5: Colligative Properties

Correct answer: B. Applying external pressure greater than osmotic pressure

Explanation: In ordinary osmosis, water passes through a semipermeable membrane from the dilute solution toward the concentrated solution. Reverse osmosis reverses this natural direction by applying an external pressure greater than the solution’s osmotic pressure on the concentrated seawater side. Water is forced through the membrane, while most dissolved salts and other impurities are retained.

Correct answer: C. Double

Explanation: For a dilute solution, osmotic pressure is given by π = iCRT. The concentration C and temperature T are the same in both cases, so the comparison depends only on i. A non-dissociated solute has i = 1, whereas complete dissociation into two ions gives i = 2. Therefore, the dissociated solution has twice the osmotic pressure of the ideal non-dissociated solution.

Correct answer: A. Glucose < NaCl < CaCl₂

Explanation: Boiling-point elevation is ΔTb = iKb m. For equal molality and the same solvent, the solution with the larger van’t Hoff factor has the greater boiling point. Glucose does not dissociate, so i = 1; NaCl completely dissociates into two ions, so i = 2; and CaCl2 gives three ions, so i = 3. Hence the order is glucose < NaCl < CaCl2.

Correct answer: B. Association

Explanation: The freezing-point depression is proportional to the number of solute particles: ΔTf = iKf m. If solute molecules associate, fewer independent particles are present and i becomes less than 1. The measured depression is therefore smaller than expected. When the ordinary formula is used without correcting for this smaller effect, the calculated molar mass becomes higher than the true molar mass. Thus, association is responsible.

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

Explanation: Osmotic pressure is the minimum external pressure that must be applied to a solution to stop the net flow of solvent through a semipermeable membrane. For dilute solutions, π = iCRT. At fixed temperature and concentration conditions, increasing the effective number of dissolved particles increases iC and therefore increases osmotic pressure. A semipermeable membrane is essential for osmosis, so options A, C, and D are incorrect.

Correct answer: C. 2.6

Explanation: BaCl₂ dissociates according to BaCl₂ → Ba²⁺ + 2Cl⁻, so one formula unit produces n = 3 particles on complete dissociation. For a degree of dissociation α, the van’t Hoff factor is i = 1 + α(n − 1). Here α = 80/100 = 0.80, so i = 1 + 0.80(3 − 1) = 1 + 1.60 = 2.60. Therefore option C is correct.

Correct answer: A. The vapour pressure of the liquid decreases, so solid–liquid equilibrium is reached at a lower temperature.

Explanation: Adding a non-volatile solute lowers the vapour pressure and chemical potential of the liquid solvent. The solid solvent, such as ice, is not affected in the same way. Consequently, the condition in which solid solvent and liquid solution are in equilibrium is achieved only at a lower temperature than for the pure solvent. This lowering of the freezing temperature is called freezing-point depression, so option A is correct.

Correct answer: B. 100 g mol⁻¹

Explanation: Molality is defined as moles of solute per kilogram of solvent. The solvent mass is 250 g = 0.250 kg. Therefore, moles of solute = molality × mass of solvent in kilograms = 0.2 × 0.250 = 0.050 mol. Molar mass = mass of solute / moles of solute = 5 g / 0.050 mol = 100 g mol⁻¹. Hence option B is correct.

Correct answer: C. 0.52 K kg mol⁻¹

Explanation: For boiling-point elevation, ΔTb = iKb m. The solute is non-dissociated, so its van’t Hoff factor is i = 1. Substituting the given values gives 0.26 K = 1 × Kb × 0.5 mol kg⁻¹. Thus Kb = 0.26/0.5 = 0.52 K kg mol⁻¹. Therefore option C is correct. The unit follows from dividing kelvin by molality.

Correct answer: C. It becomes double

Explanation: For a dilute solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as Δp/p° = xsolute, where xsolute is the solute mole fraction. Therefore, as long as the solution remains dilute and ideal, doubling xsolute doubles the relative lowering. The change is direct, not inverse or independent.

Correct answer: A. ΔTf will be less than expected

Explanation: The depression in freezing point is given by ΔTf = iKf m. For a fixed solvent, molality, and Kf, the value of ΔTf is directly proportional to the van’t Hoff factor i. Association combines several solute molecules into fewer particles, making i less than one. Consequently, the actual freezing-point depression is smaller than the value calculated for non-associated particles.

Correct answer: B. Molality for ΔTf

Explanation: Freezing-point depression is calculated from ΔTf = Kf m, so the concentration must be expressed as molality, moles of solute per kilogram of solvent. Boiling-point elevation also uses molality, but osmotic pressure uses molarity in π = CRT. Relative lowering of vapour pressure is related to mole fraction, not mass percent. Thus option B is the correct pairing.

Correct answer: C. 1.5

Explanation: For freezing-point depression, the relation is ΔT_f = iK_fm. The ideal value corresponds to i = 1, so the van't Hoff factor is obtained by dividing the observed depression by the ideal depression: i = 0.279/0.186 = 1.5. Thus, the solution produces 1.5 times the ideal number of effective particles, usually because of partial dissociation of the solute.

Correct answer: A. The osmotic pressure of root-cell sap is greater than that of the soil solution

Explanation: Water enters root cells by osmosis through selectively permeable membranes. A root-cell sap with higher osmotic pressure is generally more concentrated and has lower water potential than the surrounding soil solution. Therefore, water moves from the soil into the root. If the soil solution is more concentrated, it may draw water out of the root and cause difficulty in absorption.

Correct answer: A. Sodium chloride solution

Explanation: Boiling-point elevation follows ΔT_b = iK_bm. Glucose is a non-electrolyte and remains mainly as individual molecules, so i is approximately 1. Sodium chloride ideally dissociates into Na+ and Cl−, giving i approximately 2. At equal molality, the sodium chloride solution therefore contains more effective particles and has the greater boiling-point elevation, although real solutions may show some deviation from ideality.

Correct answer: A. Calcium chloride solution

Explanation: Freezing-point depression is given by ΔT_f = iK_fm. Glucose does not dissociate in water, so i is approximately 1. Under the stated ideal assumption, calcium chloride dissociates as CaCl2 → Ca2+ + 2Cl−, giving i = 3. At the same molality, calcium chloride therefore causes three times the ideal particle effect, a larger depression, and consequently the lower actual freezing point.

Correct answer: A. π = CRT

Explanation: For a dilute solution, osmotic pressure follows an equation analogous to the ideal-gas equation. The correct relation is π = CRT, where π is osmotic pressure, C is the molar concentration of the solute particles, R is the gas constant, and T is the absolute temperature in kelvin. The concentration must be molarity, not molality. If the solute dissociates, the more general relation is π = iCRT, where i is the van’t Hoff factor.

Correct answer: A. Less than the ideal value

Explanation: The ideal van’t Hoff factor assumes that dissociation is complete and that every formula unit produces the maximum possible number of particles. In incomplete dissociation, only some solute units split into ions, while the rest remain undissociated. Therefore, the actual number of particles is smaller than the ideal number, so the observed van’t Hoff factor is less than the ideal value. The associated colligative effect is consequently smaller than the complete-dissociation prediction.

Correct answer: A. The number of solvent molecules at the surface decreases

Explanation: A non-volatile solute does not appreciably enter the vapour phase. When it dissolves in the solvent, some surface positions are occupied by solute particles, so fewer solvent molecules are present at the surface and able to escape. The escaping tendency of the solvent therefore decreases, lowering its partial vapour pressure. For an ideal solution, Raoult’s law expresses this as p₁ = x₁p₁°, where x₁ is less than one.

Correct answer: A. At higher effective particle concentration and higher temperature

Explanation: For a dilute solution, osmotic pressure is given by π = iCRT. Thus, at comparable conditions, increasing the effective particle concentration iC or increasing the absolute temperature T increases π. A dissociated electrolyte may have a larger i because it produces more particles. Lower concentration or lower temperature produces a smaller osmotic pressure. The amount of solvent alone is not the determining factor; its effect is represented through the concentration of solute particles.

Correct answer: A. The one with lower molar mass and more particles

Explanation: Freezing-point depression is given by ΔTf = iKf m. With equal masses of solute and equal masses of water, a solute with lower molar mass provides a larger number of moles because n = mass/molar mass. Therefore its molality is greater and, if the effective particle factor is comparable, it produces a larger depression. If the solutes ionize differently, the van’t Hoff factor must also be considered; the decisive principle is the larger effective number of dissolved particles.

Correct answer: A. 0.1 M aluminium chloride

Explanation: For dilute solutions, osmotic pressure is given by π = iCRT. All four solutions have the same concentration and temperature, so the solution with the greatest van't Hoff factor will have the greatest osmotic pressure. Aluminium chloride is an electrolyte and ideally dissociates as AlCl₃ → Al³⁺ + 3Cl⁻, producing four particles, whereas glucose, urea, and sucrose remain essentially single particles.

Correct answer: A. Dissociation of the solute

Explanation: Boiling-point elevation is given by ΔT_b = iK_bm. If a solute dissociates into ions, the number of effective particles increases and the van’t Hoff factor becomes greater than one. Consequently, the observed elevation is larger than the value calculated by assuming no dissociation. Association would produce the opposite effect.

Correct answer: A. Association of the solute

Explanation: The depression in freezing point is given by ΔT_f = iK_fm. Association joins two or more solute molecules into a larger particle, reducing the number of independently moving particles. Thus, the van’t Hoff factor becomes less than one and the observed freezing-point depression is smaller than the ideal expected value. Dissociation would increase the depression.

Correct answer: A. Their effective particle concentrations are equal

Explanation: For a dilute solution, osmotic pressure is expressed as π = iCRT, where iC represents the effective concentration of solute particles. At the same temperature, equal osmotic pressures imply equal values of iC, even when the chemical solutes are different. Equal osmotic pressure does not require equal mass, colour, or solvent identity.