Tests of Carbohydrates, Fats and Proteins in Pure Samples and Detection of Their Presence in Given Food Stuffs

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Food is a necessary material which must be supplied to the body for its normal and proper functioning.
The main functions of the food are:
1. to provide energy
2. to promote growth
3. to replace worn out tissues
4. to sustain life
5. to regulate body processes like assimilation and digestion.
The essential constituents of food are:
1. Carbohydrates
2. Lipids (oils and fats)
3. Proteins
4. Minerals
5. Vitamins and
6. Water.

Carbohydrates
The name carbohydrate is used for the compounds having general formula, C_x(H₂O)_y. These are called carbohydrates because they can be treated as hydrates of carbon. For example, glucose (C₆H₁₂O₆), sucrose (C₁₂H₂₂O₁₁), etc. However, this definition of carbohydrates has lost significance because of the following two reasons:
(i) There are many compounds which have general formula C_x(H₂O)_y but do not behave as carbohydrates. For example, oxalic acid (C₂H₂O₂), formaldehyde (CH₂O), etc.
(ii) There are many compounds which do not conform to formula C_x(H₂O)_y but possess characteristic properties of carbohydrates and are treated as carbohydrates. For example, rham-nose (C₆H₁₂O₅).
A more useful definition of carbohydrates is that Carbohydrates are polyhydroxy aldehydes, polyhydroxy ketones, their derivatives and the substances which yield them on hydrolysis.
The carbohydrates which are ketones are called ketoses and those that are aldehydes are called aldoses. The general term for all the carbohydrates is glycose.
The carbohydrates which cannot be hydrolysed to simple carbohydrates are called monosaccharides. For example,glucose,fructose ,etc.

The carbohydrates which contain two to ten monosaccharide units are called oligosaccharides. For example, sucrose (C₁₂H₂₂O₁₁), maltose (C₁₂H₂₂O₁₁), raffinose (C₁₈H₃₂O₁₆), etc.
The carbohydrates which contain more than ten monosaccharide units are called polysaccharides. For example, starch, cellulose, glycogen, etc. These may be represented by general formula (C₆H₁₀O₅)n.
A more general classification of carbohydrates is into sugars and non-sugars. The sugars like glucose, fructose and canesugar are crystalline, water soluble and sweet substances. Non-sugars which include starch, cellulose, etc., are amorphous, insoluble in water and taste¬less substances. ,
The carbohydrates which can reduce Tollen’s reagent or Fehling solution are called reducing sugars. All monosaccharides are reducing sugars. Most of the disaccharides are also reducing sugars. Sucrose is a non-reducing sugar.
Carbohydrates are generally optically active because they contain chiral centres.
The carbohydrates perform two important functions in body:
(a) They act as biofuels to provide energy for functioning of living organisms.
In human system, all the carbohydrates except cellulose can serve as source of energy. Starch and various sugars which are taken as food are first hydrolysed to glucose by the enzymes present in the digestive system.
Glucose on slow oxidation to carbon dioxide and water in the presence of enzymes liberates large amount of energy which is used by the body for carrying out various functions.
C₆H₁₂O₆+ 60₂ ——> 6C0₂ + 6H₂0 + 2832 kJ
In order to fulfil the emergency requirements our body also stores some of the carbohydrates as glycogen in liver. Glycogen on hydrolysis gives glucose.
It may be noted that cellulose cannot be hydrolysed in our body because enzymes required for its hydrolysis are not present in our body. However, grazing animals are capable of hydrolysing cellulose to glucose.
(b) They act as constituents of cell membrane.

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To  Analyse the Given Salt for Acidic and Basic Radicals CO₃^2-, Zn²⁺

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

To analyse the given salt for acidic and basic radicals.

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Volumetric Analysis

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

In volumetric analysis, the quantities of the constituents present in the given unknown solution are determined by measuring the volumes of the solutions taking part in the given chemical reaction. The main process of this analysis is called titration which means the determination of the volume of a reagent required to bring a definite reaction to completion.

Apparatus Used in Volumetric Analysis
In volumetric analysis, the apparatus required is as follows:
(i) Graduated-burette, pipette, measuring flasks and measuring cylinders.
(ii) General-titration flasks, beaker, tile, glass-rod, funnel, weighing bottle, wash bottle.
(iii) A chemical balance for weighing.
It is a long, cylindrical tube of uniform bore fused at the lower end with a stop cock (Fig). It is graduated in millilitres from 0 to 50. Each division is further sub-divided into ten equal parts. Therefore, each sub-division reads 0.1 ml.

Before a burette is filled with the solution, it is thoroughly washed, so that no greasy matter is sticking inside or outside the burette. No drops should adhere to the inner wall of a clean burette. Take a small volume of solution to be taken in it, close the upper mouth of the burette with the thumb and hold in horizontal position as shown in Fig. Rotate the burette so as to wet the inner walls of the burette. Reject this solution through the stop-cock. This process is known as rinsing. Then the burette is filled with the help of a funnel inserted in the top Fig. The funnel must then be taken out after filling the burette. The solution in the burette is called titrant.
Care must be taken that no air bubbles remain in the narrow bottom tip of the burette. To remove this air, the stop-cock is opened and the liquid is allowed to run out rapidly into the beaker or flask.
Burette reading forms the most important aspect of the experiment, therefore, burette f should be read very carefully, after removing parallax.

To read the burette, hold behind the level of the liquid and in contact with the burette, a piece of white paper to illuminate the surface of the liquid. This paper, called antiparallax card, eliminates errors in reading due to parallax. In order to prepare an anti-parallax card take a rectangular piece of paper and fold it half. Give two cuts as shown in Fig. Open the fold and mount it on the burette.

It is to be remembered that in case of colourless solutions lower meniscus is read, while in case of coloured solutions, level is read from the upper meniscus. This is due to the reason that in case of coloured solutions lower meniscus is not visible clearly.
Take reading of the burette placing your eye exactly in front of meniscus (Fig.) of the solution.

Precautions

1. See that stop-cock does not leak.
2. Remove the funnel immediately after filling the burette.
3. Do not allow any air bubble to remain inside the burette.
4. Always use antiparallax card and place the eye exactly in the level of meniscus.
5. Let no drops of solution be hanging at the tip of the burette at the end point.

Pipette
This apparatus is used for accurate measurements of definite volume of solution. It consists of a long narrow tube with cylindrical bulb in the middle and a jet at its lower end.

On the upper part of the stem, there is an etched circular mark. On the bulb is marked the volume which the pipette can deliver when filled up to the circular mark [Fig].
Before a pipette is filled with the solution, it is washed and thoroughly rinsed with the solution to be measured with it. The upper part of pipette is then held by the thumb and middle finger of the right hand, the lower end is dipped into the liquid and the solution is sucked into the pipette until the liquid level is about 2 cm above the mark. The open end of pipette is then closed with index finger. The liquid is allowed to run slowly until the lower edge of meniscus just touches the mark. The solution is then allowed to run freely out of the pipette in the titration flask.

Precautions

1. Never close the pipette with the’ thumb.
2. Keep the lower end always dipping in the liquid while sucking the liquid.
3. Never pipette out hot solutions or corrosive solutions.
4. Do not blow out the last drop of the solution from the jet end.

Chemical Balance
The balance is the principal instrument used in quantitative analysis. One of the most important requirements in quantitative analysis is a sufficiently high degree of precision. The analytical balance used in quantitative analysis can be used for weighing objects not heavier than 100-200 g to a precision of 0.0002 g, i.e., 0.2 mg. The most usual design of a balance of this type is shown in Fig.
The most important part, the beam, has three knife edges made of agate or very hard steel [Fig]. The central knife edge rests on a special very smooth agate plate on the top of the balance column. The balance pans are suspended from the terminal knife edges by means of stirrups [Fig].
A pointer is fixed to the centre of the beam; as the balance swings the lower end of the pointer moves the scale, at the bottom of the column. All the three knife edges must be strictly parallel and in the same plane for correct operation of the balance. The knife edges and plates gradually wear out and the balance becomes less precise. To reduce wear and tear as much as possible the balance is provided with an arrest device whereby the balance beam can be raised and the balance “arrested”. The balance must be arrested when not in use.
The balance is enclosed in a glass case which protects it from dust, air movements, the operator’s breath, etc.
The base of the balance rests on screws 1 (Fig), whereby the knife edges and agate plates on which they rest are brought into horizontal position by means of a plumb bob attached to the balance column (at the back).
The balance pans are made of some light metal which is nickel-plated or coated with gold or platinum to prevent oxidation. Obviously substances should never be put directly on the balance pans because this spoils the balance. Therefore, substances are weighed either in special weighing bottles with ground-glass lids [Fig] or on watch glasses [Fig] or in crucibles, test tubes, etc.



For the results of weighing to be accurate the weighed object must be of the same temperature as the balance. If a hotter (or colder) object is placed on a balance pan, this has the effect of lengthening (or shortening) the corresponding arm of the beam resulting in incorrect readings.
The weights used with analytical balance are contained in a special box as shown in Fig.

Box also contains a pair of forceps for lifting the weights and putting them on and off the balance pans. The forceps should be ivory-tipped. The weights must never be touched by hand.
The weights are coated with gold or platinum to prevent corrosion and consequent changes of weight. The small weights (fractions of a gram) are made of some metal which is not corroded in air, e.g., aluminium or platinum.
The weights are arranged in the box in definite order. There are two usual systems corresponding to the numbers 5 :2 : 2 :1 or 5:2:1:1. In accordance with the first system, the box would contain weights of 50, 20, 20, 10, 5, 2, 2, 1 g and in accordance with the second, weights of 50, 20,’ 10,10,10, 5, 2,1,1,1 g. Fractions of a gram follow the same systems and are •made of different shapes so that small weights are easier to distinguish. For example, fractional weights of 0.5 and 0.05 g are made in shape of regular hexagon, weights 0.2 and 0.02 g are squares and weights 0.1 and 0.01 g are triangles. Each fractional weight has an edge bent at right angle by which it is lifted with the forceps.
By means of the weights an object can be weighed to an accuracy of 0.01 g. Thousandth and ten-thousandth fractions of a gram are weighed by means of the so called rider. The rider, as shown in Fig. 11.12, is a thin bent wire (usually of aluminium) weighing 0.01 g or 0.005 g, it is attached with the aid of the forceps by its loop on hooks. This hook is fixed to the horizontal rod 11 with the knob 6 outside the balance case. This rod is rotated or moved to place the rider at any desired point on the beam. The beam has a scale, the graduations of which differ in different balances. If the rider is moved from the zero division to the fifth (i.e., exactly over the central knife edge), this is equivalent to removal of 0.005 g from the left-hand pan or a similar increase of the load on the right-hand pan.

Setting the Balance
Before the substance can be weighed in a balance it has to be first set in proper order. The following steps are followed for setting the balance:
1. Clean the pans of the balance with a hair-brush or a clean handkerchief.
2. Level the balance by adjusting the levelling screws. See that the pointer rests at zero. Close the front door of the balance.

3. Now rotate the key arrest knob to raise the beam and see that the pointer swings or oscillates equal divisions on both sides of the zero mark as shown in Fig. If it does not oscillate equally on both the sides arrest the beam and move the adjusting screws (4) till on rotating the arrest knob, the pointer oscillates equally on both sides of the zero mark. Again arrest the beam.

Weighing the Substance

1. Take a clean and dry watch glass or weighing bottle and place it carefully on the left hand pan of the balance.
2. Pick out an appropriate gram weight from the weight box with the help of forceps and place it on the right hand pan. If the gram weight is heavier as compared to the weight of the watch glass, remove it and try lower weight. The gram weight should be slightly less than the weight of the watch glass (less than 1 gram).
3. After placing the correct gram weight start placing fractional weights.
4. Use rider for weights lighter than 10 mg.
5. Record the correct weight of empty watch glass.
6. Now add weights (gram weights and fractional weights), equal to the amount of the substance to be taken, in the right hand pan.
7. Now add required quantity of the substance to be weighed on the watch glass.
8. Take out the watch glass along with the substance.
9. Clean the balance and close it.

Precautions While Handling the Analytical Balance
In weighing it must be remembered that the analytical balance is a precise physical instrument which must be handled with great care.
To avoid damage to the balance and to ensure accurate weighing the following rules must be strictly observed:

1. Check the state of the balance before each weighing. Remove dust from the pans with a soft brush and find the zero point of the balance.
2. The unarrested balance must not be touched. The balance must be arrested before the object and weights are put on the pans or taken off them. The balance must be arrested before the rider is moved along the beam. The knob must be turned slowly and carefully.
3. Do not move the balance from its place.
4. Never overload the balance above the permitted load (usually 100 g) as this causes damage.
5. Do not place wet or dirty objects on the balance. Do not spill anything inside the balance case.
6. Do not put the object to be weighed directly on the balance pan. Do not use pieces of paper; put the substance on a watch glass, or in a weighing bottle, crucible, test tube, etc.
7. Hygroscopic substances and liquids (especially if they give off corrossive vapours) must be weighed in closed weighing bottles.
8. Do not weigh hot (or very cold) objects. The object to be weighed must reach the temperature of the balance. It must, therefore, be left for at least 20 minutes in a dessicator near the balance.
9. Always use only the side doors of the balance case when weighing. The front door, must be kept shut all the time.
10. Do not touch the balance, weights or rider with the fingers. The weights must be handled by special forceps.
11. Do not muddle the weights. Each weight must be put in its proper place in the box.
12. Remain in the balance room only while weighing.

Some Important Terms
1. Standard Solution
A solution whose concentration is known, is called a standard solution. Concentration of a solution is generally expressed in terms of normality or molarity.
2. Normality
Normality of a solution is defined as the number of gram-equivalents of solute per litre of solution. It is denoted by N. Mathematically, it may be expressed as:

.•. Number of gram equivalents of solute = Normality x Volume of solution (in litres).
A solution containing one gram-equivalent of solute per litre of solution is called normal solution.
3. Molarity
Molarity of a solution may be defined as the number of gram moles of solute per litre of the solution. It is denoted by M. Mathematically, it may be expressed as:

4. End Point
It is the point where the reaction between the two solutions is just complete.
5. Indicator
A substance which indicates the attainment of end point. Indicator undergoes a change in colour at the end point.

Equivalent Masses of Oxidizing and Reducing Agents
According to electronic concept, oxidation is the process which results in the loss of one or more electrons by atoms or ions and reduction is the process which results in the gain of one or more electrons by atoms or ions. The oxidising agent is the substance which gains one or more electrons and gets reduced. The reducing agent is the substance which loses one or more electrons and gets oxidised.
The equivalent mass of an oxidising agent is equal to its molecular mass (or formula mass) divided by the number of the electrons gained by one molecule or ion of the substance in the reaction.

Preparing a Standard Solution
A standard solution is prepared by dissolving a definite weight of substance (a primary standard), in a definite volume. A substance is classified as a primary standard if it has following characteristics:
1. It is easily available in state of high purity.
2. It is neither hygroscopic nor deliquescent.
3. It shows high solubility in water.
4. It does not dissociate or decompose during storage.
5. It should react instantaneously with another substance in stoichiometric proportion. Substances whose standard solutions cannot be prepared directly are called secondary
standard substances. These include those substances which are not available in the pure form, for example, potassium permanganate or those which are hygroscopic like NaOH, KOH, etc. The solutions of secondary standards are standardized by titrating against solution of some primary standard.
For preparing a standard solution, student must remember that he is working on precise experiments, where the slightest inaccuracy may distort the analytical results which may have taken a great deal of work and time to obtain. It is, therefore, specially important to follow strictly the usual rules concerning orderly and clean work. The apparatus required for a given determination must be procured before hand and washed thoroughly and the weighing must be done accurately.

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Prepare 250 ml of M/10 Solution of Oxalic Acid From Crystalline Oxalic Acid

Theory

Apparatus
Watch glass, analytical balance, weight box, fractional weight box, 250 ml beaker, glass rod, 250 ml measuring flask and wash bottle.

Chemical Required
Oxalic acid crystals and distilled water.

Procedure

1. Take a watch glass, wash it with distilled water and then dry it.
2. Weigh the clean and dried watch glass accurately and record its weight in the note book.
   
3. Weigh 3.150 g oxalic acid on the watch glass accurately and record this weight in the note-book.
4. Transfer gently and carefully the oxalic acid from the watch glass into a clean 250 ml beaker. Wash the watch glass with distilled water with the help of a wash bottle to transfer the particles sticking to it into the beaker [Fig].
   The volume of distilled water for this purpose should not be more than 50 ml.
5. Dissolve oxalic acid crystals in the beaker by gentle stirring with a clean glass rod.
6. When the oxalic acid in the beaker is completely dissolved, transfer carefully the entire solution from the beaker into a 250 ml measuring flask (volumetric flask) with the help of a funnel [Fig].
   
7. Wash the beaker with distilled water. Transfer the washings into the measuring flask [Fig].
8. Finally wash the funnel well with distilled water with the help of a wash bottle to transfer the solution sticking to the funnel into the measuring flask [Fig].
9. Add enough distilled water to the measuring flask carefully, up to just below the etched mark on it, with the help of a wash bottle.
10. Add the last few drops of distilled water with a pipette until the lower level of the meniscus just touches the mark on the measuring flask [Fig].
11. Stopper the measuring flask and shake gently to make the solution uniform through-out. Label it as oxalic acid solution.
    

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

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Prepare 250 ml of a N/10 Solution of Oxalic Acid from Crystalline Oxalic Acid

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Theory

Apparatus
Watch glass, analytical balance, weight box, fractional weight box, 250 ml beaker, glass rod, 250 ml measuring flask and wash bottle.

Chemical Required
Oxalic acid crystals and distilled water.

Procedure

1. Take a watch glass, wash it with distilled water and then dry it.
2. Weigh the clean and dried watch glass accurately and record its weight in the note book.
   
3. Weigh 3.150 g oxalic acid on the watch glass accurately and record this weight in the note-book.
4. Transfer gently and carefully the oxalic acid from the watch glass into a clean 250 ml beaker. Wash the watch glass with distilled water with the help of a wash bottle to transfer the particles sticking to it into the beaker [Fig].
   The volume of distilled water for this purpose should not be more than 50 ml.
5. Dissolve oxalic acid crystals in the beaker by gentle stirring with a clean glass rod.
6. When the oxalic acid in the beaker is completely dissolved, transfer carefully the entire solution from the beaker into a 250 ml measuring flask (volumetric flask) with the help of a funnel [Fig].
   
7. Wash the beaker with distilled water. Transfer the washings into the measuring flask [Fig].
8. Finally wash the funnel well with distilled water with the help of a wash bottle to transfer the solution sticking to the funnel into the measuring flask [Fig].
9. Add enough distilled water to the measuring flask carefully, up to just below the etched mark on it, with the help of a wash bottle.
10. Add the last few drops of distilled water with a pipette until the lower level of the meniscus just touches the mark on the measuring flask [Fig].
11. Stopper the measuring flask and shake gently to make the solution uniform through-out. Label it as oxalic acid solution.
    

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Qualitative Analysis Viva Questions with Answers

Viva Questions  with Answers 

Question.1.What is qualitative analysis?
Answer. The type of analysis that deals with the methods which are used to determine the constituents of a compound.

Question.2.What is a radical?
Answer. A radical may be defined as an atom or group of atoms which carries charge and behaves as a single unit in chemical reactions.

Question.3. What are acidic and basic radicals?
Answer. Radicals carrying positive charge are called basic radicals and those carrying negative charge are called acidic radicals.

Question.4. What type of bond is present in an inorganic salt?
Answer. Electrovalent bond.

Question.5. Why do inorganic salt ionise when dissolved in water?
Answer.Due to the high dielectric constant of water, the force of attraction holding the two ions in a salt decreases. Thus, the two ions separate. The ions are ‘further stabilized by solvation.

Question.6. Name the coloured basic radicals.
Answer.  Cu²⁺, Fe²⁺, Fe³⁺, Cr³⁺, Ni²⁺, Co²⁺ and Mn²⁺.

Question.7. What is the colour of iron salts?
Answer. Ferrous salts are usually light green while ferric salts are generally brown.

Question.8. Name any iron salt which is light green.
Answer. Ferrous sulphate.

Question.9. What is the colour of nickel salts?
Answer. Bluish green or green.

Question.10. What is the colour of manganese salts?
Answer. Light pink or flesh colour.

Question.11. Name the basic radicals which are absent, if the given salt is white.
Answer.  Cu²⁺, Fe²⁺, Fe³⁺, Cr³⁺, Ni²⁺, Co²⁺ and Mn²⁺.

Question.12. Why a salt containing lead turn black in colour, when placed for a long time in laboratory ?
Answer. Due to the formation of black lead-sulphide by the action of H₂S in atmosphere.

Question.13. Name the salts which produce crackling sound when heated.
Answer. Lead nitrate, barium nitrate, potassium bromide, sodium chloride.

Question.14. What is sublimation?
Answer. It is the process by which a salt directly changes into gaseous phase without melting, when heated. On cooling vapours condense back to the solid state.

Question.15. Tell the importance of preliminary tests in qualitative analysis.
Answer. Sometimes, preliminary tests give authentic information about an ion in the salt. For example, golden yellow colour in flame test shows the presence of sodium. In a charcoal cavity test, brown residue shows the presence of cadmium in a salt and so on.

Question.16. How is dry heating test performed and what information you get if the residue changes to yellow when hot?
Answer. In dry heating test, the salt is heated in a dry test tube. Yellow residue when hot shows the presence of zinc.

Question.17. What is the expected information when copper sulphate is heated in a dry test tube?
Answer. A white residue is formed and water condenses on the colder walls of the test tube.

Question.18. Name the radical which produces CO₂ on heating.
Answer. Carbonate.

Question.19. What is the colour of residue when zinc salt is heated?
Answer. A residue yellow when hot and white when cold is formed.

Question.20. What is the colour of residue when cadmium salt is heated?
Answer. A residue brown when hot, brown when cold.

Question.21. If the residue in dry heating test is white, name’the radicals which are absent.
Answer.  Cu²⁺, Fe²⁺, Ni²⁺, Mn²⁺ Co²⁺, Cr³⁺, Cd²⁺, Zn²⁺ and Pb²⁺.

Question.22. How is charcoal cavity test performed? Describe the chemistry for the formation of  incrustation as well as metallic bead.
Answer. The salt is mixed with the double the quantity of sodium carbonate and the mixture is heated in the charocal cavity in luminous flame (reducing flame).

Question.23. Which flame is used in charcoal cavity test? How is it obtained?
Answer. A reducing flame is used in charcoal cavity test. It is obtained by closing the air holes of the Bunsen burner.

Question.24. Why should we avoid excess of cobalt nitrate in cobalt nitrate test?
Answer. Excess of cobalt nitrate is avoided because it forms black cobalt oxide in the oxidising flame. This colour masks the other colours which might be produced during the test.

Question.25. In the flame test, sodium imparts yellow colour to the flame while magnesium does not impart any colour. Why?
Answer. In case of magnesium, the energy of flame is unable to promote the electron to higher energy level, hence, no colour is imparted to the flame.

Question.26. What is the chemistry of flame test.
Answer. In flame test, the valence electron of the atom gets excited and jumps to the higher level. When the electron jumps back to the ground state, the radiation is emitted whose frequency falls in the visible region.

Question.27. What is the function of blue glass in flame test?
Answer. The blue glass can absorb a part or whole of the coloured light in certain cases. Therefore, the flame appears to be of different colour when viewed through blue glass. This helps in identification of some basic radicals.

Question.28. Why do we use cone. HCl in preparing a paste of the salt for flame test?
Answer. In order to convert metal salts into metal chlorides which are more volatile than other salts.

Question.29. Why can’t we use glass rod instead of platinum wire for performing flame test?
Answer. This is because glass contains sodium silicate which imparts its own golden yellow colour to the flame.

Question.30. Why is platinum metal preferred to other metals for flame test?
Answer. Because platinum does not react with acids and does not itself impart any characteristic colour to the flame.

Question.31. Why do barium salts not impart colour to the flame immediately?
Answer.Because barium chloride is less volatile, it imparts colour to the flame after some time.

Question.32. Why should we avoid the use of platinum wire for testing lead salts?
Answer. Because lead combines with platinum and the wire gets corroded.

Question.33. Why should only a particle or two of the given salt should be touched with the bead in borax bead test?
Answer. If salt is used in excess an opaque bead is formed.

Question.34. Why borax bead test is not applicable in case of white salts?
Answer. White salts do not form coloured meta-borates. .

Question.35. What is Nessler’s Reagent? 
Answer. It is a solution of mercuric iodide in potassium iodide. Its formula is K₂[HgI₄].

Question.36. Name the acid radicals detected with dil. H₂SO₄?
Answer.

CO₃^2-,S^2-,SO₃^2-,NO^2-      

Question.37. Why dil. H₂SO₄ is preferred while testing acid radicals over dil. HCl ?
Answer. When the salt is treated with HCl, during reaction HCl gas is also given out along with the gas evolved by the salt. So the actual gas cannot be identified whereas with H₂SO₄, no such problem arises.

Question.38. Name the acid radicals detected by cone. H₂SO₄.
Answer.

Question.39. Name the radicals which are tested with the help of water extract.
Answer. NO₃^–, NO₂^– and CH3COO^–.

Question.40. Name the radicals which are confirmed with the help of sodium carbonate extract.
Answer.S^2-, Cl^–, Br^–,I^–, PO₄^3-, SO₄^2-, SO₃^2-, C₂O₄^2-.

Question.41. How is sodium carbonate extract prepared?
Answer. The salt is mixed with double the amount of solid Na₂CO₃ and about 20 ml of distilled water. It is then boiled till it is reduced to one-third, and then filtered. The filtrate is sodium carbonate extract or (S.E.).

Question.42. What is water extract?
Answer. The given salt or mixture is shaken well with distilled water and the solution is filtered. The filtrate is water extract.

Question.43. CO₂ and SO₂ both turn lime water milky. How will you distinguish between them?
Answer. By passing through acidified K₂Cr₂O₇ solution. SO₂ turns green while CO₂ has no effect.

Question.44. NO₂ and Br₂ both are brown in colour. How will you distinguish between them?
Answer. By passing through FeSO₄ solution. NO, turns FeSO₄ soln. black while Br₂ has no effect.

Question.45. How will you test the presence of carbonate?
Answer. Treat a small quantity of the mixture with dil. H₂SO₄. CO₂ gas is evolved. When the gas is passed through lime water, it is turned milky.

Question.46. What is lime water?
Answer. A solution of Ca(OH)₂ in water is called lime water.

Question.47. What will happen if excess of CO₂ is passed through lime water?
Answer. The white ppt. of CaCO₃ changes into soluble calcium bicarbonate and the milkiness, therefore, disappears.

Question.48. How do you test for sulphide?
Answer. Warm the salt with dil. H₂SO₄. H,S gas is evolved. It turns a paper dipped in lead acetate black.

Question.49. Is there any gas other than CO₂, which turns lime water milky?
Answer. Yes, it is SO₂ gas.

Question.50. All nitrates on heating with cone. H₂SO₄ in presence of paper pallet evolve NO₂ gas. What is the function of paper pallet?
Answer.

Question.51. How will you test whether the given solution in a bottle is lime water?
Answer. Take 2 ml of the solution in a test tube and blow into it by means of a glass tubing. Milkiness indicates that the solution is lime water.

Question.52. How is ring test performed for nitrates?
Answer. To the salt solution, freshly prepared ferrous sulphate solution is added and then sulphuric acid (cone.) is added along the walls of the tube. A dark brown ring is formed at the junction of the two solutions.

Question.53. Why the hot reaction mixture in case of cone. H₂SO₄  ( test is not thrown into the sink?
Answer. In order to avoid spurting, due to which H₂SO₄ may fly and spoil clothes and may result into serious injuries.

Question.54. What is Tollen’s reagent?
Answer. Ammonical AgNO₃ solution is called Tollen’s reagent.

Question.55. Give formula of Diphenylamine reagent.
Answer. (C₆H₅)₂ NH.

Question.56. Why a dark brown ring is formed at the junction of two layers in ring test for nitrates?
Answer. H₂SO₄ being heavier forms the lower layer and reacts only with a small amount of nitrate and FeSO₄ at its surface, therefore, a brown ring appears only at the junction of the two layers.

Question.57. Why acetic acid is added before adding lead acetate solution?
Answer. In order to prevent the hydrolysis of lead acetate which would yield white precipitate of lead hydroxide.

Question.58. What is the formula of Sodium nitroprusside?
Answer. Na₂[Fe(CN)₅ NO].

Question.59. What is chromyl chloride test?
Answer. Heat a small amount of the mixture with cone.H₂SO₄ and solid K₂Cr₂O₇ in a dry test tube. Deep brownish red vapours of chromyl chloride are formed. Pass these vapours in water. A yellow sol. of H₂CrO₄ is formed. Add to this solution NaOH, acetic acid and lead acetate, a yellow ppt. confirms chloride in the mixture.

Question.60. What is the chemistry of carbon disulphide test for a bromide or iodide?
Answer. To a part of the soda extract add dil. HCl. Now to this add small amount of CS₂and excess of chlorine water and shake the solution well. Chlorine displaces bromine or iodine from the bromide or iodide, which dissolves in carbon disulphide to produce orange or violet colouration.

Question.61. Why do bromides and iodides not respond to chromyl chloride test?
Answer. Because chromyl bromide (CrO₂Br₂) and chromyl iodide (CrO₂ I₂) compounds are not formed, instead of these bromine and iodine are evolved.

Question.62. Describe the chemistry of match stick test.
Answer.In match stick test, the sulphate is reduced to sulphide by carbon of match stick which then gives violet colour with sodium nitroprusside solution.

Question.63. Why does iodine give a blue colour with starch solution?
Answer. The blue colour is due to the physical adsorption of iodine upon starch.

Question.64. Why O.S. is not prepared in cone. HNO₃?
Answer. HNO₃ is an oxidising agent which on decomposition gives oxygen. A yellow ppt. of sulphur is obtained in presence of HNO₃ when H₂S is passed.

Question.65. Name group reagents for different groups.
Answer.

Question.66. Why is it essential to add dil. HCl before proceeding to the test for the basic radicals of group II?
Answer. In the precipitation of group II cations as their sulphides. H₂S is used in the presence of dil. HCl. H₂S is itself a weak acid and dissociates as follows:

Hydrochloric acid being a strong acid is largely ionised to  H⁺. Thus hydrogen ion concentration is increased and consequently the concentration of sulphide ions produced by the ionisation of H₂S is sufficiently decreased due to common ion effect. As a result of which the sulphide ion concentration is sufficient only to exceed the solubility product of the sulphides of group II cations.
Since the solubility products (K_sp) for the sulphides of groups III and IV cations are very high, those cations are not precipitated out under the above conditions.

Question.67. Why is the O.S. boiled with cone. HNO, in III group?
Answer. In the presence of NH₄Cl, Fe(OH)2 is not precipitated because of its high solubility product. For this reason F^e++ salts are oxidised to F^e++ salts by boiling with cone. HNO₃ before adding NH₄Cl and NH₄OH; otherwise F^e++ would not be ppted in III group.

Question.68.Why is NH₄Cl added along with NH₄OH in III group?
Answer. It is done in order to decrease the concentration of OH^– ions by suppressing the ionisation of NH₄OH by common ion effect. If NH₄OH alone is used in that case, the concentration of OH^– is enough to ppt. the hydroxide of IV, V and VI groups.

Question.69.What is blue lake?
Answer. It is blue particles (blue litmus adsorbed on white ppt. of Al(OH)₃ floating in colourless solution.

Question.70.H₂S gas is passed in presence of NH₄OH in group IV. Explain why?
Answer. When H₂S gas is passed in alkaline medium or NH₄OH, the OH⁺ ions from the dissociation of H₂S gas combine with hydroxyl ions (OH^–) from the dissociation of NH₄OH to form nearly unionised H₂O.

The removal of H⁺ ions from the solution causes more of H₂S to dissociate, thereby increasing the concentration of S^2- ions to such an extent that the ionic product of IV group metal sulphides exceeds their solubility product. Hence they are precipitated.

Question.71.Presence of NH₄Cl is quite essential before the addition of (NH4)₂ CO₃ in group V. Explain why? 
Answer. Ammonium chloride suppresses the ionisation of NH₄OH and (NH4)₂CO₃ due to common ion effect which results in the decrease in the concentration of OH^– and CO₃^2- ions. So the ionic product does not exceed the solubility product of Mg(OH)₂ or MgCO, and thereby they are not precipitated in V group.

Question.72.Na,CO, cannot be used in place of (NH4)₂ CO, in the group V. Explain why?
Answer. Na₂CO₃ is highly ionised electrolyte, which produces very high cone, of CO₃^2- ions. As a result ionic product of MgCO₃ may increase its Ksp and “it may get precipitated along with the radicals of V group.

Question.73.An aqueous solution of HCl has cone. KP M. What is the approximate value of pH of this solution?
Answer. Slightly less than 7.

Question.74.How will your prepare chlorine water?
Answer. Take cone. HCl in a test tube and add KMnO₄ soln. dropwise till the pink colour starts persisting. Now add a few drops of cone. HCl so that pink colour disappears. The colourless solution thus obtained is chlorine water.

Question.75.Can we use ammonium sulphate in place of ammonium chloride in group III precipitation?
Answer. No, ammonium sulphate cannot be used because it would cause precipitation of group V radicals as their sulphates in group III.

Question.76.Name a cation which is not obtained from a metal?
Answer. Ammonium ion (NH_4⁺).

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Preparation of 250 ml of M/20  Solution of Mohr’s Salt 

Theory

Apparatus
Watch glass, weight box, fractional weight box, 250 ml beaker, glass rod, 250 ml measuring flask and wash bottle.

Chemical Required
Mohr’s salt, cone. H₂SO₄  and distilled water.

Procedure

1. Weigh the clean and dry watch glass and record its weight in the note-book.
2. Weigh accurately 4.9 g of Mohr’s salt crystals on the watch glass and record the weight in the note-book.
3.  Transfer carefully the weighed Mohr’s salt from the watch glass into a clean 250 ml beaker. Add to this beaker about 5 ml of cone, sulphuric acid to check the hydrolysis of ferrous sulphate.
4. Wash the watch glass thoroughly with distilled water to transfer the sticking salt completely into the beaker. Dissolve the salt in the beaker with gentle stirring.
5. Transfer the entire solution carefully into the 250 ml measuring flask through a funnel.
6. Wash the beaker with distilled water and transfer the washing’s into the measuring flask.
7. Add enough distilled water to the measuring flask carefully up to just below the etched mark on its neck with the help of wash bottle.
8. Add the last few drops of distilled water with a pipette until the lower level of the meniscus just touches the mark on the measuring flask.
9. Stopper the measuring flask and shake it gently to make the solution homogeneous (i.e., uniform throughout) and label it as M/20 Mohr’s salt solution.

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Preparation of 250 ml of N/20  Solution of Mohr’s Salt

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Theory

Apparatus
Watch glass, weight box, fractional weight box, 250 ml beaker, glass rod, 250 ml measuring flask and wash bottle.

Chemical Required
Mohr’s salt, cone. H₂SO₄  and distilled water.

Procedure

1. Weigh the clean and dry watch glass and record its weight in the note-book.
2. Weigh accurately 4.9 g of Mohr’s salt crystals on the watch glass and record the weight in the note-book.
3.  Transfer carefully the weighed Mohr’s salt from the watch glass into a clean 250 ml beaker. Add to this beaker about 5 ml of cone, sulphuric acid to check the hydrolysis of ferrous sulphate.
4. Wash the watch glass thoroughly with distilled water to transfer the sticking salt completely into the beaker. Dissolve the salt in the beaker with gentle stirring.
5. Transfer the entire solution carefully into the 250 ml measuring flask through a funnel.
6. Wash the beaker with distilled water and transfer the washing’s into the measuring flask.
7. Add enough distilled water to the measuring flask carefully up to just below the etched mark on its neck with the help of wash bottle.
8. Add the last few drops of distilled water with a pipette until the lower level of the meniscus just touches the mark on the measuring flask.
9. Stopper the measuring flask and shake it gently to make the solution homogeneous (i.e., uniform throughout) and label it as M/20 Mohr’s salt solution.

Law of Equivalents
According to this law, the number of equivalents of the substance to be titrated (titre) is equal to the number of equivalents of the titrant used. Derivation of the normality equation. Consider an acid alkali neutralization reaction. Let Vx cm3 of aq add solution of normality requires V2 cm3 of base of N2 normality for complete neutralization.

Process of Titration
The process of titration is employed to find out the volume of one solution required to react completely with a certain known volume of solution of some other substance. This is the most important step in volumetric analysis. The process of titration is carried out as under:

1. Support a cleaned and rinsed burette with a burette clamp. Close the stopcock and, with the help of a funnel, fill the burette to just above the zero mark. Open the stop cock briefly to remove any air bubbles in the tip.
2. Place a glazed white tile below the burette and place the titration flask on the glazed tile below the burette nozzle. Adjust the height of the burette so that the nozzle tip just enters the mouth of the titration flask.
3. Note initial reading of the burette and run out the solution from the burette (one ml at a time}. During titration, operate the stopcock with your left hand and constantly swirl the flask with the right hand. (See Fig).
   
4. Continue running more of the solution from the burette into the titration flask. The solution should fall directly into the solution of titration flask. It should not fall on the walls of flask.
5. Stop addition of the solution when the end point is reached and take final reading of the burette. The difference between the final and initial readings gives rough volume of the solution used for completion of the reaction.
6. The solution from the titration flask is thrown in the sink and the titration flask is washed thoroughly first by keeping it under tap water and then with a little of distilled water. Do not rinse the titration flash.
7. Pour more solution in the burette.
8. Pipette out 20 ml of the solution into the titration flask and add a drop or two of the indicator solution.
9.  Take initial reading of the burette. Run solution from the burette into the titration flask slowly with constant shaking. Continue running of the solution till the volume added is 1 ml less than the rough volume found out in the first titration. Now add solution from the burette drop wise. (Add a drop, close the pinch cock, shake and find out if the end point has been attained).
10. Continue adding solution drop wise from the burette, till by addition of last single drop, the end point is attained.
11. Note down the final reading of the burette. The difference between the final and initial readings of the burette gives the exact volume of the solution required for completion of the reaction.
12. Check the correctness of the end point by adding one drop of solution (taken in the titration flask) with the help of a pipette. Restoration of original colour confirms the correctness of the end point.
13. Perform 5-6 titrations so that at least three concordant readings (difference not more than 0.05 ml) are obtained.
    

Experiments on Potassium Permanganate Titrations -Permanganometric Titrations

Important Instructions for KMn04 Titrations

1. KMnO₄ solution is always taken in the burette.
2.  Avoid the use of a burette having a rubber tap as KMnO₄ attacks rubber.
3. Add about an equal volume of dil. H₂SO₄ (- 4N) to the solution to be titrated (say a full test tube for 20 ml of the solution) before adding KMnO₄. HCl cannot be used as it gets oxidized to Cl₂ by KMnO₄. HNO₃ also can not be used as it itself is a strong oxidizing agent.
4. If oxalic acid or some oxalate is to be titrated, add required amount of dil. H₂SO₄ and heat the flask to 60°—70°C on a wire gauge. In order to get some idea about the temperature of the solution touch the flask to the back side of your hand (Fig). When it becomes just unbearable to touch, the required temperature is reached. The purpose of heating is to increase the rate of reaction which otherwise is slow at room temperature.
   
5. In case of ferrous salts, no warming is required.
6.  Read the upper meniscus while taking burette reading with KMnO₄ solution.
7. In case, on addition of KMnO₄ a brown ppt. appears, this shows that either H₂SO₄ has not been added or has been added in insufficient amount. In such a case, throw away the solution and titrate again.
8. Potassium permanganate does not dissolve into water readily. It is dissolved by the process of extraction. Transfer the weighed KMnO₄ into a beaker and add into it 20-30 ml of distilled water and stir. Transfer the solution into a measuring flask. Add more distilled water (20-30 ml) into the beaker and repeat the operation till the permanganate completely dissolves. Add more distilled water into the measuring flask till the lower meniscus of the solution is in line with the mark on the neck. Stopper the measuring flask and shake to get the solution of uniform strength.

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Volumetric Analysis Viva Questions with Answers

Viva Questions with answers

Question.1.What is a standard solution ?
Answer. A solution whose strength is known is called a standard solution.

Question.2.What is a normal solution ?
Answer. A solution containing one gram-equivalent mass of the solute per litre of the solution is called a normal solution.

Question.3.What is the equivalent mass of KMnO₄ when it acts as oxidizing agent in acidic medium ?
Answer. KMnO₄ loses 5 electrons per molecule, when it acts as oxidizing agent in the presence of acids. Therefore, its equivalent mass is one-fifth of its molecular mass.

Question.4.Is sodium hydroxide a primary standard ?
Answer. No.

Question.5.Are ‘molality’ and “molarity’’ same ?
Answer. No, molality of a solution is defined as the number of moles of solute present in 1000 grams of the solution whereas molarity tells us about the number of moles of the solute present per litre of the solution.

Question.6.What would be the normality of 0.10M KMnO₄ ?
Answer. It will be 0.1 x 5 = 0.5 N.

Question.7.What volume of 10M HCl must be diluted with water to get 1L of 1M HCl ?
Answer. 0.1L.

Question.8.What is the basicity of H₂SO₄ ?
Answer. 2.

Question.9.What is the relationship between normality (N), molarity (M), molecular mass and equivalent mass ?
Answer. Normality x Eq. Mass = Molarity x Mol. Mass.

Question.10.Why front door of the balance is closed during weighing ?
Answer. Opening the front door causes vibrations in the pan due to operator’s breath which leads to inaccurate results.

Question.11.What is the maximum weight that can be weighed in a chemical balance ?
Answer.100 grams.

Question.12.What is the weight of a rider ?
Answer. 10 mg.

Question.13.What is the use of a rider ?
Answer. A rider is used for weights less than 10 mg.

Question.14. What is the principle of volumetric analysis?
Answer. In volumetric analysis, the concentration of a solution is determined by allowing a known volume of the solution to react, quantitatively with another solution of known concentration.

Question.15. What is titration ?
Answer. The process of adding one solution from the burette to another in the conical flask in order to complete the chemical reaction involved, is known as titration.

Question.16. What is indicator ?
Answer. Indicator is a chemical substance which changes colour at the end point.

Question.17. What is end point ?
Answer. The stage during titration at which the reaction is just complete is known as the end point of titration.

Question.18. Why a titration flask should not be rinsed ?
Answer. This is because during rinsing-some liquid will remain sticking to the titration flask therefore the pipetted volume taken in the titration flask will increase.

Question.19. What are primary and secondary standard substances?
Answer. A substance is known as primary standard if it is available in high degree of purity, if it is stable and unaffected by air, if it does not gain or lose moisture in air, if it is readily soluble and its solution in water remains as such for long time.
On the other hand, a substance which does not possess the above characteristics is called a secondary standard substance. Primary standards are crystalline oxalic add, anhydrous Na₂CO₃ , Mohr’s salt, etc.

Question.20. Burette and pipette must be rinsed with the solution with which they are filled, why ?
Answer. The burette and pipette are rinsed with the solution with which they are filled in order to remove any water sticking to their sides, which otherwise would decrease the cone, of the solutions to be taken in them.

Question.21. It is customary to read lower meniscus in case of colourless and transparent solutions and upper meniscus in case of highly coloured solutions, why ?
Answer. Because it is easy to read the lower meniscus in case of colourless solutions, while the upper meniscus in case of coloured solutions. In case of coloured solutions lower meniscus is not visible clearly.

Question.22. What is a molar solution ?
Answer. A molar solution is a solution, a litre of which contains one gm-mole of the substance. This is symbolised as 1M.

Question.23. Why the last drop of solution must not be blown out of a pipette?
Answer. Since the drops left in the jet end is extra of the volume measured by the pipette.

Question.24. Pipette should never be held from its bulb, why ?
Answer. The body temperature may expand the glass and introduce an error in the measurement volume.

Question.25. What is acidimetry and alkalimetry ?
Answer. It is the branch of volumetric analysis involving chemical reaction between an acid and a base.

Question.26. What is permanganometry ?
Answer. Redox titrations involving KMnO₄ as the oxidising agent are called permanganometric titrations.

Question.27. Which is an oxidising agent and a reducing agent in the reaction between KMnO₄ and FeSO₄?
Answer. KMnO₄ acts as oxidising agent and FeSO₄ acts as reducing agent.

Question.28. What is the indicator used in KMnO₄ titration ?
Answer. No indicator is used because KMnO₄ acts as a self-indicator.

Question.29. Why does KMnO₄ act itself as an indicator ?
Answer. In the presence of dilute sulphuric acid, KMnO₄ reacts with reducing agent (oxalic acid or . ferrous sulphate). When all the reducing agent has been oxidised, the excess of KMnO₄ is not decomposed and imparts pink colour to the solution.

Question.30. What is the end point in KMnO₄ titrations ?
Answer. From colourless to permanent light pink.

Question.31. Why is Mohr’s salt preferred as a primary standard over ferrous sulphate in volumetric
analysis ?
Answer. This is because of the fact that Mohr’s salt is stable and is not readily oxidised by air. Ferrous sulphate gets oxidised to ferric sulphate.

Question.32. Why are a few drops of dilute sulphuric acid added while preparing a standard solution
of Mohr’s salt ?
Answer. Few drops of H₂SO₄ are added to prevent the hydrolysis of ferrous sulphate.

Question.33. Why a burette with rubber pinch cock should not be used in KMnO₄ titrations ?
Answer. Because KMnO₄ attacks rubber.

Question.34. Sometimes a brown ppt. is observed in KMnO₄ titrations. Why ?
Answer. It is due to insufficient quantity of dil. sulphuric acid. Brown coloured ppt. (MnO₂.H₂0) is formed due to the incomplete oxidation of KMnO₄.

Question.35. Why should you heat the oxalic acid solution to about 60-70°C before titrating with KMnO₄ solution ?
Answer. In cold, the reaction is very slow due to the slow formation of Mn²⁺ ions. Oxalic acid is heated to speed up the liberation of Mn²⁺ ions which then autocatalyses the reaction and thus the reaction proceeds rapidly. This also serves the purpose of expelling the carbondioxide evolved during the reaction which otherwise does not allow the reaction to go to completion.

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Determine the equivalent mass and number of molecules of water of crystallisation in a sample of Mohr’s salt FeSO₄ (NH₄)₂ SO₄.nH₂0. Provided N/20 KMnO₄

Determine the equivalent mass and number of molecules of water of crystallisation in a sample of Mohr’s salt FeSO₄ (NH₄)₂ SO₄.nH₂0. Provided N/20 KMnO₄

Chemical equations

Theory
Normality of ferrous ammonium sulphate can be determined by directly titrating it against standard (N/20) KMnO₄ solution.
Eq. mass = Strength/normality
Substituting the value of strength and value of normality as calculated above, the equivalent mass of Mohr’s salt can be calculated. Suppose it comes out to be E.
Since Eq. mass of mohr’s salt is equal to its molecular mass, therefore molecular mass of Mohr’s salt is also equal to E.

Indicator
KMnO₄ is a self-indicator.

Endpoint
Colourless to permanent pink (KMnO₄in burette).

Procedure

1. Rinse and fill the burette with N/20 KMnO₄ . solution.
2. Weigh exactly 4.90 g of Mohr’s salt and dissolve in water to prepare exactly 250 ml of solution using a 250 ml measuring flask. Rinse the pipette with the prepared Mohr’s salt solution and pipette out 20.0 ml of it in a washed titration flask.
3. Add one test tube (~ 20 ml) full of dilute sulphuric acid (~ 4 N) to the solution in titration flask.
4. Note the initial reading of the burette.
5. Add KMnO₄ solution into the titration flask from the burette till a permanent light
   pink colour is imparted to the solution in the titration flask on addition of a last single drop of KMnO₄ solution.
6. Note the final reading of the burette.
7. Repeat the above steps 4-5 times to get three concordant readings.

Observations

Calculations

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Prepare M/20 Solution of Ferrous Ammonium Sulphate (Mohr’s salt). Using this Solution Find out the Molarity and Strength of the Given KMnO₄ Solution

Chemical Equations

Indicator
KMnO₄ is a self-indicator.

End Point
Colourless to permanent pink colour (KMnO₄ in burette).

Procedure

1. Prepare 250 ml of M/20 Mohr’s salt solution by dissolving 4.9 g of Mohr’s salt in water 20 as described in experiment 11.3. Rinse the pipette with the M/20 Mohr’s salt solution and pipette out 20.0 ml of it in a washed titration flask.
2. Rinse and fill the burette with the given KMn04 solution.
3. Add one test-tube (~ 20 ml) full of dilute sulphuric acid (~ 2 M) to the solution in titration flask.
4. Note the initial reading of the burette.
5. Now add KMnO₄ solution from the burette till a permanent light pink colour is imparted to the solution in the titration flask on addition of last single drop of KMnO₄ solution.
6. Note the final reading of the burette.
7. Repeat the above steps 4-5 times to get a set of three concordant readings.

Observations
Weight of watch glass =……. g
Weight of watch glass + Mohr’s salt =…………..g
Weight of Mohr’s salt = 4.90 g
Volume of Mohr’s salt solution prepared = 250 ml
Molarity of Mohr’s salt solution = M/20
Volume of Mohr’s salt solution taken for each titration = 20.0 ml

Calculations
(a) Molarity of the KMnO₄ solution.
From the overall balanced chemical equation, it is clear that 2 moles of KMnO₄ reacts with 10 moles of Mohr’s salt.

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Prepare a Solution of Ferrous Ammonium Sulphate (Mohr’s salt) Containing Exactly 17.0 g of the Salt in one litre. With the help of this Solution, Determine the Molarity and the Concentration of KMnO₄ in the Given Solution

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Chemical Equations

Indicator
KMnO₄ is a self-indicator.

End Point
Colourless to permanent pink colour (KMnO₄ in burette).

Procedure
1. Weigh exactly 4.250 g of Mohr’s salt on a watch glass and dissolve in water to prepare exactly 250 ml of solution with the help of a 250 ml measuring flask. Rinse the pipette with the prepared Mohr’s salt solution and pipette out 20.0 ml of it in a washed titration flask.
2. Rinse and fill the burette with the given KMnO₄ solution.
3. Add one test-tube (~ 20 ml) full of dilute sulphuric acid (~ 2 M) to the solution in titration flask.
4. Note the initial reading of the burette.
5. Now add KMnO₄ solution from the burette till a permanent light pink colour is imparted to the solution in the titration flask on addition of last single drop of KMnO₄ solution.
6. Note the final reading of the burette.
7. Repeat the above steps 4-5 times to get a set of three concordant readings.

Observations
Weight of watch glass =……. g
Weight of watch glass + Mohr’s salt =…………..g
Weight of Mohr’s salt = 4.250 g
Volume of Mohr’s salt solution prepared = 250 ml
Volume of Mohr’s salt solution taken for each titration = 20.0 ml

Calculations
Concentration of Mohr’s salt, ferrous ammonium sulphate, FeSO₄.(NH₄)₂ SO₄.6H₂0 in the prepared solution = 17.0 g/litre.

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Prepare M/20 Ferrous Ammonium Sulphate (Mohr’s salt) Solution. Find out the Percentage Purity of Impure KMnO₄ Sample 2.0 g of  Which have been Dissolved per Litre 

Chemical Equations

Indicator
KMnO₄ is a self-indicator.

End Point
Colourless to permanent pink colour (KMnO₄ in burette).

Procedure
1. Prepare 250 ml of M/20 Mohr’s salt solution by dissolving 4.9 g of Mohr’s salt in water as described in experiment 11.4. Rinse the pipette with the M/20 Mohr’s salt solution and pipette out 20.0 ml of it in a washed titration flask.
2. Rinse and fill the burette with the given KMnO₄ solution.
3. Add one test-tube (~ 20 ml) full of dilute sulphuric acid (~ 2 M) to the solution in titration flask.
4. Note the initial reading of the burette.
5. Now add KMnO₄ solution from the burette till a permanent light pink colour is imparted to the solution in the titration flask on addition of last single drop of KMnO₄ solution.
6. Note the final reading of the burette.
7. Repeat the above steps 4-5 times to get a set of three concordant readings.

Observations
Weight of watch glass =……. g
Weight of watch glass + Mohr’s salt =…………..g
Weight of Mohr’s salt = 4.9 g
Volume of Mohr’s salt solution prepared = 250 ml
Solution taken in burette = KMnO₄ solution
Volume of Mohr’s salt solution taken for each titration = 20.0 ml

Calculations
Calculation of molarity of KMnO₄ solution
From the overall balanced chemical equation, it is clear that 2 moles of KMnO₄ react with 10 moles of Mohr’s salt.

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Determine the Equivalent Mass and Number of Molecules of Water of Crystallisation in a Sample of Mohr’s salt, FeSO₄(NH₄)₂ SO₄ . nH₂0. Provided KMnO₄

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Chemical Equations

Theory
Prepare a solution of Mohr’s salt with known strength (g/litre). Molarity of ferrous ammonium sulphate can be determined by directly titrating it against standard (M/100) KMnO₄ solution.
Molecular mass = strength/molarity.
Substituting the value of strength and value of molarity as calculated above, the molecular mass of Mohr’s salt can be calculated. Suppose it comes out to be M.
Theoretical molecular mass of Mohr’s salt, FeSO₄.(NH₄)₂SO₄. nH₂0
= 284 + 18n
∴ 284 + 18n = M
whence, n can be calculated.
In case of Mohr’s salt equivalent mass is equal to its molecular mass. Therefore, Equivalent mass of Mohr’s salt = M.

Indicator
KMnO₄ is a self-indicator.

End Point
Colourless to permanent pink colour (KMnO₄ in burette).

Procedure

1. Weigh exactly 4.90 g of Mohr’s salt and dissolve in water to prepare exactly 250 ml of solution, using a 250 ml measuring flask. Rinse the pipette with the prepared Mohr’s
   salt solution and pipette out 20.0 ml of it in a washed titration flask.
2. Rinse and fill the burette with M/100 KMnO₄ solution.
3. Add one test-tube (~ 20 ml) full of dilute sulphuric acid (~ 4 M) to the solution in titration flask.
4. Note the initial reading of the burette.
5. Now add KMnO₄ solution from the burette till a permanent light pink colour is imparted to the solution in the titration flask on addition of last single drop of KMnO₄ solution.
6. Note the final reading of the burette.
7. Repeat the above steps 4-5 times to get a set of three concordant readings.

Observations
Weight of watch glass =……. g
Weight of watch glass + Mohr’s salt =…………..g
Weight of Mohr’s salt = 4.90 g
Volume of Mohr’s salt solution prepared = 250 ml
Volume of Mohr’s salt solution taken for each titration = 20.0 ml
Molarity of KMnO₄ solution =M/100

Calculations

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Prepare M/50 Solution of Oxalic Acid. With its help, Determine 50 the Molarity and Strength of the Given Solution of Potassium Permanganate (KMnO₄)

Chemical Equations

Indicator
KMnO₄ is a self-indicator.

End Point
Colourless to permanent pink colour (KMnO₄ in burette).

Procedure

1. Weigh 1.260 g of oxalic acid crystals and dissove them in water to prepare 500 ml of M/50 oxalic acid solution using a 500 ml measuring flask. Rinse the pipette with the M/50 oxalic acid solution and pipette out 20 ml of it in a washed titration flask.
2. Rinse and fill the burette with M/100 KMnO₄ solution.
3. Add one test-tube (~ 20 ml) full of dilute sulphuric acid (~ 2 M) to the solution in titration flask.
4. Note the initial reading of the burette.
5. Now add KMnO₄ solution from the burette till a permanent light pink colour is imparted to the solution in the titration flask on addition of last single drop of KMnO₄ solution.
6. Note the final reading of the burette.
7. Repeat the above steps 4-5 times to get a set of three concordant readings.

Observations
Weight of watch glass =……. g
Weight of watch glass + Mohr’s salt =…………..g
Weight of Mohr’s salt = 1.260 g
Volume of Mohr’s salt solution prepared = 500 ml
Volume of Mohr’s salt solution taken for each titration = 20.0 ml
Molarity of KMnO₄ solution =M/50

Calculations
(a) Molarity of the KMnO₄ solution
From the overall balanced chemical equation it is clear that 2 moles of KMnO₄ react with 5 moles of oxalic acid.

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Find out the Percentage Purity of Impure Sample of Oxalic Acid.You are Supplied M/100 KMnO₄ Solution

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Chemical Equations

Indicator
KMnO₄ is a self-indicator.

End Point
Colourless to permanent pink colour (KMnO₄ in burette).

Procedure

1. Weigh exactly 2.0 g of oxalic acid and dissolve in water to prepare 500 ml of its solution using a 500 ml measuring flask. Rinse the pipette with the oxalic acid solution and pipette out 20 ml of it in a washed titration flask.
2. Rinse and fill the burette with M/100 KMnO₄ solution.
3. Add one test-tube (~ 20 ml) full of dilute sulphuric acid (~ 2 M) to the solution in titration flask.
4. Note the initial reading of the burette.
5. Heat the flask to 60-70°C and add KMnO₄ solution from the burette till a permanent light pink colour just appears in the solution in the titration flask.
6. Note the final reading of the burette.
7. Repeat the above steps 4-5 times to get a set of three concordant readings.

Observations
Weight of watch glass =……. g
Weight of watch glass + Mohr’s salt =…………..g
Weight of Mohr’s salt = 2.00 g
Volume of Mohr’s salt solution prepared = 500 ml
Solution taken in burette = M/100 KMnO₄
Volume of Mohr’s salt solution taken for each titration = 20.0 ml

Calculations
(a) Molarity of the KMnO₄ solution
From the overall balanced chemical equation it is clear that 2 moles of KMnO₄ react with 5 moles of oxalic acid.

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The Given Solution has been Prepared by Dissolving 1.6 g of an Alkali metal Permanganate per litre of Solution. Determine Volumetrically the Atomic mass of the Alkali metal. Prepare M/20 Mohr’s salt Solution for Titration

Chemical Equations
Molecular equations. Let A represent the alkali metal and AMnO₄ represent alkali metal permanganate,

Indicator
AMnO₄ will act as a self-indicator.

End Point
Colourless to permanent pink colour (AMnO₄ in burette).

Procedure

1. Weigh exactly 4.90 g of Mohr’s salt on a watch glass and dissolve in water to prepare exactly 250 ml of its solution using a 250 ml measured flask. Rinse the pipette with
   the M/20 Mohr’s salt solution and pipette out 20.0 ml of it in a washed titration flask.
2. Rinse and fill the burette with the given AMnO₄ solution.
3. Add one test-tube (~ 20 ml) full of dilute sulphuric acid (~ 4M) to the solution in titration flask.
4. Note the initial reading of the burette.
5. Now add AMn04 solution from the burette till a permanent light pink colour is just imparted to the solution in the titration flask.
6. Note the final reading of the burette.
7. Repeat the above steps 4-5 times to get a set of three concordant readings.

Observations
Weight of watch glass =……. g
Weight of watch glass + Mohr’s salt =…………..g
Weight of Mohr’s salt = 4.90 g
Volume of solution prepared = 250 ml
Molarity of Mohr’s salt solution = M/20
Volume of Mohr’s salt solution taken for each titration = 20.0 ml

Calculations
Prom the balanced chemical equation, it can be seen that 2 moles of AMnO₄ react with 10 moles of Mohr’s salt.

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Determine the percentage composition of a mixture of  sodium oxalate
  and oxalic acid . Provided M/100 KMnO₄ solution.

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Chemical Equations

Indicator
KMnO₄ is a self-indicator.

End Point
Colourless to permanent pink (KMnO₄ in burette).

Procedure
1. Weigh exactly 1.0 g of the given mixture of sodium oxalate and oxalic acid and dissolve in water to prepare exactly 250 ml of solution using a 250 ml measuring flask. Rinse the pipette with the given oxalate solution and pipette out 20.0 ml of it in a washed titration flask.
2. Rinse and fill the burette with the KMnO₄. solution.
3. Add one test-tube (~ 20 ml) full of dilute sulphuric acid (- 2 M) to the solution in titration flask.
4. Note the initial reading of the burette.
5. Heat the solution of titration flask to 60-70°C and run down KMnO₄ solution from the burette till a permanent light pink colour is imparted to the solution in the titration flask on addition of a last single drop of KMnO₄ solution.
6. Note the final reading of the burette.
7. Repeat the above steps 4—5 times to get three concordant reading.

Observations
Weight of watch glass =……. g
Weight of watch glass + Mohr’s salt =…………..g
Weight of mixture = 1.0 g
Volume of solution prepared = 250 ml
Molarity of KMnO₄ solution =1/100
Volume of oxalate solution taken for each titration = 20.0 ml.

Calculations

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You are Provided with a Partially Oxidised Sample of Ferrous Sulphate (FeSO₄.7H₂0) Crystals. Prepare a Solution by Dissolving 14.0 g of these Crystals per litre and Determine the Percentage Oxidation of the given Sample. Given M/100 KMnO₄ Solution

Chemical Equations

Theory
Since the given sample contains partially oxidized ferrous sulphate, it contains both ferrous ions, Fe²⁺ (unoxidised) and ferric ions Fe³⁺ (oxidised). The strength of partially oxidised sample is known. The solution of partially oxidised FeSO₄ of known strength is titrated against standard KMnO₄ solution to determine the molarity and strength of the unoxidised ferrous sulphate. From this the percentage oxidation of the sample can be calculated.

Indicator
KMnO₄ is a self-indicator.

End Point
Colourless to permanent pink (KMnO₄ in burette).

Procedure

1. Weigh exactly 3.50 g of the given sample of ferrous sulphate on a watch glass and dissolve in water to prepare exactly 250 ml of solution using a 250 ml measuring flask. Rinse and fill the pipette with prepared ferrous sulphate solution and pipette out 20.0 ml of it in a washed titration flask.
2. Rinse and fill the burette with the M/100 KMnO₄ solution.
3. Add one test-tube (~ 20 ml) full of dilute sulphuric acid (- 2 M) to the solution in titration flask.
4. Note the initial reading of the burette.
5. Now add KMnO₄ solution from the burette till a permanent light pink colour is imparted to the solution in the titration flask on addition of a last single drop of KMnO₄ solution.
6. Note the final reading of the burette.
7. Repeat the above steps 4—5 times to get three concordant reading.

Observations
Weight of watch glass =……. g
Weight of watch glass + Mohr’s salt =…………..g
Weight of mixture = 3.50 g
Volume of solution prepared = 250 ml
Molarity of KMnO₄ solution =M/100
Volume of oxalate solution taken for each titration = 20.0 ml.

Calculations
Molarity of the standard KMnO₄ solution = M/100

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Calculate the Percentage of Fe²⁺ ions in a Sample of Ferrous Sulphate. Prepare a Solution of the given Sample having Strength Exactly Equal to 14.0 g/litre. Provided M/100 KMnO₄

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Chemical Equations

Theory
Since the given sample contains partially oxidized ferrous sulphate, it contains both ferrous ions, Fe²⁺ (unoxidised) and ferric ions Fe³⁺ (oxidised). The strength of partially oxidised sample is known. The solution of partially oxidised FeSO₄ of known strength is titrated against standard KMnO₄ solution to determine the molarity and strength of the unoxidised ferrous sulphate. From this the percentage oxidation of the sample can be calculated.

Indicator
KMnO₄ is a self-indicator.

End Point
Colourless to permanent pink (KMnO₄ in burette).

Procedure

1. Weigh exactly 3.50 g of the given sample of ferrous sulphate on a watch glass and dissolve in water to prepare exactly 250 ml of solution using a 250 ml measuring flask. Rinse and fill the pipette with prepared ferrous sulphate solution and pipette out 20.0 ml of it in a washed titration flask.
2. Rinse and fill the burette with the M/100 KMnO₄ solution.
3. Add one test-tube (~ 20 ml) full of dilute sulphuric acid (- 2 M) to the solution in titration flask.
4. Note the initial reading of the burette.
5. Now add KMnO₄ solution from the burette till a permanent light pink colour is imparted to the solution in the titration flask on addition of a last single drop of KMnO₄ solution.
6. Note the final reading of the burette.
7. Repeat the above steps 4—5 times to get three concordant reading.

Observations
Weight of watch glass =……. g
Weight of watch glass + Mohr’s salt =…………..g
Weight of mixture = 3.50 g
Volume of solution prepared = 250 ml
Molarity of KMnO₄ solution =M/100
Volume of oxalate solution taken for each titration = 20.0 ml.

Calculations
Volume of M/100 KMnO₄ solution required for the oxidation of 20.0 ml of the prepared ferrous sulphate solution = x ml.
From the equations it is clear that 2 moles of KMnO₄ react with 10 moles of ferrous sulphate.

Exercises

1. Prepare a standard solution of M/50 FeSO₄(NH₄)₂SO₄.6H₂0 (Mohr’s salt). Using this solution find out the molarity of the given solution of KMn04.
2. Prepare M/50 solution of oxalic acid. Using this solution find out the molarity and strength of the given solution of KMnO₄.
3. Prepare a solution of ferrous ammonium sulphate containing exactly 4.9 g of the salt per 250 ml of solution. Using this solution determine the concentration of KMnO₄ in g/litre in the given solution.
4. Prepare M/20 solution of oxalic acid. Using this solution find out percentage purity of impure sample of KMnO₄, 3.5 g of which have been dissolved per litre.
5. Prepare M/50 ferrous ammonium sulphate solution. With its help, find out the percentage purity of impure sample of KMnO₄, 3.6 g of which have been dissolved per litre.
6. Prepare M/20 oxalic acid solution. You are provided two solutions of KMnO₄, A and B. Find out volumetrically which solution, (A or B) is more concentrated. Report the strength of more concentrated solution in g/litre.
7. You are provided with a solution of alkali metal permanganate, AMn04 containing 3.15 g of it per litre of the solution. Prepare M/20 oxalic acid solution and using this solution determine the atomic mass of the alkali metal ‘A’.
8. Determine volumetrically the percentage purity of a given sample of sodium oxalate. Provided M/50 KMnO₄ solution.

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