5: Colligative Properties

Correct answer: A. Salt lowers the freezing point of water

Explanation: When salt dissolves in the thin liquid layer on ice, it produces a solution whose freezing point is lower than 0°C. At temperatures where pure water would remain frozen, the ice-salt mixture can therefore remain liquid and more ice melts to establish equilibrium. This is a practical example of freezing-point depression, not burning or complete elimination of vapour pressure.

Correct answer: A. 0.5 mol kg⁻¹

Explanation: Molality is defined as the number of moles of solute present in one kilogram of solvent: m = moles of solute ÷ mass of solvent in kilograms. Here, the amount of solute is 0.5 mol and the solvent mass is 1 kg. Therefore, m = 0.5 ÷ 1 = 0.5 mol kg⁻¹. The statement that the solute is non-dissociating does not change the calculation of molality; it would matter only when calculating particle-dependent colligative effects.

Correct answer: A. ΔTf = Kf m

Explanation: The general expression for depression in freezing point is ΔTf = iKf m, where i is the van’t Hoff factor, Kf is the cryoscopic constant of the solvent, and m is the molality. A non-dissociating solute neither splits nor associates appreciably, so i = 1. Substitution therefore gives ΔTf = Kf m. The units also support multiplication: Kf has units K kg mol⁻¹ and molality has units mol kg⁻¹, producing kelvin.

Correct answer: A. Very small

Explanation: For a dilute solution containing a non-volatile solute, Raoult’s law states that the relative lowering of vapour pressure is equal to the mole fraction of the solute: (p° − p)/p° = xsolute. Therefore, when xsolute is very small, the relative lowering is also very small. It remains a positive quantity for an ordinary dilute solution, is not determined by colour, and is not automatically equal to one.

Correct answer: A. Van’t Hoff factor (i)

Explanation: Abnormal molar mass occurs when a solute associates to form larger particles or dissociates to form more particles in solution. The van’t Hoff factor, i, corrects the actual number of solute particles. It is included in expressions such as ΔTb = iKb m, ΔTf = iKf m, and π = iCRT. Therefore, option A is correct.

Correct answer: D. Al₂(SO₄)₃

Explanation: The depression in freezing point is given by ΔTf = iKf m. Since all solutions have the same molality and solvent, Kf and m are constant; the result depends on i. Glucose and urea do not ionise, so i is approximately 1. NaCl gives two ions, whereas Al₂(SO₄)₃ gives 2Al³⁺ and 3SO₄²⁻, five ions in total. Hence option D gives the greatest depression.

Correct answer: D. Al₂(SO₄)₃

Explanation: Freezing-point depression follows ΔTf = iKf m. Because molality and solvent are identical, the solution with the largest van’t Hoff factor produces the greatest depression. Glucose has i ≈ 1, NaCl gives 2 ions, CaCl₂ gives 3 ions, and Al₂(SO₄)₃ gives 5 ions on complete ionisation. Therefore, Al₂(SO₄)₃ produces the greatest depression and option D is correct.

Correct answer: C. 0.04

Explanation: For a solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as (p° − p)/p° = xsolute, where p° is the vapour pressure of the pure solvent and p is the vapour pressure of the solution. Since the solute mole fraction is 0.04, the relative lowering is 0.04. Thus, option C is correct.

Correct answer: A. Glucose

Explanation: Freezing-point depression is given by ΔTf = iKf m. At equal molality and for the same solvent, the solution with the smallest van’t Hoff factor has the smallest depression and therefore the highest freezing point. Glucose is a non-electrolyte and does not dissociate, so i = 1. NaCl, CaCl₂, and AlCl₃ form two, three, and four ions respectively under ideal conditions.

Correct answer: B. The effective number of particles decreases

Explanation: Association means that two or more solute molecules or particles combine to form one larger associated species. Consequently, the number of independently moving solute particles decreases. Since colligative properties depend on the number of particles, the observed effect becomes smaller and the van’t Hoff factor is generally less than one. Therefore, option B is correct.

Correct answer: A. Because the vapour pressure of the solution decreases

Explanation: A liquid boils when its vapour pressure becomes equal to the external pressure. Adding a non-volatile solute lowers the mole fraction and vapour pressure of the solvent. Therefore, the solution must be heated to a higher temperature before its vapour pressure reaches the external pressure. This increase is called elevation of boiling point, so A is correct.

Correct answer: A. Relative lowering of vapour pressure

Explanation: A colligative property depends on the number of dissolved particles rather than their identity. Relative lowering of vapour pressure is a colligative property and, for a dilute solution, is related to the solute mole fraction. Colour, odour, and chemical reactivity depend strongly on the chemical nature and structure of the solute. Hence, option A is correct.

Correct answer: C. AlCl₃

Explanation: The van’t Hoff factor represents the number of solute particles produced per formula unit when dissociation is complete. KCl gives 2 ions, BaCl₂ gives 3 ions, and AlCl₃ gives 4 ions: one Al³⁺ and three Cl⁻. Glucose is a non-electrolyte and gives one particle. Thus AlCl₃ has the largest factor, i = 4, so option C is correct.

Correct answer: A. Because there is no net flow of water across the membrane

Explanation: An isotonic solution has the same effective osmotic concentration as the fluid inside the red blood cell. Water molecules can cross the semipermeable membrane in both directions, but the rates are equal, so there is no net movement of water. Consequently, the cell neither swells nor shrinks, and option A is correct.

Correct answer: A. It increases when the number of solute particles increases

Explanation: Freezing-point depression is a colligative property, expressed for a dilute solution as ΔTf = iKf m. Thus, at a fixed solvent and molality, increasing the number of dissolved particles increases the value of ΔTf. Dissociation raises the effective particle number, whereas association lowers it. Colour and physical state are not the controlling factors, so option A is correct.

Correct answer: B. (p⁰ − p)/p⁰

Explanation: The absolute lowering of vapour pressure is the difference between the vapour pressure of the pure solvent and that of the solution: Δp = p⁰ − p. Relative lowering is obtained by dividing this decrease by the original vapour pressure of the pure solvent. Hence, relative lowering = (p⁰ − p)/p⁰, which is option B.

Correct answer: B. It increases

Explanation: The elevation in boiling point is given by ΔTb = iKb m. If the solute, solvent, and temperature conditions are otherwise unchanged, i and Kb remain constant. Therefore, ΔTb is directly proportional to molality m. Increasing molality increases the number of dissolved particles per kilogram of solvent and consequently produces a larger boiling-point elevation.

Correct answer: D. 0.4 M

Explanation: Complete dissociation of one formula unit of K₃PO₄ produces three K⁺ ions and one PO₄³⁻ ion, giving four particles in total. Therefore, the van’t Hoff factor is i = 4. The effective particle concentration is iC = 4 × 0.1 M = 0.4 M. This particle concentration is the one relevant to osmotic pressure.

Correct answer: A. From the pure solvent to the solution

Explanation: A semipermeable membrane permits solvent molecules to pass but prevents solute particles from crossing. Initially, the pure-solvent side has a higher solvent chemical potential and the solution side has a lower solvent chemical potential because of the dissolved solute. Consequently, solvent molecules move from the pure solvent toward the solution. This net movement continues until equilibrium or an opposing pressure is established.

Correct answer: A. The vapour pressure decreases

Explanation: For an ideal solution containing a non-volatile solute, Raoult’s law gives psolvent = xsolvent p°solvent. Increasing the amount of solute lowers the mole fraction of the solvent, xsolvent. Since the solute itself does not contribute appreciably to the vapour phase, fewer solvent molecules escape from the liquid surface and the solvent vapour pressure decreases. The lowering depends on the relative number of solute particles.

Correct answer: A. It will be larger

Explanation: The elevation in boiling point is calculated using the relation ΔT_b = iK_bm, where i is the van’t Hoff factor, K_b is the ebullioscopic constant, and m is molality. When i and m remain unchanged, ΔT_b is directly proportional to K_b. Therefore, a solvent with a larger K_b produces a larger boiling-point elevation for the same solution conditions. Hence, option A is correct.

Correct answer: C. 0.08

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. Since the solute mole fraction is given as 0.08, the relative lowering is directly 0.08. The solvent mole fraction, 0.92, would not be used in this expression. Therefore, option C is correct.

Correct answer: B. Seawater has a lower freezing point and higher boiling point

Explanation: Dissolved salts increase the number of solute particles in seawater. Colligative effects depend on the number of particles: dissolved solute lowers the solvent’s freezing point and raises its boiling point. Seawater also has a nonzero osmotic pressure because salts create a difference in chemical potential across a semipermeable membrane. Therefore, option B is correct.

Correct answer: B. It depends on the number of solute particles present in the solution.

Explanation: Elevation in boiling point is a colligative property, so its magnitude depends mainly on the number of dissolved particles, not their chemical identity. A non-volatile solute lowers the vapour pressure of the solvent. Therefore, the solution must be heated to a higher temperature before its vapour pressure equals atmospheric pressure and boiling begins. The relation is ΔTb = iKb m, where i represents the effective number of particles.

Correct answer: A. Osmotic pressure

Explanation: Osmotic pressure is a colligative property. For a dilute ideal solution, it is represented by π = iCRT, so it depends on temperature, concentration, and the effective number of solute particles. It does not primarily depend on the chemical identity of those particles. In contrast, colour, odour, and chemical reactivity are characteristic properties determined by the substance’s molecular structure and chemical nature. Hence osmotic pressure is the only suitable answer.