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Class 12 Chemistry Part 1

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Previous year questions: Solutions

Solutions32 min readLesson 8 of 8

Q1 (CBSE Board 2023, 3 marks)

Benzoic acid (molar mass 122 g mol^-1) associates into a dimer when dissolved in benzene. A solution of 6.1 g of benzoic acid in 100 mL of benzene has an osmotic pressure of 6.5 atm at 27 degrees Celsius. Calculate the percentage association of benzoic acid. R = 0.0821 L atm K^-1 mol^-1.

Answer: The amount of benzoic acid is 6.1 divided by 122, which is 0.050 mol. The volume is 0.100 L, so the formula concentration is 0.500 mol L^-1. For complete association the observed osmotic pressure would be built from 0.250 mol of dimer particles; the observed value lies between the two limits. The van't Hoff factor is the observed osmotic pressure divided by the calculated value for no association: i = 6.5 divided by (0.500 times 0.0821 times 300), which is 0.528. For dimerisation n is 2, so i = 1 + (1/2 - 1) alpha and alpha = 2(1 - i) = 2 times 0.472 = 0.944. Benzoic acid is therefore about 94 percent associated in benzene.

Q2 (CBSE Board 2023, 1 mark)

Assertion A: Osmotic pressure is a colligative property. Reason R: Osmotic pressure is proportional to the molality. Choose the correct option: (a) both true and R explains A, (b) both true but R does not explain A, (c) A is true and R is false, (d) A is false and R is true.

Answer: Option (c). The assertion is correct because osmotic pressure depends on the number of solute particles and not on their chemical identity. The reason is false. Osmotic pressure is proportional to the molarity or the amount of solute per unit volume, not to the molality. The equation is the product of concentration, R and T, so at a fixed temperature it is the concentration that matters.

Q3 (CBSE Board 2024, 5 marks)

(a) At the same temperature, carbon dioxide is more soluble in water than oxygen. Which gas has the higher value of K_H and why? (b) How does the size of a blood cell change when it is placed in an aqueous solution containing more than 0.9 percent sodium chloride by mass? (c) A 1 molal aqueous solution of an electrolyte A2B3 is 60 percent ionised. Calculate the boiling point of the solution. K_b for water is 0.52 K kg mol^-1.

Answer: (a) The gas with the lower solubility at the same partial pressure has the higher Henry constant, so oxygen has the higher K_H. Carbon dioxide is more soluble because its molecules interact more strongly with water (it also reacts partly with water), so a smaller K_H is enough to describe the same pressure. (b) The solution is hypertonic with respect to the cell contents, so water leaves the cell by exosmosis and the cell shrinks. (c) A2B3 gives five ions, so n is 5, and i = 1 + (5 - 1) times 0.60, which is 3.4. The elevation is i K_b m = 3.4 times 0.52 times 1, which is 1.77 K. The solution therefore boils at 373.15 plus 1.77, or 374.92 K.

Q4 (CBSE Board 2024, 3 marks)

At 283 K a solvent has a vapour pressure of 100 mm Hg. One mole of a strong electrolyte AB is dissolved in 50 moles of this solvent and the electrolyte dissociates completely. Find the vapour pressure of the resulting dilute solution at 283 K.

Answer: Complete dissociation of AB gives two ions, so the van't Hoff factor is 2. The effective amount of solute particles is 2 mol. The amount of solvent is 50 mol, so the mole fraction of the solvent in the solution is 50 divided by 52, which is 0.9615. By Raoult's law, the vapour pressure of the solution is the mole fraction of solvent times the vapour pressure of the pure solvent, which is 100 times 0.9615, or about 96.2 mm Hg. The fall is larger than it would have been for a non-electrolyte because each formula unit of AB supplies two particles.

Q5 (CBSE Board 2025, 1 mark)

The value of Henry's constant K_H is: (A) greater for gases with higher solubility, (B) greater for gases with lower solubility, (C) constant for all gases, (D) not related to the solubility of gases.

Answer: Option B. For a fixed partial pressure, the dissolved mole fraction is p divided by K_H, so a larger K_H means a smaller amount dissolves. The Henry constant therefore increases as the solubility of the gas decreases. The constant is different for different gases and for different solvents, and it changes with temperature, so option C is also wrong.

Q6 (CBSE Board 2026, 4 marks)

A passage explains osmosis, osmotic pressure, isotonic, hypotonic and hypertonic solutions, and notes that the osmotic pressure of blood cells matches that of a 0.9 percent sodium chloride solution. (a) Calculate the amount of CaCl2 (i = 2.59) that must be dissolved in 2.46 L of water so that the osmotic pressure is 0.70 atm at 27 degrees Celsius. R = 0.082 L atm K^-1 mol^-1 and the molar mass of CaCl2 is 111 g mol^-1. (b) When raisins are kept in water they swell. Name the phenomenon. (c) Name the phenomenon responsible for desalination of sea water.

Answer: (a) The molar concentration follows from the van't Hoff form of the osmotic pressure equation: C = 0.70 divided by (2.59 times 0.082 times 300), which is 0.0110 mol L^-1. The amount in 2.46 L is 0.0110 times 2.46, which is 0.0270 mol. Multiplying by the molar mass gives 0.0270 times 111, which is 3.0 g of CaCl2. (b) The raisins swell by endosmosis, the movement of water into the more concentrated solution inside the raisin. (c) Desalination of sea water is carried out by reverse osmosis, in which an applied pressure larger than the osmotic pressure pushes water through a semipermeable membrane.

Q7 (CBSE SQP 2022-23, 3 marks)

State Henry's law and explain why scuba diving tanks are filled with air diluted with helium. Argon exerts a partial pressure of 6 bar above water; calculate its solubility in water at 298 K, given that K_H for argon in water is 40 kbar.

Answer: At a constant temperature, the partial pressure of a gas above a solution is directly proportional to the mole fraction of that gas in the solution: p = K_H x. At the high pressure of a dive, a large amount of nitrogen would dissolve in the blood and would escape as bubbles during a rapid ascent, causing decompression sickness. Helium has a much larger Henry constant than nitrogen and is therefore much less soluble; replacing part of the nitrogen with helium reduces the amount dissolved and makes the ascent safer. For argon, x = p / K_H = 6 bar divided by 40,000 bar, which is 1.5 times 10^-4. The solubility is therefore 1.5 times 10^-4 in mole fraction units.

Q8 (CBSE SQP 2023-24, 5 marks)

(a) How does temperature affect the solubility of glucose in water? (b) Two samples of 10 mL each, one fresh water and one ocean water, are cooled. Sample P freezes at 0 degrees Celsius and sample Q at minus 1.3 degrees Celsius. Identify the ocean water with a reason. (c) Calculate the van't Hoff factor of an aqueous K3[Fe(CN)6] solution whose degree of dissociation is 0.852, and find its boiling point when the concentration is 1 molal. K_b for water is 0.52 K kg mol^-1.

Answer: (a) Dissolution of glucose in water absorbs heat, so it is an endothermic process. An increase in temperature shifts the dissolution equilibrium towards the dissolved state, so the solubility of glucose increases with temperature. (b) Ocean water contains dissolved salts, which lower its vapour pressure and hence its freezing point. Sample Q freezes below 0 degrees Celsius, so Q is the ocean water and P is the fresh water. (c) One formula unit of K3[Fe(CN)6] gives four ions, so n is 4. i = 1 + (n - 1) alpha = 1 + 3 times 0.852 = 3.556. The elevation is 3.556 times 0.52 times 1, which is 1.85 K, so the solution boils at 373.15 plus 1.85, or about 375.0 K.

Q9 (CBSE SQP 2025-26, 3 marks)

Calculate the vapour pressure of a 1 molal aqueous solution of MgCl2 at 25 degrees Celsius, assuming that MgCl2 is 70 mole percent dissociated. The vapour pressure of pure water at 25 degrees Celsius is 23.8 mm Hg.

Answer: One formula unit of MgCl2 gives three ions, so n is 3. With alpha equal to 0.70, i = 1 + (3 - 1) times 0.70 = 2.4. A 1 molal solution has 1 mol of MgCl2 in 1 kg of water, and 1 kg of water is 1000 divided by 18, or 55.5 mol. In van't Hoff terms the solute contributes i times 1, which is 2.4 mol of particles. The mole fraction of water is 55.5 divided by 58.0, which is 0.957. The vapour pressure of the solution is 0.957 times 23.8, which is about 22.8 mm Hg.

Q10 (CBSE Additional PQ 2023-24, 4 marks)

A passage explains that the Henry constant is not a true constant because it depends on temperature; for many solutes it rises with temperature at low temperatures, passes through a maximum and then falls. For the n-octane and water pair the maximum occurs near 90 degrees Celsius. (a) Which of these is not a true constant like the Henry constant: speed of light, Planck's constant, rate constant, gas constant? (b) How does the solubility of n-octane in water change when the temperature rises from 20 to 60 degrees Celsius at constant pressure? (c) What is the ratio of the solubility of n-octane at 20 degrees Celsius to that at 60 degrees Celsius: greater than one, less than one, or equal to one?

Answer: (a) The rate constant is not a true constant, because it changes with temperature through the Arrhenius equation. (b) For n-octane in water the maximum of K_H lies near 90 degrees Celsius, so between 20 and 60 degrees Celsius K_H is still increasing with temperature. At constant pressure the mole fraction dissolved is p divided by K_H, so a larger K_H gives a smaller solubility. The solubility of n-octane therefore decreases as the temperature rises. (c) Since the solubility at 20 degrees Celsius is larger than at 60 degrees Celsius, the ratio of the two solubilities is greater than one.