CBSE Class 12 Chemistry Solutions Questions Answers

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SOLUTIONS (QUESTIONS AND ANSWERS)

CHAPTER 2 - SOLUTIONS

1 MARK QUESTIONS

What is molarity?

Answer: The number of moles of solute dissolved in one litre (1 dm\(^3\)) of solution is known as molarity.

What do you understand by saying that the molality of a solution is 0.2?

Answer: It means that 0.2 mol of the solute is dissolved in 1 kg of the solvent.

Why does the vapour pressure of a liquid remain constant at constant temperature?

Answer: At equilibrium, the rate of evaporation = the rate of condensation. Hence the vapour pressure of a liquid is constant at constant temperature.

Define azeotropes.

Answer: Constant boiling mixtures are called azeotropes.

Which substance is usually added to water in a car radiator to act as an antifreeze?

Answer: Ethylene glycol is usually added to water in the car radiator to act as an antifreeze.

Which liquids form ideal solutions?

Answer: Liquids having similar structure and polarities.

Which property of a solution depends only upon the number of moles of solute dissolved and not on the nature of the solute?

Answer: Colligative properties.

Write one example each of a solid-in-gas and a liquid-in-gas solution.

Answer: Solid in gas: camphor in nitrogen gas. Liquid in gas: chloroform mixed with N₂ gas.

What is molal elevation constant or ebullioscopic constant?

Answer: The elevation in boiling point which takes place when the molality of the solution is unity is known as the ebullioscopic or molal elevation constant.

Define van't Hoff factor.

Answer: The ratio of the observed colligative property to the theoretical (calculated) value is called the van't Hoff factor.

Two liquids A and B boil at 120 \(^\circ\)C and 160 \(^\circ\)C respectively. Which of them has the higher vapour pressure at 70 \(^\circ\)C?

Answer: Liquid A. The lower the boiling point, the more volatile the liquid is, so liquid A will have the higher vapour pressure at 70 \(^\circ\)C.

What happens when blood cells are placed in pure water?

Answer: Water molecules move into the blood cells through the cell walls. So the blood cells swell and may even burst.

What is the effect of temperature on the molality of a solution?

Answer: No effect (molality involves masses, which do not change with temperature).

Write Henry's law.

Answer: The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas at a given temperature.

What is an antifreeze?

Answer: An antifreeze is a substance which is added to water to lower its freezing point, e.g. ethylene glycol.

Why is cutting onions taken from the fridge more comfortable than cutting onions lying at room temperature?

Answer: The vapour pressure is low at lower temperature. So fewer vapours of the tear-producing chemicals are produced.

What will be the van't Hoff factor for a 0.1 M ideal solution?

Answer: The van't Hoff factor = 1, because an ideal solution does not undergo dissociation or association.

What is the optimum concentration of fluoride ions for cleaning of teeth?

Answer: 1.5 ppm. [If it is more than 1.5 ppm it can be poisonous, and if less than 1.5 ppm it is ineffective.]

What role does molecular interaction play in the solution of alcohol and water?

Answer: It leads to positive deviation from ideal behaviour.

Henry's law constants for two gases are 21.5 and 49.5 atm. Which gas is more soluble?

Answer: The gas with \(K_H\) = 21.5 atm is more soluble, because \(K_H\) is inversely proportional to solubility (the lower the \(K_H\), the higher the solubility).

SHORT ANSWER TYPE QUESTIONS OF 2 MARKS

State Raoult's law. Prove that it is a special case of Henry's law.

Answer: Raoult's law states that the partial pressure of a volatile component of a solution is directly proportional to its mole fraction. It is a special case of Henry's law because it becomes the same when \(K_H\) (Henry's constant) is equal to the vapour pressure of the pure solvent (\(p^0\)).

Raoult's law: \(p_1=x_1p_1^0\). Henry's law: \(p=K_Hx\). Both say that the partial pressure is proportional to the mole fraction; they become identical when \(K_H=p_1^0\).

List two conditions that ideal solutions must satisfy.

Answer: (a) \(\Delta_{mix}H\) and \(\Delta_{mix}V\) of ideal solutions should be zero. (b) They should obey Raoult's law over the entire range of concentration.

Explain ideal and non-ideal solutions with respect to intermolecular interactions in a binary solution of A and B.

Answer: For a binary solution of A and B, it would be ideal if the A-B interactions are equal to the A-A and B-B interactions, and it would be non-ideal if they are different. The deviation from ideal behaviour will be positive if the A-B interactions are weaker than the A-A and B-B interactions. The deviation will be negative if the A-B interactions are stronger than the A-A and B-B interactions.

(a) What are minimum boiling and maximum boiling azeotropes? (b) Can azeotropes be separated by fractional distillation?

Answer: (a) Minimum boiling azeotropes are the non-ideal solutions showing positive deviation, while maximum boiling azeotropes are those which show negative deviation. Because of positive deviation their vapour pressures are comparatively higher and so they boil at lower temperatures, while in the case of negative deviation the vapour pressures are lower and so higher temperatures are required for boiling them. (b) No, azeotropes cannot be separated by fractional distillation.

(a) When a non-volatile solute is added to a solvent, there is an increase in the boiling point of the solution. Explain. (b) Define ebullioscopic constant and give its units.

Answer: (a) When a non-volatile solute is added to a volatile solvent, the vapour pressure of the pure solvent decreases, because a part of the surface is occupied by the non-volatile solute, which cannot volatilise. As a result, the vapour pressure of the solution decreases and hence the solution requires a comparatively higher temperature to boil, causing an elevation of boiling point. (b) The ebullioscopic constant is defined as the elevation in boiling point of a solution of a non-volatile solute when its molality is unity. Its unit is K kg mol\(^{-1}\).

How did van't Hoff explain the abnormal molecular masses of electrolytes like KCl in water and non-electrolytes like benzoic acid in benzene?

Answer: The molecular mass of KCl in aqueous medium has been observed to be almost half of the expected value. This is explained by the dissociation of KCl into K⁺ and Cl^- ions: the actual number of particles becomes double, and so do the colligative properties; but since molecular mass is inversely proportional to the colligative property, it becomes almost half. In the case of benzoic acid in benzene, association of molecules takes place as they dimerise; their number becomes almost half and so the molecular mass doubles.

When a pressure higher than the osmotic pressure is applied on the surface of the solution separated from a solvent by a semi-permeable membrane, what will happen?

Answer: Reverse osmosis will take place. The solvent molecules will move from the solution to the solvent side, and the level of the solution will decrease.

The freezing point depression of a 0.1 M sodium chloride solution is nearly twice that of a 0.1 M glucose solution. Explain.

Answer: Sodium chloride, being an ionic compound, ionises in aqueous solution as NaCl \(\rightarrow\) Na⁺ + Cl^-. The concentration of solute particles in this case becomes approximately 0.2 M, which is twice the concentration of the glucose solution. Consequently, the freezing point depression of the NaCl solution is also approximately twice that of the glucose solution.

The depression in freezing point is a colligative property. Explain.

Answer: The freezing point depression depends upon the molal concentration of the solute (the number of solute particles) and does not depend upon the nature of the solute. It is therefore a colligative property.

Equimolar solutions of glucose and common salt are not isotonic. Why?

Answer: Glucose is a non-electrolyte; when added to water it does not break up into ions. Common salt is an electrolyte; when added to water it breaks up to give sodium and chloride ions. The number of particles in the common salt solution is nearly double the number of particles in the glucose solution, so the osmotic pressure of the common salt solution is nearly twice that of the glucose solution.

QUESTIONS FOR 3 MARKS

A 5% solution of sucrose (C₁₂H₂₂O₁₁) is isotonic with a 3% solution of an unknown organic substance. Calculate the molecular mass of the unknown substance.

Answer: 205.2 g mol\(^{-1}\)

Isotonic solutions have the same osmotic pressure, so at the same temperature they have the same molar concentration. Take 100 mL of each solution (the mass of sucrose is 5 g; that of the unknown substance is 3 g; molar mass of sucrose = 342 g mol\(^{-1}\)).

\[\frac{5}{342}=\frac{3}{M}\ \Rightarrow\ M=\frac{3\times342}{5}=205.2\ \text{g mol}^{-1}\]

A solution of barium chloride is prepared by dissolving 3.100 g of it in 250 g of water. The solution boils at 100.083 \(^\circ\)C. Calculate the van't Hoff factor and the molality of this solution. (\(K_b\) for water = 0.52 K kg mol\(^{-1}\), molar mass of BaCl₂ = 208.3 g mol\(^{-1}\))

Answer: Molality = 0.0595 m; van't Hoff factor \(i\) = 2.68

Observed \(\Delta T_b\) = 100.083 \(-\) 100 = 0.083 \(^\circ\)C.

\[\text{Molality}=\frac{w_{solute}\times1000}{M_{solute}\times w_{solvent}}=\frac{3.1\times1000}{208.3\times250}=0.0595\]

\[\text{Calculated }\Delta T_b=K_b\,m=0.52\times0.0595=0.03095\ ^\circ\text{C}\]

\[i=\frac{\text{observed }\Delta T_b}{\text{calculated }\Delta T_b}=\frac{0.083}{0.03095}=2.68\]

Why is a semi-permeable membrane so important in the phenomenon of osmosis? What are isotonic, hypotonic and hypertonic solutions? Does osmosis take place in all three types of solutions?

Answer: A semi-permeable membrane is very important in osmosis because it permits the movement of only the solvent molecules through it. Two solutions having the same osmotic pressure at a given temperature are called isotonic solutions. If a given solution has a lower osmotic pressure it is called hypotonic, and it is hypertonic if its osmotic pressure is higher than that of the solution on the other side of the semi-permeable membrane. Osmosis takes place only between hypotonic and hypertonic solutions (that is, when the two solutions differ in concentration); it does not take place between isotonic solutions.

Which will have more osmotic pressure and why? A solution prepared by dissolving 6 g/L of CH₃COOH, or a solution prepared by dissolving 7.45 g/L of KCl?

Answer: The KCl solution has the higher osmotic pressure.

\[\text{Moles of CH}_3\text{COOH}=\frac{6}{60}=0.1;\qquad \text{moles of KCl}=\frac{7.45}{74.5}=0.1\]

The molar concentration of both solutions is the same. KCl ionises into K⁺ and Cl^-, whereas CH₃COOH does not ionise appreciably. Osmotic pressure is a colligative property, and its value depends on the number of particles. Since KCl produces more ions, the osmotic pressure of the KCl solution will be more than that of the CH₃COOH solution.

What are the bends? If a diver had the "bends", describe how this can be treated.

Answer: Bends is a painful and dangerous condition caused by nitrogen bubbles in the blood. It is treated by keeping the diver in a decompression chamber, where the pressure is lowered gradually.

A scuba diver's cylinder has a mixture of helium, nitrogen and oxygen. As divers go down, the high pressure increases the solubility of these gases in the blood. When they come up, the pressure decreases and nitrogen is released as its solubility decreases. The bubbles of nitrogen gas can block capillaries, causing a condition called bends, which is both painful and dangerous. To avoid the formation of bends in the blood, the divers are kept in decompression chambers where the pressure is lowered gradually, releasing the gas from the blood slowly.

At 300 K, a solution containing 18 g of glucose per litre has an osmotic pressure of 4.98 bar. If the osmotic pressure of another solution is 1.52 bar at the same temperature, what would be its concentration?

Answer: 0.0305 mol L\(^{-1}\)

For solution A: \(\pi V=nRT\), so 4.98 \(\times\) 1 L = (18/180) \(\times\) \(R\) \(\times\) \(T\). For solution B: 1.52 \(\times\) 1 L = \(n\) \(\times\) \(R\) \(\times\) \(T\). Dividing:

\[\frac{1.52}{4.98}=\frac{n}{0.1}\ \Rightarrow\ n=\frac{1.52\times0.1}{4.98}=0.0305\ \text{mol}\]

So the concentration is 0.0305 mol L\(^{-1}\) (as the volume is 1 L).

Calculate the amount of NaCl that must be added to 1000 mL of water so as to reduce its freezing point by two kelvin. For water \(K_f\) = 1.86 K kg mol\(^{-1}\); the density of water is 1.0 g mL\(^{-1}\) and NaCl is completely dissociated.

Answer: 31.45 g

Mass of water = density \(\times\) volume = 1 \(\times\) 1000 = 1000 g = 1 kg. NaCl is completely dissociated into two ions, so \(i\) = 2. Let \(z\) g of NaCl be added (molar mass 58.5 g mol\(^{-1}\)).

\[\Delta T_f=iK_fm\ \Rightarrow\ 2=2\times1.86\times\frac{z/58.5}{1}\ \Rightarrow\ z=\frac{58.5}{1.86}=31.45\ \text{g}\]

Predict the boiling point of a solution prepared by dissolving 25.0 g of urea and 25.0 g of thiourea in 100 g of water. Given for water \(K_b\) = 0.52 K kg mol\(^{-1}\) and the boiling point of pure water = 373.15 K.

Answer: 377.03 K (\(\Delta T_b\) = 3.88 K)

\[n_{urea}=\frac{25}{60}=0.417;\qquad n_{thiourea}=\frac{25}{76}=0.329\]

\[\text{Molality}=\frac{0.417+0.329}{100}\times1000=7.46\ \text{m}\]

\[\Delta T_b=K_bm=0.52\times7.46=3.88\ \text{K}\]

\[T_b(\text{solution})=373.15+3.88=377.03\ \text{K}\]

Predict the boiling point of a solution prepared by dissolving 3.42 g of cane sugar in 100 g of water. Given for water \(K_b\) = 0.52 K kg mol\(^{-1}\) and the boiling point of pure water = 373.15 K.

Answer: 373.202 K (\(\Delta T_b\) = 0.052 K)

\[\Delta T_b=K_bm=0.52\times\frac{3.42\ \text{g}/342\ \text{g mol}^{-1}}{100\ \text{g}}\times1000=0.52\times0.1=0.052\ \text{K}\]

\[T_b(\text{solution})=373.15+0.052=373.202\ \text{K}\]

QUESTIONS FOR 5 MARKS

(a) Give the difference between molarity and molality for a solution. How does a change in temperature influence their values? (b) Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr₂ in 200 g of water. (Molar mass of MgBr₂ = 184 g; \(K_f\) for water = 1.86 K kg mol\(^{-1}\))

Answer: (a) Molarity is the number of moles of solute dissolved per litre of solution; molality is the number of moles of solute dissolved in 1 kg (1000 g) of solvent. Molarity decreases with an increase in temperature, while molality is independent of temperature, because molality involves mass, which does not change with temperature, while molarity involves volume, which is temperature dependent. (b) The freezing point of the solution is 271.56 K (about \(-1.59\ ^\circ\)C).

\[M=\frac{\text{number of moles of solute}}{\text{volume of solution in litres (dm}^3)};\qquad m=\frac{\text{number of moles of solute}}{\text{mass of solvent in kg}}\]

(b) Given: \(w_2\) = 10.50 g; \(w_1\) = 200 g; \(M_2\) (MgBr₂) = 184 g mol\(^{-1}\). MgBr₂ dissociates completely as MgBr₂ \(\rightarrow\) Mg²⁺ + 2Br^-, so \(i\) = 3.

\[\Delta T_f=i\times\frac{1000\times K_f\times w_2}{w_1\times M_2}=3\times\frac{1000\times1.86\times10.50}{200\times184}=3\times0.53=1.59\ \text{K}\]

\[T_f=T_f^0-\Delta T_f=273.15-1.59=271.56\ \text{K}\]

OR

(a) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain. (b) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water. (\(K_b\) for water = 0.512 K kg mol\(^{-1}\); molar mass of NaCl = 58.44 g)

Answer: (a) Osmosis is the process of flow of solvent molecules from the pure solvent to the solution, or from a solution of lower concentration to a solution of higher concentration, through a semi-permeable membrane. Osmotic pressure is the pressure required to just stop the flow of solvent due to osmosis. Yes, the osmotic pressure of a solution is a colligative property. (b) The boiling point of the solution is about 374.20 K.

(a) The osmotic pressure is expressed as

\[\pi=\frac{n}{V}RT\]

where \(\pi\) = osmotic pressure, \(n\) = number of moles of solute, \(V\) = volume of solution, \(T\) = temperature. From the equation it is clear that the osmotic pressure depends upon the number of moles of solute \(n\), irrespective of the nature of the solute. Hence osmotic pressure is a colligative property.

(b) Given: \(K_b\) = 0.512 K kg mol\(^{-1}\); \(w_2\) = 15.00 g; \(w_1\) = 250.0 g; \(M_2\) = 58.44 g mol\(^{-1}\). NaCl dissociates completely into Na⁺ and Cl^-, so \(i\) = 2.

\[\Delta T_b=i\times\frac{1000\times K_b\times w_2}{w_1\times M_2}=2\times\frac{1000\times0.512\times15.00}{250.0\times58.44}=2\times0.526=1.05\ \text{K}\]

\[T_b=T_b^0+\Delta T_b=373.15+1.05=374.20\ \text{K}\]

A 1.00 molal aqueous solution of trichloroacetic acid (CCl₃COOH) is heated to its boiling point. The solution has the boiling point of 100.18 \(^\circ\)C. Determine the van't Hoff factor for trichloroacetic acid. (\(K_b\) for water = 0.512 K kg mol\(^{-1}\))

Answer: \(i\) = 0.35 (as obtained from the data given in the question)

\(\Delta T\) = 100.18 \(-\) 100 = 0.18 \(^\circ\)C = 0.18 K. Now \(i=\dfrac{\Delta T}{K_b\,m}=\dfrac{0.18\ \text{K}}{(0.512\ \text{K kg mol}^{-1})(1.00\ \text{mol kg}^{-1})}=0.35\).

OR

Define the following terms: (i) mole fraction; (ii) isotonic solutions; (iii) van't Hoff factor; (iv) ideal solution.

Answer: (i) Mole fraction: the ratio of the number of moles of a constituent to the total number of moles of the solution. (ii) Isotonic solutions: solutions which have the same osmotic pressure are called isotonic solutions. (iii) Van't Hoff factor: the ratio of the experimental value of a colligative property to the calculated value of the property when the solution behaves ideally (assuming no association or dissociation). (iv) Ideal solution: a solution that obeys Raoult's law over the entire range of concentration, shows no change in enthalpy and volume on mixing (\(\Delta_{mix}H=0\), \(\Delta_{mix}V=0\)), and in which the A-B intermolecular attractions are the same as the A-A and B-B attractions.

A solution is prepared by dissolving 30 g of a non-volatile non-electrolyte solute in 90 g of water. The vapour pressure of the solution was 2.8 kPa at 298 K. When 18 g of water was further added to it, the vapour pressure became 2.9 kPa at 298 K. Calculate the molar mass of the solute.

Answer: Molar mass of the solute = 23 g mol\(^{-1}\)

Let \(n\) be the moles of solute and \(p^0\) the vapour pressure of pure water. Moles of water: 90/18 = 5 mol in the first case, and 108/18 = 6 mol in the second case. Using \(p=p^0x_A\), where A = water:

\[2.8=p^0\,\frac{5}{5+n}\ \ \ldots(i)\qquad\qquad 2.9=p^0\,\frac{6}{6+n}\ \ \ldots(ii)\]

On dividing (i) by (ii):

\[\frac{28}{29}=\frac{5(6+n)}{6(5+n)}\ \Rightarrow\ 168(5+n)=145(6+n)\ \Rightarrow\ 23n=30\ \Rightarrow\ n=\frac{30}{23}\]

Mass of solute = 30 g, so molar mass = 30 g \(\div\) (30/23) mol = 23 g mol\(^{-1}\).

(a) Define the following terms: (i) mole fraction; (ii) van't Hoff factor. (b) 100 mg of a protein is dissolved in enough water to make 10.0 mL of a solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25 \(^\circ\)C, what is the molar mass of the protein? (\(R\) = 0.0821 L atm mol\(^{-1}\) K\(^{-1}\) and 760 mm Hg = 1 atm)

Answer: (a) (i) The mole fraction of a component is the ratio of the number of moles of the component to the total number of moles of all the components. (ii) The van't Hoff factor is the ratio of the normal molar mass to the abnormal molar mass; equivalently, the ratio of the observed value of a colligative property to the calculated value assuming no association or dissociation. (b) Molar mass of the protein = 1.398 \(\times\) 10\(^4\) g mol\(^{-1}\) (about 13,980 g mol\(^{-1}\)).

(b) Mass of protein = 100 mg = 0.1 g; \(V\) = 10 mL = 0.01 L; \(T\) = 298 K.

\[\pi=\frac{13.3}{760}\ \text{atm};\qquad \pi V=nRT=\frac{w}{M}RT\]

\[\frac{13.3}{760}\times0.01=\frac{0.1}{M}\times0.0821\times298\ \Rightarrow\ M\approx13980\ \text{g mol}^{-1}\]

OR

(a) What is meant by: (i) colligative properties; (ii) molality of a solution? (b) What concentration of nitrogen should be present in a glass of water at room temperature? Assume a temperature of 25 \(^\circ\)C, a total pressure of 1 atmosphere and a mole fraction of nitrogen in air of 0.78. [\(K_H\) for nitrogen = 8.42 \(\times\) 10\(^{-7}\) M/mm Hg]

Answer: (a) (i) All the properties which depend on the number of solute particles, irrespective of their nature, relative to the total number of particles present in the solution are known as colligative properties. (ii) Molality of a solution is the number of moles of solute present in 1 kilogram of solvent. (b) The concentration of nitrogen is 4.99 \(\times\) 10\(^{-4}\) M.

(b) Partial pressure of nitrogen:

\[p_{N_2}=x_{N_2}\times P_{total}=0.78\times760\ \text{mm Hg}=592.8\ \text{mm Hg}\]

Since \(K_H\) is given in M per mm Hg, the concentration of dissolved nitrogen is \(K_H\times p_{N_2}\):

\[\text{Concentration}=8.42\times10^{-7}\ \text{M/mm Hg}\times592.8\ \text{mm Hg}=4.99\times10^{-4}\ \text{M}\]

(a) A weak electrolyte AB is 5% dissociated in aqueous solution. What is the freezing point of a 0.100 molal aqueous solution of AB? (\(K_f\) for water = 1.86 K kg mol\(^{-1}\)) (b) A 0.02 molal solution of acetic acid is 3% dissociated at 25 \(^\circ\)C. Calculate the osmotic pressure of the solution.

Answer: (a) Freezing point = 272.95 K (\(\Delta T_f\) = 0.195 K). (b) Osmotic pressure = 0.504 atm.

(a) AB \(\rightleftharpoons\) A⁺ + B^-. Initially: 1, 0, 0; at equilibrium: \(1-\alpha\), \(\alpha\), \(\alpha\). With \(\alpha\) = 5/100 = 0.05, the total particles = \(1+\alpha\) = 1.05, so \(i\) = 1.05 and the effective molality = 0.100 \(\times\) 1.05 = 0.105 m.

\[\Delta T_f=K_f\times0.105=1.86\times0.105=0.1953\ \text{K};\qquad T_f=273.15-0.1953=272.95\ \text{K}\]

(b) CH₃COOH \(\rightleftharpoons\) CH₃COO^- + H⁺. Initially: 1, 0, 0; at equilibrium: \(1-\alpha\), \(\alpha\), \(\alpha\). Van't Hoff factor:

\[i=\frac{1+\alpha}{1}=1+0.03=1.03\]

For a dilute solution the molarity is taken as approximately equal to the molality (0.02 mol L\(^{-1}\)), and \(T\) = 25 \(^\circ\)C = 298 K:

\[\pi=iCRT=1.03\times0.02\times0.0821\times298=0.504\ \text{atm}\]