5: Colligative Properties

Correct answer: A. Because the solute lowers the liquid solvent’s vapour pressure, so solid–liquid equilibrium is reached at a lower temperature

Explanation: A non-volatile solute lowers the vapour pressure and chemical potential of the liquid solvent. Consequently, the liquid solution must be cooled to a lower temperature before its chemical potential becomes equal to that of the solid solvent. At that lower temperature, solid–liquid equilibrium is established and freezing begins. This decrease in freezing temperature is called depression in freezing point and is a colligative property because it depends on the number of dissolved particles.

Correct answer: C. Twice that of glucose

Explanation: Osmotic pressure is given by π = iCRT. Glucose is a non-electrolyte and does not dissociate appreciably in water, so its van’t Hoff factor is approximately 1. NaCl completely dissociates into Na⁺ and Cl⁻, giving i = 2. Since both solutions have the same concentration and temperature, the NaCl solution has twice as many dissolved particles and therefore twice the osmotic pressure of the glucose solution.

Correct answer: B. Water may move out of the root

Explanation: A highly concentrated soil solution may become hypertonic relative to the cell sap inside root cells. Across the selectively permeable plasma membrane, water moves toward the side with the higher effective solute concentration. Therefore, water absorption by the root decreases, and water may move out of root cells. This can cause loss of turgor and plasmolysis in severe cases. The exact direction depends on the relative concentrations, but option B is correct under the stated condition.

Correct answer: C. 1.5

Explanation: For freezing-point depression, ΔT_f = iK_fm. The value calculated for a non-dissociated solute corresponds to i = 1, so the observed effect is compared with the expected effect by using i = observed depression ÷ normal depression. Thus, i = 0.279 ÷ 0.186 = 1.5. The value greater than one indicates that the solute produces more effective particles, such as through dissociation. Therefore, option C is correct.

Correct answer: D. Osmotic pressure

Explanation: Osmotic pressure is especially useful for very dilute solutions because it can be measured at or near room temperature and remains appreciable even when other colligative changes are extremely small. For a dilute solution, π = CRT; by substituting the concentration expressed through the solute mass and molar mass, the unknown molar mass can be calculated accurately.

Correct answer: A. The observed molar mass is greater than the expected value

Explanation: Association occurs when two or more solute molecules combine to form fewer particles in solution. Consequently, the van’t Hoff factor becomes less than 1, the colligative effect decreases, and the molar mass calculated from that effect becomes greater than the true expected value. Therefore, a higher observed molar mass indicates association.

Correct answer: D. Al₂(SO₄)₃

Explanation: Freezing-point depression is ΔT_f = iK_fm. Since the solvent and molality are the same, the largest van’t Hoff factor gives the greatest depression. Glucose and urea do not dissociate, so i=1; NaCl gives two ions, so i=2; Al₂(SO₄)₃ gives two Al³⁺ and three SO₄²⁻ ions, so i=5. Hence option D is correct.

Correct answer: A. The solute dissociates into ions in solution

Explanation: Dissociation increases the number of particles, so i becomes greater than 1. Since the molar mass calculated from a colligative effect is related by M_observed = M_true/i, division by a value greater than 1 makes the observed molar mass smaller than the true value. Association gives i<1 and produces the opposite result.

Correct answer: B. On association of the solute

Explanation: Association causes several solute molecules to combine into fewer larger particles. Consequently, the observed colligative effect becomes smaller than the effect expected from the original number of molecules. If molar mass is calculated without correcting for this reduced particle number, the calculated or observed molar mass becomes higher than the true value. In terms of the van’t Hoff factor, association gives i < 1. Therefore, option B is correct.

Correct answer: B. Boiling-point elevation and molality

Explanation: For dilute solutions, boiling-point elevation and freezing-point depression are written as ΔT_b = iK_bm and ΔT_f = iK_fm, so molality is the appropriate concentration term. Osmotic pressure is usually expressed as π = iCRT, using molarity, while relative lowering of vapour pressure is related to the solute mole fraction. Hence option B is the only correct match.

Correct answer: B. 0.492 atm

Explanation: For a non-dissociated solute, the van’t Hoff factor is i = 1, so osmotic pressure is calculated by π = iCRT. The solution volume is 500 mL = 0.5 L, therefore its molarity is C = 0.01/0.5 = 0.020 mol L⁻¹. Using R = 0.082 L atm mol⁻¹ K⁻¹ and T = 300 K, π = 1 × 0.020 × 0.082 × 300 = 0.492 atm. Hence, option B is correct.

Correct answer: D. Osmotic pressure

Explanation: Osmotic pressure is especially useful for finding the molar mass of proteins and other macromolecules because it can be measured in very dilute solutions at or near room temperature. The solute need not be heated, so decomposition of a heat-sensitive substance is avoided. Boiling-point and freezing-point methods require heating or cooling and are less suitable. Thus option D is correct.

Correct answer: B. 1 + 2x

Explanation: One formula unit of AB₂ produces three ions, A²⁺ and two B⁻ ions, when completely dissociated. For a substance that dissociates into n particles with degree of dissociation α, the van’t Hoff factor is i = 1 + α(n − 1). Here n = 3 and α = x, so i = 1 + x(3 − 1) = 1 + 2x. Thus, option B is the only correct expression.

Correct answer: D. 0.25 m

Explanation: Complete dissociation of one formula unit of Al₂(SO₄)₃ gives two Al³⁺ ions and three SO₄²⁻ ions, for a total of five dissolved particles. Effective molality equals the original molality multiplied by the number of particles formed: m_effective = i × m = 5 × 0.05 = 0.25 m. Therefore, the effective molality is 0.25 m, making option D correct.

Correct answer: B. 0.90

Explanation: For association into dimers, two original solute particles combine to form one particle. If the fraction associated is α, the van’t Hoff factor is i = 1 − α/2, because the number of particles decreases by half for the associated portion. With α = 20% = 0.20, i = 1 − 0.20/2 = 1 − 0.10 = 0.90. Therefore, option B is correct.

Correct answer: A. Ethanoic acid in benzene

Explanation: Ethanoic acid molecules associate through hydrogen bonding in a nonpolar solvent such as benzene and commonly form dimers. Two solute molecules becoming one particle decreases the effective number of particles, so i becomes less than 1. Sodium chloride and barium chloride generally dissociate into ions in water, while glucose remains molecular. Thus option A is correct.

Correct answer: C. 0.80

Explanation: For a solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as (p° − p)/p° = x₂, where x₂ is the mole fraction of the solute. Thus x₂ = 0.20. In a binary solution, x₁ + x₂ = 1, so the solvent mole fraction is x₁ = 1 − 0.20 = 0.80. Therefore, option C is correct.

Correct answer: C. Al₂(SO₄)₃

Explanation: The van’t Hoff factor is approximately equal to the number of independent particles produced by one formula unit during complete dissociation. KCl produces 2 ions, CaCl₂ produces 3 ions, and glucose remains as one molecule. Al₂(SO₄)₃ dissociates as 2Al³⁺ + 3SO₄²⁻, producing 5 particles. Hence it has the largest van’t Hoff factor, i = 5.

Correct answer: C. 4.92 atm

Explanation: For an ideal dilute solution, osmotic pressure follows π = iCRT. If the concentration and solute remain unchanged, iC is constant, so π is directly proportional to absolute temperature. The temperature increases from 300 K to 600 K, which is a factor of two. Therefore, π₂ = 2.46 × (600/300) = 4.92 atm, making option C correct.

Correct answer: B. NaCl has twice the osmotic pressure

Explanation: Osmotic pressure is given by π = iCRT. Glucose is a non-electrolyte and remains molecular in solution, so i = 1. Completely dissociated NaCl produces two ions, Na⁺ and Cl⁻, so i = 2. Since both solutions have the same C and temperature, the NaCl solution has twice the osmotic pressure of the glucose solution. The one-litre volume does not change this ratio.

Correct answer: B. 0.930 K

Explanation: The depression in freezing point for an electrolyte solution is calculated using ΔT_f = iK_fm. Substituting the given values gives ΔT_f = 2 × 1.86 K kg mol⁻¹ × 0.25 mol kg⁻¹. The units cancel appropriately, and the result is 0.930 K. Therefore, the freezing point is lowered by 0.930 K, so option B is correct.

Correct answer: B. Vapour pressure decreases and boiling point increases.

Explanation: Increasing the number of dissolved particles strengthens the colligative effects. For a solution containing a non-volatile solute, the solvent vapour pressure decreases. Because boiling occurs when vapour pressure reaches the external pressure, the lower vapour pressure requires a higher temperature, so the boiling point rises. The same increase in particles generally raises osmotic pressure and lowers the freezing point; therefore only option B gives the correct pair.

Correct answer: C. 1.5

Explanation: Use the freezing-point depression equation ΔTf = iKf m. Rearranging gives i = ΔTf/(Kf m). Substituting the given values, i = 0.279/(1.86 × 0.1) = 0.279/0.186 = 1.5. The van’t Hoff factor is greater than one, which is consistent with a solute that produces more effective particles in solution, such as an electrolyte undergoing dissociation. Therefore, option C is correct.

Correct answer: B. At the same temperature, it increases when the number of solute particles in the solution increases.

Explanation: Osmotic pressure is the pressure required to stop the net movement of solvent through a semipermeable membrane. It is a colligative property and, for a dilute solution, follows π = iCRT. Thus, at constant temperature, increasing the concentration or effective number of solute particles increases osmotic pressure. It is not determined only by solute identity or solvent density. Therefore, option B correctly states the relationship.

Correct answer: C. 0.078 K

Explanation: For an electrolyte solution, the boiling-point elevation is calculated using ΔTb = iKb m. Substituting the actual van’t Hoff factor and the given molality gives ΔTb = 1.5 × 0.52 × 0.1 K = 0.078 K. The actual value of i must be used because incomplete dissociation makes the effective particle number different from the ideal value for complete dissociation. Hence option C is correct.