convexo-concave
   

TERMS ASSOCIATED WITH SPHERICAL LENSES

1. Aperture: The diameter of the circular edge of the lens, is called the aperture of the lens.
   In diagram  AB is the aperture of the lens.
2. Principal axis: The straight line passing through the two centres of curvature of the two spherical surfaces of the lens (or through one centre of curvature of one spherical surface and normal to the other plane surface), is called the principal axis of the lens.
   
3. Optical centre: It is a point on the principal axis of the lens, such that a ray of light passing through it goes undeviated.
   In diagram, O is the optical centre of the lens.
4. First principal focus: It is a point on the principal axis of the lens, such that the rays actually diverging from it (in case of a convex lens) or appearing to be going towards it (in case of a concave lens), after refraction from the lens, go parallel to the principal axis.
   In diagram,F₁ is the first principal focus of the lens. (For object at F₁ image at infinity).
5. Second principal focus: It is a point on the principal axis of the lens, such that the rays incident on the lens parallel to the principal axis after refraction from the lens, actually meet at this point (in case of a convex lens) or appear to come from it (in case of a concave lens).
   In diagram, F₂ is the second principal focus of the lens (For image at F₂ object at infinity).
6. Focal length: The distance between the optical centre of the lens and the principal focus (first or second) of the lens, is called focal length of the lens. It is represented by the symbol f. In diagram,OF₁ = OF₂ = f.
7. Principal section: A section of the lens cut by a plane passing through the principal focus and optical centre of the lens, is called principal section of the lens. It contains the principal axis.
   In diagram, the shaded portion is the principal section of the lens cut by the plane of the book page.

THREE SPECIAL RAYS
The special rays are :

1. Incident on the lens parallel to principal axis. After refraction from the lens, it actually passes through second principal focus F₂ (in case of a convex lens) or appears to come from the second principal focus F₂ (in case of a concave lens).
   
2. Incident on the lens through first principal focus F₁ (in case of a convex lens) or in direction of first principal focus F₁ (in case of a concave lens).
   After refraction from the lens, it goes parallel to the principal axis.
3. Incident on the lens in direction of optical centre. It passes undeviated through the lens.

SIGN CONVENTION
(a) Definition: It is a convention, which fixes the sign of different distances measured. The sign convention followed is the New Cartesian sign convention.
(b) Rules: It gives following rules:

1. All distances are measured from the optical centre of the lens (along the principal axis).
2. The distances measured in the same direction as the direction of incident light, are taken as positive.
3. The distances measured opposite to the direction of incident light, are taken as negative.
4. The distances measured above the principal axis are taken as positive but distances measured below the principal axis are taken as negative.
   

(c) Facts: According to above mentioned rules of sign convention,

1. Focal length for a convex lens is taken positive and the same for concave lens is taken negative.
2. The distance of an object is always negative.
3. The distance for real image is positive, while that for a virtual image is negative.
4. The size of object is positive and size of real image is negative while size of virtual image is positive.

LENS FORMULA
The equation relating the object distance (u), the image distance (u) and the lens focal length (f), is called lens formula. It is also called Gaussian formula.

ASSUMPTIONS MADE
Following assumptions are made in derivation of the lens formula.

1. The lens is thin.
2. The lens has a small aperture.
3. The point object lies on to the principal axis and placed perpendicular.
4. The incident rays make small angles with the lens surface or the principal axis.

POSITION, NATURE AND SIZE OF IMAGE WHEN OBJECT IS PUT IN DIFFERENT POSITION IN FRONT OF A CONVEX LENS
                                                                        It is described below in tabular form.

POWER OF A LENS
(a) Definition: It is the capacity or ability of a lens to deviate (converge or diverge) the path of rays passing through it. A lens producing more converging or more diverging is said to have more power and vice-versa.
It is represented by the symbol P.
(b) Relation with focal length: A lens of less focal length produces more converging or diverging rays and is said to have more power.

(c) Unit: Unit of power is dioptre (D). One dioptre is the power of a lens of focal length 1 metre.

(d) Sign: A converging lens has positive focal length and positive power.
A diverging lens has negative focal length and negative power.

LENS COMBINATION
(a) Definition: Two or more thin lenses, placed in contact together to have a common principal axis, form a lens combination.
(b) Focal length: If f₁, f₂,….., f_n be the focal length of individual lens and F be the focal
length of the combination.

Note: The lenses forming the combination must be thin to have their optical centres coinciding at one point to represent optical centre of the combination.
(c) Power: If P₁, P₂,….., P_n be the power of individual lenses and P be the power of the
combination.

(d) Magnification: If m₁, m₂,….., m_n are the magnification of individual lenses and m is the equivalent magnification of the combination then,

CHROMATIC ABERRATION OF A LENS
(a) Definition: The defect or drawback of a lens due to which it makes a coloured image of an object illuminated with white light, is called chromatic aberration. It is due to dispersion of white light by lens (just like a prism does).
(b) Remedy: It is removed by combining a convex and a concave lens of suitable focal length and material.
The combination of two lenses is called an achromatic combination (achromic doublet).

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Thermochemistry

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Enthalpy of Dissolution
It is well known that when a solute is dissolved in a solvent, heat is either absorbed or evolved. Thus, dissolution of a solute in a solvent is accompanied by enthalpy change (∆H) of the system. If heat is absorbed (i.e., the solution gets cooled), ∆H is given a positive sign. If heat is evolved (i.e., the solution gets warmed), ∆H is given negative sign.
The enthalpy change per mole of a solute dissolved varies with the concentration of the solution. Therefore, it is necessary to express the enthalpy change with reference to the concentration of the solution. The enthalpy of dissolution is defined as the enthalpy change per mole of a solute when it is dissolved in a pure solvent to give a solution of specified concentration. For example, when one mole of anhydrous calcium chloride is dissolved in 400 moles of water, 78.60 kJ heat is evolved. A thermochemical equation to express this can be written as under:

Enthalpy of Neutralisation
Enthalpy of neutralisation of an acid at a given temperature is defined as enthalpy change (∆H) accompanying the neutralisation of one gram equivalent of the acid by a base in dilute solutions at that temperature. Enthalpy of neutralisation of an acid may also be defined as the enthalpy change accompanying the formation of one mole of water by reaction between the acid and a base in dilute solutions. The neutralisation of hydrochloric acid by sodium hydroxide in dilute solutions at 298 K is represented by the thermochemical equation.
HCl (aq) + NaOH (aq) ——–> NaCl (aq) + H₂0 (l); ∆H = – 57.32 kJ
Thus, the enthalpy of neutralisation of hydrochloric acid by sodium hydroxide at 298 K is – 57.32 kJ.
Similarly, enthalpy change accompanying neutralisation of one gram equivalent of a base by an acid in dilute solutions at a given temperature is known as the enthalpy of neutralisation of the base at that temperature. In the above example, the enthalpy of neutralisation of sodium hydroxide with hydrochloric acid is also – 57.32 kJ at 298 K.
The neutralisation of hydrochloric acid by sodium hydroxide in dilute solutions, when the acid, alkali and the salt formed are completely dissociated, may be represented as
H⁺ (aq) + Cl^– (aq) + Na⁺ (aq) + OH^– (aq) ——–> Na⁺ (aq) + Cl^– (aq) + H₂O(l)
or H⁺ (aq) + OH^– (aq) ——–>H₂0(l)
Neutralisation of strong acid and strong base involves the combination of H⁺ and OH^– ions to form unionised water. It is, therefore, expected that the enthalpy of neutralisation of every strong acid by a strong base should be same and this is largely so, as is evident from the data in Table .

Calorimeter Constant of Calorimeter
Measurement of heat changes are carried out in calorimeters. During measurement of heat changes, calorimeter also absorbs some heat which is expressed in terms of calorimeter constant. Calorimeter constant is defined as the amount of heat required to raise the temperature of the calorimeter by one degree Celsius. It is denoted by W and has units joules per degree Celsius or joules per kelvin.

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To Find the Focal Length of a Convex Lens by Plotting Graphs Between U and V or Between 1/u and 1/v

Aim
To find, the focal length of a convex lens by plotting graphs between u and v or between 1/u and 1/v.

Apparatus
An optical bench with three uprights (central upright fixed, two outer uprights with lateral movement), a convex lens with lens holder, two optical needles, (one thin, one thick) a knitting needle and a half metre scale.

Theory
The relation between u, v and f for a convex lens is

where,
f = focal length of convex lens
u = distance of object needle from optical centre of the lens
v = distance of image needle from optical centre of the lens.
Note. According to sign-convention, u has negative value and v has positive value. Hence, f comes positive.

Ray diagram

Procedure
To determine rough focal length

1. Mount the concave mirror in mirror holder.
2. Go out in the open and face the mirror towards distant tree or building.
3. Obtain the image of the tree or the building on a white painted wall (screen) and move the mirror forward and backward to get a sharp image on the wall.
4. Measure the distance between the mirror and the wall (screen). This will be equal to the rough focal length of the mirror.
   To set the lens
5. Clamp the holder with lens in a fixed upright and keep the upright at 50 cm mark.
6. Adjust the lens such that its surface is vertical and perpendicular to the length of the optical bench.
7. Keep the upright fixed in this position throughout.
   To set the object needle
8. Take the thin optical needle as object needle (O). Mount it in outer laterally move¬able upright near zero end.
9. Move the object needle upright and clamp it at a distance (in full cms) nearly 1.5 times the obtained rough focal length of the lens.
10. Adjust height of the object needle to make its tip lie on horizontal line through the optical centre of the lens.
11. Note the position of the index mark on the base of the object needle upright.
    To set the image needle
12. With left eye closed, see with the right open eye from the other end of the optical bench. An inverted and enlarged image of the object needle will be seen. Tip of the image must lie in the middle of the lens.
13. Mount the thick optical needle (image needle) in the fourth upright near the other end of the optical bench.
14. Adjust the height of the image needle so that its tip is seen in line with the tip of the image when seen with right open eye.
15. Move the eye towards right. The tips will get separated. The image tip and the image needle tip have parallax.
16. Remove the parallax tip to tip.
17. Note the position of the index mark on base of the image needle upright.
18. Record the position of the index marks on the base of upright of the lens, the object needle and the image needle in the table against observation 2.
    To determine index correction
19. Find the index correction for distance between optical centre of lens and tip of the object needle and also for distance between optical centre of lens and tip of the image needle as described.
    To get more observations
20. Move object needle upright towards mirror in steps of 1 cm to get observation 2 and 1. Repeat the experiment.
21. Move object needle upright away from mirror (from position of observation 2) in steps of 1 cm to get observations 4, 5 and 6. Repeat the experiment.
22. Record all the observations as given ahead.
    (Note. Same as in Experiment 1).

Observations
Rough focal length of the given convex lens = …….cm
Actual length of the knitting needle x=…….cm
Observed distance between the object needle and the lens
when knitting needle is placed between them y =…….cm
Observed distance between the image needle and the
lens when knitting needle is placed between them z =…….cm
Index correction for the object distance u, x -y =…….cm
Index correction for the image distance v, x-z =…….cm

                                     Table for u,v;  1/u and 1/v

Calculations
Calculations of focal length by graphical methods:
(i) u-v Graph. Select a suitable but the same scale to represent u along X’-axis and v along Y-axis. According to sign conventions, in this case, u is negative and v is positive. Plot the various points for different sets of values of u and v from observation table second quadrant. The graph comes out to be a rectangular hyperbola as shown in graph between u and v.
Draw a line OA making an angle of 45° with either axis (i.e., bisecting ∠YOX’) and meeting the curve at point A. Draw AB and AC perpendicular on X’- and Y-axes, respectively.
The values of u and v will be same for point A. So the coordinates of point A must

Explanation
Same as for concave mirror:
(iii) Another u-v Graph. Select a suitable but the same scale to represent u along
X’-axis and v along Y-axis. Mark the points at distances u₁, u₂, u₃,…… etc. along the OX’-axis
and the corresponding points at distances v₁, v₂, v₃,…… etc. along OY- axis for different sets of observations from the table.
Draw straight lines joining u₁ with v₁; u₂ with v₂; u₃ with v₃;……. etc. These lines will intersect at point K as shown in the following graph.
Draw KL and KM perpendiculars on X’- and Y-axes, respectively

Explanation
Same as for concave mirror:
Note. It will be better to choose any four suitable sets of (a, v) from the observation table. All the six sets of observations may complicate the graph.

Result
The focal length of the given convex lens as determined from

Precautions

1. Tips of the object and image needles should lie at the same height as the centre of the lens.
2. Parallax should be removed from tip to tip by keeping eye at a distance at least 30 cm away from the needle.
3. The object needle should be placed at such a distance that only real, inverted image of it is formed.
4. Index correction for u and v should be applied.

Sources of error

1. The uprights may not be the vertical.
2. Parallax removal may not be perfect.

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Determine the Calorimeter Constant (W) of Calorimeter (Polythene Bottle)

Theory
In order to determine the calorimeter constant, a known volume of hot water at a known temperature is added to a known volume of water taken in the calorimeter at room temperature. Since energy is conserved, the heat gained by the calorimeter and the cold water must be equal to the heat lost by hot water. If t₁ , t₂ and t₃ are the temperatures of cold water, hot water and mixture respectively and m₁, m₂ and m₃  are the masses of calorimeter, cold water and hot water respectively, then we can write

Procedure

1. Put 100 ml of distilled water in polythene bottle with a thermometer and stirrer Fig.
   
2. Note the temperature (t₁°C).
3. Heat some water in a beaker to a temperature 20-30°C higher than that of room temperature.
4. Put 100 ml of this warm water in another beaker.
5. Note the temperature of this water. Let it be t₂°C.
6.  Add warm water from the beaker into the polythene bottle without any loss of time.
7. Stir the contents.
8. Read the temperature attained after mixing. Let it be t₃°C.

Observations
Volume of water taken in bottle = 100 ml
Temperature of water = t₁°C
Volume of warm water added = 100 ml
Temperature of warm water = t₂°C
Temperature after mixing = t₃°C

Calculations
Heat given out by hot water = Heat taken by bottle and cold water.

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

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Determine the Enthalpy of Dissolution of Given Solid Copper Sulphate (CuS0₄.5H₂o) in Water at Room Temperature 

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Theory
In this experiment, the enthalpy of dissolution is measured by the use of calorimetric techniques. A known volume of the water is taken in a polythene bottle as shown in Fig. Its temperature is noted and then known weight of the solute is added to it. The solution is stirred gently and change in temperature is recorded. From the change in temperature, heat absorbed or evolved can be calculated. In this experiment one mole of solute is dissolved per 400 moles of water. For maintaining this ratio 7.0 g of CuS0₄.5H₂0 is dissolved in 200 mL of water.

Requirements
(а) Apparatus. 250 ml or 500 ml polythene bottle fitted with a rubber cork with two holes, one for thermometer (1/10 th degree) and other for stirrer, two beakers, stirrer, measuring cylinder, etc.
(b) Chemicals. Hydrated copper sulphate, distilled water.

Procedure
A. Determination of Calorimeter Constant

1. Put 100 ml of distilled water in polythene bottle with a thermometer and stirrer Fig.
   
2. Note the temperature (t₁°C).
3. Heat some water in a beaker to a temperature 20-30°C higher than that of room temperature.
4. Put 100 ml of this warm water in another beaker.
5. Note the temperature of this water. Let it be t₂°C.
6.  Add warm water from the beaker into the polythene bottle without any loss of time.
7. Stir the contents.
8. Read the temperature attained after mixing. Let it be t₃°C.

B. Determination of Enthalpy of Dissolution

1. Put 200 ml of distilled water into the polythene bottle.
2. Now fit a cork with two holes into the mouth of the polythene bottle. Insert a thermometer into one hole with its bulb about 1 cm above the bottom of the bottle. Put the stirrer into the second hole.
3. Note down the temperature (t₁).
4. Take a known weight of finely powdered substance.
5. Transfer the known weight (say w g) of finely powdered hydrated copper sulphate quickly by removing the rubber cork and putting it back into its position without any loss of time.
6. Stir it with the help of a stirrer till hydrated copper sulphate is dissolved. However, the rate of stirring should he kept as low as efficiency permits to minimise the energy introduced by stirring (vigorous stirring does cause some increase in temperature).
7. Note down the temperature (t₂) when the substance just dissolves.

Observations
Weight of the hydrated copper sulphate dissolved =w g
Volume of water taken into the bottle = 200 ml
= 200 g (assuming density = 1 g/ml)
Temperature of water =t₁°C
Temperature of water after dissolving hydrated copper sulphate = t₂°C
Calorimeter constant of the polythene bottle = W J/°C

Calculations
Assuming density and specific heat of the solution to be same as that of water, heat evolved or absorbed for dissolution of w g of the solute

Result
Enthalpy of dissolution of copper sulphate is…… J/mol.
Note: If t₂ > t₁ heat is evolved during dissolution and ∆_sol H has negative sign.
Similarly we can find out the enthalpy of dissolution of potassium nitrate. For that dissolve 5.5 g of KNO₃ in 200 mL of water. Here, the mole ratio of solute and solvent is 1 : 200.

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Determine the Enthalpy of Neutralisation of Hydrochloric Acid with Sodium Hydroxide Solution

Theory
Heat is evolved during neutralisation of an acid with an alkali. Known volumes of the standard
solutions of an acid and alkali are mixed and the change in temperature is observed and from this, the enthalpy of neutralisation is calculated. Enthalpy of neutralisation is the heat evolved when one gram equivalent of the acid is completely neutralised by a base in dilute solution.

Requirements
(а) Apparatus. A wide-mouthed polythene bottle (to serve as calorimeter), a rubber (cork having two holes, thermometer (1/10th degree), stirrer fitted with a cork on the handle, and a 100 ml measuring cylinder.
(b) Chemicals. 1.0 M hydrochloric acid and 1.0 M sodium hydroxide solution.

Procedure
A. Determination of Calorimeter Constant

1. Put 100 ml of distilled water in polythene bottle with a thermometer and stirrer Fig.
   
2. Note the temperature (t₁°C).
3. Heat some water in a beaker to a temperature 20-30°C higher than that of room temperature.
4. Put 100 ml of this warm water in another beaker.
5. Note the temperature of this water. Let it be t₂°C.
6.  Add warm water from the beaker into the polythene bottle without any loss of time.
7. Stir the contents.
8. Read the temperature attained after mixing. Let it be t₃°C.

B. Determination of Enthalpy of Neutralisation

1. Clean and dry the polythene bottle.
2. Place 100 ml of 1.0 M hydrochloric acid solution in it.
3. Record the temperature of the acid solution.
4. Similarly, note the initial temperature of the sodium hydroxide solution taken in a separate vessel.
5. Both the solutions should have the same temperature, otherwise wait for some time so that they attain the same temperature.
6. Transfer 100 ml of sodium hydroxide solution into the acid solution quickly.
7. Immediately fit the cork having the thermometer and the stirrer in the mouth of the polythene bottle (Fig.) and stir well.
8. Note the temperature after small intervals till it becomes constant.
9. Record the highest temperature (to 0.1°) reached.

Observations
Initial temperature of the acid and base = t₁°C
Final temperature after neutralisation = t₂°C
Change in temperature, ∆t = ( t₂ – t₁)°C.
Mass of the mixture solution after neutralisation = 200 g*
Calorimeter constant of calorimeter  = WJ/°C

Calculations

Result
The enthalpy of neutralisation of HCl with NaOH is……… kJ.
Percentage error =………….
Note: Enthalpy of neutralisation of all strong acids with strong bases and vice versa is – 57.3 kJ. It may be noted that 1000 mL of 1 M HCl contains 1 mole (or 1 equivalent) of HCl.

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Determine the Enthalpy Change During the Interaction (Hydrogen Bond Formation) Between Acetone and Chloroform 

Theory
When acetone is mixed with chloroform, heat is evolved due to formation of hydrogen bonds between chloroform and acetone :

Heat evolved during this interaction can be determined experimentally by mixing the two liquids and measuring the heat change by using a calorimeter.

Requirements
(a) Apparatus. A wide mouthed polythene bottle fitted with a thermometer ( 1/10 th degree) and a stirrer (to serve as calorimeter), 100 ml measuring cylinder.
(b) Chemicals. Pure acetone and pure chloroform.

Procedure
A. Determination of Calorimeter Constant

1. Put 100 ml of distilled water in polythene bottle with a thermometer and stirrer Fig.
   
2. Note the temperature (t₁°C).
3. Heat some water in a beaker to a temperature 20-30°C higher than that of room temperature.
4. Put 100 ml of this warm water in another beaker.
5. Note the temperature of this water. Let it be t₂°C.
6.  Add warm water from the beaker into the polythene bottle without any loss of time.
7. Stir the contents.
8. Read the temperature attained after mixing. Let it be t₃°C.

B. Determination of Enthalpy of Interaction of Acetone and Chloroform

1. Take a clean and dry polythene bottle calorimeter.
2. Place 100 ml acetone in it.
3. Note the temperature of acetone.
4. Take 100 ml of chloroform in a beaker and note its temperature. Both the solutions should have same temperature otherwise wait for sometime so that they attain same temperature.
5. Transfer the chloroform into the calorimeter and immediately fit the cork (or lid) having thermometer and stirrer. Stir gently.
6. Note the temperature after small intervals till it becomes constant.
7. Record the highest temperature reached.

Observations

Result
Enthalpy change during mixing of 100 ml of acetone with 100 ml of chloroform = – X Joules.

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To find the focal length of a concave lens using a convex lens

Aim
To find the focal length of a concave lens using a convex lens.

Apparatus
An optical bench with four upright (two fixed uprights in middle, two outer uprights with lateral movement), a convex lens (less focal length), a concave lens (more focal length), two lens holders, two optical needles (one thin, one thick), a knitting needle and a half metre  scale.

A Short Description about the Arrangement
As a concave lens always forms a virtual image, its focal length can not be found directly as for a convex lens. For this purpose, indirect method is used, as described below.
An object needle O is placed on one side of a convex lens L₁ and its real inverted image I is located (by image needle) on the other side as shown in ray diagram.
The concave lens L₂ is placed between convex lens L₁ and image needle I. The concave lens diverges the rays and the image is now formed at I’ as shown in ray diagram.
For concave lens, I is the virtual object and I’ is the real image. Hence, O₂I = u and O₂I’ = v.
Focal length can be calculated, using lens formula

Theory

Ray diagram

Procedure
To determine rough focal length of convex lens

1. Mount the convex lens in lens holder.
2. Go out in the open and face the lens towards distant tree or building.
3. Obtain the image of the tree or the building on a white painted wall (screen) and move the lens forward and backward to get a sharp image on the wall.
4. Measure the distance between the lens and the wall (screen). This will be equal to the rough focal length of the mirror.
   To set the convex lens
5. Follow steps 2 to 4 of Experiment 2 To set the object needle
6. Follow steps 5 to 8 of Experiment 2 To set the image needle at I
7. Follow steps 21 to 27 of Experiment 2 To set the concave lens
8. Clamp the holder with concave lens on fixed upright on the I side of the convex lens.
9. Fix this upright at some distance away from the convex lens.
10. Set the concave lens surface in same manner as convex lens surface with principal axes of the lenses coinciding.
    To set the image needle at I’
11. Repeat steps 4 and 5 of the experiment.
    To get more observations
12. Follow steps 29, 30 and 31 of Experiment 2.

Observations

Calculations
1. Find difference of positions of L2 and I and write it as observed u in column 3a.
2. Find difference of positions of L2 and I’ and write it as observed v in column 36.
3. Apply index correction and write corrected values of u and v in columns 4a and 46.
4. Calculate f = uv/u-v and write in column 5.
5. Take mean of different values oif as recorded in column 5.

Result
The focal length of the given concave lens = -……cm

Precautions

1. Focal length of the convex lens should be less than the focal length of concave lens so that the combination is convex.
2. The lenses must be clean. .
3. Other precautions are same as given in Experiment 3.

Viva Voce

Question. 1. Define a spherical lens.
Answer. Read Art. 8.01 (a).

Question.2. Describe different types of lenses.
Answer. Read Art. 8.01 (b).

Question.3. Describe different types of convex lenses.
Answer. Read Art. 8.01 (c).

Question.4. Describe different types of concave lenses.
Answer. Read Art. 8.01 (d).

Question.5. Define different terms associated with spherical lenses.
Answer. Read Art. 8.02 (1-7).

Question.6. Mention three special rays.
Answer. Read Art. 8.03.

Question.7. Define sign convention.
Answer. Read Art. 8.04 (a).

Question.8. Give rules of sign convention.
Answer. Read Art. 8.04 (b).

Question. 9. Give facts obtained from sign convention.
Answer. Read Art. 8.04 (c).

Question.10. Define and give lens formula.
Answer. Read Art. 8.05.

Question.11. Describe various assumptions made in derivation of lens formula.
Answer. Read Art. 8.06.

Question.12. Give position, nature and size of image when object is put in different positions in front of a convex lens.
Answer. Read Art. 8.07.

Question.13. Define power of a lens. Give its unit and sign.
Answer. Read Art. 8.08.

Question.14. Define a lens combination. Give expression for. its focal length and power.
Answer. Read Art. 8.09.

Question.15. Define chromatic aberration.
Answer. Read Art. 8.10 (a).

Question.16. Describe the difference between the images formed by a convex and a concave lens. .
Answer. A concave lens always forms a virtual, erect and diminished image. Image formed by a convex lens is generally real and inverted and on bringing the object near the lens the size of image goes on increasing. However, when the object is placed in front of a convex lens between its optical centre and principal focus, the image formed is virtual, erect and magnified.

Question.17. Which convex lens has more focal length, thick or thin?
Answer. A thin convex lens has more focal length. ^ ..

Question.18. Can you find rough focal length of a concave lens?
Answer. No, because it does not form a real image to be obtained on a screen.

Question.19. What is the type of the eye lens?
Answer. The eye lens is convex.

Question.20. What are the practical uses of lenses?
Answer. Lenses are used in spectacles, microscopes, telescopes and other optical instruments.

Question.21. How can a convex lens be used as a magnifier?
Answer. For this purpose the lens is put very close to the eye in between the eye and the object to be magnified.

Question. 22. How will you distinguish between a glass slab, a convex lens and a concave lens without touching it?
Answer. The glass piece is put over a printed page and the virtual image of the printed matter is seen. The magnification of the image is judged.
If the image has same size as the object, the glass piece is a glass slab.
If the image is magnified, the glass piece is a convex lens.
If the image is diminished, the glass piece is a concave lens.

Question. 23. Define optical centre of a len.
Answer. It is a fixed point inside the lens on its principal axis, through which fight rays passing undeviated.

Question.24. What is the principal axis of a lens?
Answer. The straight fine passing through the centres of curvature of the curved surfaces of the lens is called the principal axis of the lens.

Question.25. What is the principal focus of a lens?
Answer. It is fixed point on the principal axis of a lens where a beam of fight incident parallel to its principal axis converges or appears to diverge after passing through the convex lens or concave lens.

Question. 26. What is the focal length of a lens?
Answer. It is the distance between optical centre and principal focus of a lens. Its S.I. unit is metre.

Question.27. Define S.I unit of power.
Answer. The Dioptre is the S.I. unit of power. One dipotre is the power of lens whose focal length is one metre.

Question. 28. What are the sign for the power of a convex lens and concave lens?
Answer. The power of a convex lens is positive since its focal length is positive while the power of a concave lens is negative since its focal lens is negative.

Question. 29. What is a lens maker formula?
Answer. It is relation between focal length, radii of curvature, refractive index of material of lens and refractive index of surroundings.

Question.30. What are the factors affecting the power of lens?
Answer.

1. Refractive index of lens material
2. Refractive index of surroundings i.e., change of medium
3. Radii of curvature
4. Wavelength of light
5. Thickness of lens.

Question.31. How the power of lens charge w.r.t. the two surrounding medium?
Answer. The power of a lens is maximum for vacuum or air and it decreases with increase in two refractive index of medium.

Question.32. How the power of lens charge w.r.t. to wavelength of light?
Answer. The power of a lens is different for different colour of light. The power of a lens is maximum of violet and minimum for red colour light.

Question. 33. Does power depend upon aperature of a lens?
Answer. No.

Question.34. Under what condition, the nature of lens change?
Answer. The refractive index of surrounding medium is greater them that of material of lens. The convex lens act as concave lens and vice-versa.

Question. 35. Under what condition, a lens does not show the refraction.
Answer.When refractive index of surrounding medium is equal to refractive index of material of lens.

Question.36. Why goggles (Sun glasses) have zero power?
Answer. The surfaces are curved in same direction and of same radius

Question. 37. What type of lens is an air bubble inside water?  
Answer. Concave lens.

Question.38.Define refractive index.
Answer.It is the property of a transparent medium which resist the propagation of light in that medium. It is measured in term of speed of light in a medium w.r.t. speed of light in vacuum.

Question.39.What is relative refractive index?
Answer.Relative refractive index of medium 2 w.r.t. medium 1 is the ratio of the speed of light in medium 1 to the speed of light in medium 2

It does not have emit and dimensions.

Question.40.What is absolute refractive index?
Answer. Absolute refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in that medium.

Question.41.Is the absolute refractive can be less than unit?
Answer. No.

Question.42.What is the power of combination of a convex and concave lens of the same focal length?
Answer. Zero.

Question.43.Why is the rough focal length of concave lens not determine?
Answer. It makes virtual image for all positions of objects.

Question.44.How chromatic aberration can be minimized?
Answer. It can be minimized by taking thin and small aperature lens.

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Thermochemistry Viva Questions with Answers

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Question.1. Define enthalpy of neutralization.
Answer. Enthalpy of neutralization of an acid or a base is the enthalpy change when one gram equivalent of the acid is neutralised by a base or vice versa.

Question.2. What is the enthalpy of neutralisation of a strong acid and a strong base ?
Answer. When one gram equivalent of a strong acid is neutralised by one gram equivalent of a strong base or vice versa, the enthalpy change is always equal to – 57.3 kJ.

Question.3. Why the enthalpy of neutralisation of a strong acid with a strong base is always the same ?
Answer. This is because it always involves the combination of one gram equivalent of H+ ions with OH” ions to form unionised water molecules.

Question.4. Define enthalpy of solution.
Answer. It is the enthalpy change taking place when one mole of a substance is dissolved in a specified number of moles of solvent at a given temperature and pressure.

Question.5. Why is copper sulphate taken in powdered form ?
Answer. To facilitate its dissolution in minimum time and thus preventing heat loss to the surroundings.

Question.6. Will enthalpy of solution of hydrated copper sulphate and anhydrous copper sulphate be same ?
Answer. No, in case of anhydrous copper sulphate enthalpy change will not only correspond to the dissolution process but also to hydration process, i.e., we get enthalpy of hydration plus enthalpy of solution.

Question.7. Why is temperature recorded with a thermometer calibrated to 1/10 th degree ?
Answer. For more accurate results.

Question.8. Is the enthalpy of neutralisation of acetic acid the same as that of HCl. If not why ?
Answer. Acetic acid is a weak acid and is not completely ionised. Some heat is used up for the ionisation of acetic acid. Hence the net heat evolved is less and not the same as that of HCl which is completely ionised.

Question.9. Is the dissolution of hydrated copper sulphate an exothermic or endothermic process ?
Answer. Endothermic process.

Question.10. 50 ml of a liquid A are mixed with 50 ml of liquid B. The volume of resulting solution is found to be 99.5 ml. What do you conclude about nature of solution ?
Answer. The solution shows a negative deviation from Raoult’s law, A—B interactions are stronger than A—A and B—B interactions.

Question.11. When a liquid A is mixed with liquid B, the resulting solution is found to be cooler. What do you conclude about nature of solution ?
Answer. The solution shows a positive deviation. Absorption of heat takes place. A—B interactions are weaker than A—A and B—B interactions.

Question.12. What type of deviation is expected of a solution obtained by adding cone. H₂So₄to water ?
Answer. The solution shows negative deviation. Heat is liberated. A—B interactions are stronger than A—A and B—B interactions.

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Separate the Coloured Components Present in the Mixture of Red and Blue Inks by Ascending Paper Chromatography and Find their R_f Values 

Apparatus
Gas jar, glass rod, filter paper strip (What man No. 1 filter paper), jar cover, fine capillary tube.

Requirement
A mixture of red and blue inks, alcohol and distilled water.

Procedure

1. Take a What man filter paper strip (20 x 2 cm) and draw a line with pencil above 3 cm from one end. Draw another line lengthwise from the centre of the paper as shown in Fig.
   
2. With the help of fine capillary tube, put a drop of the mixture of red and blue inks at the point P. Let it dry in air. Put another drop on the same spot and dry again. Repeat 2-3 times, so that the spot is rich in mixture.
3. Suspend the filter paper vertically in a gas jar containing the solvent (eluent) with the help of a glass rod in such a way that the pencil line (and the spot) remains about 2 cm .above the solvent level (50% alcohol + distilled water).
4. Cover the jar and keep it undisturbed. Notice the rising solvent along with the mixture of red and blue inks. After the solvent has risen about 15 cm you will notice two different spots of blue and red colours on the filter paper.
5. Take the filter paper out of the jar and mark the distance that the solvent has risen on the paper with a pencil. This is called the solvent front.
6. Dry the paper. Put pencil marks in the centre of the blue and red spots.
7. Measure the distance of the two spots from the original line and the distance of the solvent from the original line.
8. Calculate the Revalues of the blue and red inks by using the formula :
   

Observations and Calculations

Precautions

1. Use good quality pencil for drawing the reference line so that the mark does not dissolve in the solvent in which the chromatography is carried out.
2. Always make use of a fine capillary tube.
3. Keep the jar undisturbed and covered during the experiment.
4. A spot should be small and rich in mixture.
5. Allow the spot to dry before putting the strip in the jar.
6. Keep the strip erect. Do not let it to be curled.
7. Do not allow the spot to dip in the solvent.

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Separate the Coloured Components Present in the Given Grass/Flower by Ascending Paper Chromatography and Determine their R_f Values 

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Apparatus
Gas jar, glass rod, filter paper strip (What man No. 1 filter paper), jar cover, fine capillary tube.

Requirement
A mixture of red and blue inks, alcohol and distilled water.

Procedure

1. Take a What man filter paper strip (20 x 2 cm) and draw a line with pencil above 3 cm from one end. Draw another line lengthwise from the centre of the paper as shown in Fig.
   
2. With the help of fine capillary tube, put a drop of the mixture of red and blue inks at the point P. Let it dry in air. Put another drop on the same spot and dry again. Repeat 2-3 times, so that the spot is rich in mixture.
3. Suspend the filter paper vertically in a gas jar containing the solvent (eluent) with the help of a glass rod in such a way that the pencil line (and the spot) remains about 2 cm .above the solvent level (50% alcohol + distilled water).
4. Cover the jar and keep it undisturbed. Notice the rising solvent along with the mixture of red and blue inks. After the solvent has risen about 15 cm you will notice two different spots of blue and red colours on the filter paper.
5. Take the filter paper out of the jar and mark the distance that the solvent has risen on the paper with a pencil. This is called the solvent front.
6. Dry the paper. Put pencil marks in the centre of the blue and red spots.
7. Measure the distance of the two spots from the original line and the distance of the solvent from the original line.
8. Calculate the Revalues of the blue and red inks by using the formula :
   

Observations and Calculations

Note: In the above experiment, crush fresh flowers or grass in a mortar and extract the juice with acetone. Use this juice for making the spot.

Precautions

1. Use good quality pencil for drawing the reference line so that the mark does not dissolve in the solvent in which the chromatography is carried out.
2. Always make use of a fine capillary tube.
3. Keep the jar undisturbed and covered during the experiment.
4. A spot should be small and rich in mixture.
5. Allow the spot to dry before putting the strip in the jar.
6. Keep the strip erect. Do not let it to be curled.
7. Do not allow the spot to dip in the solvent.

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Separate Co²⁺ and Ni²⁺ Ions Present in the Given Mixture by Using Ascending Paper Chromatography and Determine their R_f Values

Apparatus
Gas jar, glass rod, filter paper strip (What man No. 1 filter paper), jar cover, fine capillary tube.

Chemicals Requirement
Sample solution containing cobalt (II) and nickel (II) ions, acetone, concentrated aqueous ammonia, Rubeanic acid spray reagent.

Procedure

1. Take a What man filter paper strip (20 x 2 cm) and draw a line with pencil above 3 cm from one end. Draw another line lengthwise from the centre of the paper as shown in Fig.
   
2. With the help of fine capillary tube, put a drop of the mixture of red and blue inks at the point P. Let it dry in air. Put another drop on the same spot and dry again. Repeat 2-3 times, so that the spot is rich in mixture.
3. Suspend the filter paper vertically in a gas jar containing the solvent (eluent) with the help of a glass rod in such a way that the pencil line (and the spot) remains about 2 cm .above the solvent level (50% alcohol + distilled water).
4. Cover the jar and keep it undisturbed. Notice the rising solvent along with the mixture of red and blue inks. After the solvent has risen about 15 cm you will notice two different spots of blue and red colours on the filter paper.
5. Take the filter paper out of the jar and mark the distance that the solvent has risen on the paper with a pencil. This is called the solvent front.
6. Dry the paper. Put pencil marks in the centre of the blue and red spots.
7. Measure the distance of the two spots from the original line and the distance of the solvent from the original line.
8. Calculate the Revalues of the blue and red inks by using the formula :
   

After elution and drying, place the paper in a large, dry, covered beaker containing a smaller beaker of concentrated aqueous ammonia. After about two minutes, remove the paper and spray it on both sides with rubeanic acid reagent. Allow it to dry. Nickel becomes visible as blue purple band while cobalt becomes visible as yellow orange band. Evaluate R_f values of the two ions.

Observations and Calculations

Result
Revalue of Ni²⁺ =………
Revalue of Con²⁺ =……..
The above experiment can be carried by using a mixture of
(i) Iron (II) and cobalt (II) (ii) Iron (II) and nickel (II)
(iii) Copper (II) and iron (II) (iv) Copper (II) and nickel (II)
(v) Iron (II) and zinc (II) (vi) Lead (II) and Cadmium (II).

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Chromatography Viva Questions with Answers

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Question.1. What is chromatography ?
Answer. It is technique for rapid and efficient separation of components of a mixture and purification of compounds. It is based on differential migration of the various components of a mixture through a stationary phase under the influence of a moving phase.

Question.2. What is the basis (principle) of chromatographic process ?
Answer. It is based on the differential migration of the individual components of a mixture through a – stationary phase under the influence of a moving phase.

Question.3. What type of solvents are generally employed in chromatography ?
Answer. Generally solvents having low viscosities are employed in chromatography. This is due to the fact that the rate of flow of a solvent varies inversely as its viscosity.

Question.4. Name some chromatographic techniques.
Answer. Paper chromatography, column chromatography, thin layer chromatography, gas chromatography.

Question.5. What are the moving and stationary phases in paper chromatography ?
Answer. Water absorbed on cellulose constituting the paper serves as the stationary phase and organic solvent as moving phase.

Question.6. What is meant by the term developing in chromatography ?
Answer. During chromatography, if the components to be separated are colourless, then these separated components on chromatogram are not visible. Their presence is detected by development, which involves spraying a suitable reagent (called developing reagent) on the chromatogram, or placing the chromatogram in iodine chamber, when various components become visible. This process is called developing of chromatogram.

Question.7. How does the liquid rise through the filter paper ?
Answer. By means of capillary action.

Question.8. What is meant by the term R_f value ?
Answer. R_f (retention factor) of a substance is defined as the ratio of the distance moved up by the solute from the point of its application to the distance moved up by the solvent from the same point.

Question.9. On what factors does the R_f value of a compound depend ?
Answer.

1. Nature of the compound.
2. Nature of the solvent.
3. Temperature.

Question.10. Give the biochemical uses of chromatography.
Answer. It helps in the separation of amino acids, proteins, peptides, nucleic acids, etc.

Question.11. Name the scientist who introduced chromatographic technique.
Answer. Russian botanist M. Tswett (1906).

Question.12. What are the advantages of chromatography over other techniques ?
Answer.

1. It can be used for a mixture containing any number of components.
2. Very small quantities of the substances can be effectively detected and separated from a mixture.

Question.13. What- is loading (or spotting) ?
Answer. The application of the mixture as a spot on the original line on the filter paper strip or addition of mixture to the column, is called loading (or spotting).

Question.14. What are the essential characteristics of the substance used as a developer ?
Answer.

1. It should be volatile.
2. It should impart colour to the different spots.
3. It should not react with various compounds which are being separated.

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Preparation of Inorganic Compounds

A double salt is a substance obtained by the combination of two different salts which crystallise together as a single substance but ionize as two distinct salts when dissolved in water. The constituent salts are always taken in some definite molecular proportions. Alums are common examples of double salts.

Alums are isomorphous crystalline solids which are soluble in water. Due to hydrolysis, their aqueous solutions have acidic character.
Another example of double salts is Mohr’s salt. Its formula is FeS04.(NH₄)2S0₄ .6H₂0. It is used as primary standard in volumetric analysis. Its crystals do not lose water of crystalisation by efflorescence nor it is oxidised in air. It is stable salt unlike green vitriol (FeS0₄ .7H₂0) which gets oxidised by air.
Before we discuss preparation of some of these double salts, let us first review the process of crystallisation.

Process of Crystallisation
The process of crystallisation involves following steps :
1. Preparation of Solution of the Impure Sample
1. Take a clean beaker (250 ml) and add powdered impure sample under consideration in it (~ 6.0 gm).
2. Add distilled water (25-30 ml) and stir contents gently with the help of a glass rod
giving circular motion as shown in Fig.
3. The solution in the beaker is heated (60°-70°C) on a wire gauze (Fig).

4. Stir the solution continuously and add more of impure substance till no more of it dissolves.
2. Filtration of Hot Solution
1. Take a circular filter paper. First fold it one-half, then fold it one-fourth as shown in Fig. Open the filter paper, three folds on one side and one fold on the other side to get a cone (Fig).

2. Take a funnel and fit the filter paper cone into the funnel so that the upper half of the cone fits well into the funnel but lower part remains slightly away from the funnel.
3. Wet the filter paper cone with a spray of water from a wash bottle pressing the upper part of the filter paper cone gently against the wall of the funnel with the thumb (Fig).

4. Place the funnel on a funnel stand and place a clean china dish below the funnel for the collection of the filtrate. To avoid splashing of the filtrate, adjust the funnel so that its stem touches the wall of the dish.
5. Hold a glass rod in slanting position in your hand or with a precaution that the lower end of the rod should reach into the filter paper cone but it does not touch it. Pour the solution along the glass rod as shown in Fig. The filtrate passes through the filter paper and is collected into the china dish placed below. The insoluble impurities are left behind on the filter paper.

3. Concentration of Filtrate
1. Place the dish containing the clear filtrate over wire gauze, kept over a tripod stand and heat it gently (Do not boil). Stir the solution with a glass rod (Fig). This is done to ensure uniform evaporation and to prevent formation of solid crust.

2. When the volume of the solution is reduced to one-half, take out a drop of the concentrated solution on one end of glass rod and cool it by blowing air (Fig). Formation of thin crust indicates that the crystallisation point has reached.

3. Stop heating by removing the burner.
4. Cooling the Concentrated Solution
1. Pour the concentrated solution into a crystallising dish. (It is a thin walled shallow glass dish with a flat bottom and vertical sides. It has a spout to pour off the mother liquor).
2. Cover the dish with a watch glass and keep it undisturbed (Fig).
3. As the solution cools, crystals separate out. The concentrated solution is cooled slowly for better yield of the crystals.
Sometimes the china dish containing the concentrated solution is cooled by placing on a beaker filled to the brim with cold water. Cooling may also be done by keeping the china dish in open air depending upon the weather conditions.

5. Separation and Drying of Crystals
1. Decant off the mother liquor and wash the crystals with a thin stream of cold water with the help of a wash bottle.

2. Dry the crystals by pressing them gently between the sheets of filter paper Fig. The crystals can be dried by spreading them on a porous plate for sometime or by placing the crystals in vacuum desiccator.
Crystals have definite geometry and therefore a definite shape. Figure shows some of these shapes. Copper sulphate crystals are formed in triclinic shape, potash alum comes out in octahedral geometry. Potassium nitrate crystals are needle like and ferrous sulphate have monoclinic shape.

Shapes of crystals of some common substances are given in Table
Table Shapes of Crystals of Some Common Substances

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To Prepare a Pure Sample of Ferrous Ammonium Sulphate (Mohr’s salt), [FeSO₄ . (NH₄)₂ SO₄.6HO₂0]

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Theory
Mohr’s salt is prepared by dissolving an equimolar mixture of hydrated ferrous sulphate and ammonium sulphate in water containing a little of sulphuric acid, and then subjecting the resulting solution to crystallisation when light green crystals of ferrous ammonium sulphate. FeS0₄ . (NH₄ )2S0₄.6H₂0 separate out.

Requirements
Two beakers (250 ml), china-dish, funnel, funnel-stand, glass-rod, wash-bottle, tripod stand and wire-gauze. Ferrous sulphate crystals, ammonium sulphate crystals, dilute sulphuric acid and ethyl alcohol.

Procedure

1. Take a 250 ml beaker and wash it with water. Transfer 7.0 g ferrous sulphate and 3.5 g ammonium sulphate crystals to it. Add about 2-3 ml of dilute sulphuric acid to prevent the hydrolysis of ferrous sulphate.
2. In another beaker boil about 20 ml of water for about 5 minutes to expel dissolved air.
3. Add the boiling hot water to the contents in the first beaker in small instalments at a time. Stir with a glass rod until the salts have completely dissolved.
4.  Filter the solution to remove undissolved impurities and transfer the filtrate to a china-dish.
5. Heat the solution in the china-dish for some time to concentrate it to the crystallisation point.
6. Place the china-dish containing saturated solution over a beaker full of cold water. On cooling crystals of Mohr’s salt separate out.
7. Decant off the mother liquor quickly. Wash the crystals in the china-dish with a small quantity of alcohol to remove any sulphuric acid sticking to the crystals.
8. Dry the crystals by placing them between filter paper pads.

Observations
Weight of crystals obtained =……….. g
Expected yield = ………..g
Colour of the crystals = …………..
Shape of the crystals =…………
Note: The crystals of Mohr’s salt are monoclinic in shape.

Precautions

1. Cool the solution slowly to get good crystals.
2. Do not disturb the solution while it is being cooled.
3. Do not heat the solution for a long time as it may oxidize ferrous ions to ferric ions.

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To determine refractive index of a glass slab using a travelling microscope

Aim
To determine refractive index of a glass slab using a travelling microscope.

Apparatus
Three “glass slabs of different thickness but same material, a travelling microscope, lycopodium powder. A slab is a piece of transparent material with rectangular faces. All faces are transparent and opposite faces are parallel. The dimension along with the light travels inside the slab is called its thickness.
A Short Description of a Travelling Microscope
It is a compound microscope fitted vertically on a vertical scale. It can be moved up and down, carrying a Vernier scale moving along the main scale.
In any position, the reading is taken by combining main scale and vernier scale reading.

Theory

Diagram

Procedure
Adjustment of travelling microscope

1. Place the travelling microscope (M) on the table near a window so that sufficient light falls on it.
2. Adjust the levelling screws so that the base of the microscope becomes horizontal.
3. Make microscope horizontal. Adjust the position of the eye piece so that the cross wires are clearly visible.
4. Determine the vernier constant of the vertical scale of the microscope.
   Other steps
5. Make a black-ink cross-mark on the base of the microscope. The mark will serve as
   point P. ,
6. Make the microscope vertical and focus it on the cross at P, so that there is no parallax between the cross-wires and the image of the mark P.
7. Note the main scale and the vernier scale readings (R₁) on the vertical scale.
8. Place the glass slab of least thickness over the mark P.
9. Raise the microscope upwards and focus it on the image P₁ of the cross-mark.
10. Note the reading (R₂) on the vertical scale as before (Step 7).
11. Sprinkle a few particles of lycopodium powder on the surface of the slab.
12. Raise the microscope further upward and focus it on the particle near S.
13. Note the reading (R₃) on the vertical scale again (Step 7).
14. Repeat above steps with other glass slab of more thicknesses.
15. Record observations in tabular form as given below.

Observations and calculations
Vernier constant (least count) for vertical scale of microscope =……cm.

Result

Precautions

1. In microscope, the parallax should be properly removed.
2. The microscope should be moved in upper direction only to avoid back lash error.

Sources of error
The microscope scale may not be properly calibrated.

Viva Voce

Question. 1. Define a slab.
Answer. Read Art. 9.07.

Question. 2. Define thickness of a slab.
Answer. Read Art. 9.07.

Question. 3. Define lateral displacement.
Answer. Read Art. 9.09.

Question. 4. Why a slab does not deviate and disperse light, where as a prism does?
Answer. In a slab, the refracting faces are parallel. The emergent ray is parallel to the incident ray. There is no deviation and dispersion.
In a prism, the refracting faces are not parallel. The emergent ray is not parallel to incident ray. There is a deviation and hence dispersion.

Question. 5. Why lycopodium power is spread over the glass surface?
Answer. To focus the microscope accurately, otherwise the bottom surface will be focussed because of transparency of glass slab.

Question. 6. What is normal shift?
Answer. It is the difference between actual depth and apparent depth. Its S.I. unit is metre.

Question. 7. What is cause of normal shift?
Answer. Due to refraction of light.

Question. 8. On what factors, apparent depth depends?
Answer.

1. nature of medium (R.I.)
2. thickness of medium (actual depth)
3. colour of light.

Question. 9. In general for which colour we take the refractive index of a material in lens and glass slabs.
Answer. Yellow colour. Since it is the mean colour of visible spectrum.

Question. 10. What may be refractive index for hollow glass slab?
Answer. n = 1.

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To Prepare a Pure Sample of Potash Alum (Fitkari), [K₂SO₄.Al₂ (SO₄)₃. 24H₂0]

Theory
Potash alum is prepared by dissolving an equimolar mixture of hydrated aluminium sulphate and potassium sulphate in minimum amount of water containing a little of sulphuric acid and then subjecting the resulting solution to crystallisation, when octahedral crystals of potash alum separate out.

Requirements
Two beakers (250 ml), china-dish, funnel, funnel-stand, glass-rod, wash-bottle, tripod stand and wire-gauze. Potassium sulphate, aluminium sulphate and dil. sulphuric acid.

Procedure

1. Take a 250 ml beaker. Wash it with water and then transfer 2.5 g potassium sulphate crystals to it. Add about 20 ml of water. Stir to dissolve the crystals. Warm if required.
2. Take the other 250 ml beaker, wash it with water and then transfer 10 g aluminium sulphate crystals to it. Add about 20 ml of water and 1 ml of dilute sulphuric add to prevent hydrolysis of aluminium sulphate. Heat for about 5 minutes. If milkiness still persists, filter the solution.
3. Mix the two solutions in a china-dish and place the china-dish on a wire-gauze placed over a burner. Stir the solution with a glass-rod. Concentrate the solution till the crystallisation point is reached. Place the dish over a beaker containing cold water.
4. Soon the crystals of potash alum separate out. Decant off the mother liquor and wash the crystals with a small quantity of ice-cold water.
5. Dry the crystals by placing them between filter paper pads or by spreading them over porous plate.

Observations
Weight of crystals obtained = …………g
Expected yield = …………g
Colour of the crystals =…..
Shape of the crystals = ………
Note: The crystals of potash alum are octahedral in shape.

Precautions

1. Cool the solution slowly to get good crystals.
2. Do not disturb the solution while it is being cooled.

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

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To find the refractive index of a liquid by using convex lens and plane mirror

Aim
To find the refractive index of a liquid by using convex lens and plane mirror.

Apparatus
A convex lens, a plane mirror, clean transparent liquid in a beaker, an optical needle, (a thick knitting needle passed through a rubber cork), an iron stand with base and clamp arrangement, plumb line, plane glass slab, a spherometer, half metre scale etc.

Theory 
If f₁ and f₂ be the focal length of glass convex lens and liquid lens and F be the focal length of their combination then,

Liquid lens formed is a planoeconcave lens with R₁= R (radius of curvature of convex lens surface), R₂ =∞

Diagram

Procedure
(a) For focal length of convex lens

1. Take any one convex lens and find its rough focal length.
2. Take a plane mirror and place it on the horizontal base of the iron stand.
3. Place the convex lens on the plane mirror.
4. Screw tight the optical needle in the clamp of the stand and hold it horizontally above the lens at distance equal to its rough focal length.
5. Bring the tip of the needle at the vertical principal axis of the lens, so that tip of the needle appears touching the tip of its image.
6. Move the needle up and down and remove parallax between tips of the needle and its image.
7. Measure distance between tip and upper surface of the lens by using a plumb line and half metre scale.
8. Also measure distance between tip and the surface of its plane mirror.

(b) For focal length of the combination

1. Take a few drops of transparent liquid on the plane mirror and put the convex lens over it with its same face above as before (A piano concave liquid lens is formed between plane mirror and convex lens).
2. Repeat steps 6, 7 and 8.
3. Record your observations as given below.

(c) For radius of curvature of convex lens surface

Observations

1. Rough focal length of convex lens =……… cm.
2.                                                   Table for distance of needle tip from lens and mirror
   

Calculations

Precautions

1. The liquid taken should be transparent.
2. Only few drops of liquid should be taken so that its layer is not thick.
3. The parallax should be removed tip to tip.

Sources of error

1. Liquid may not be quite transparent.
2. The parallax may not be fully removed.

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