5: Colligative Properties

Correct answer: A. Elevation of boiling point

Explanation: A non-volatile solute lowers the vapour pressure of the solvent. Since boiling requires vapour pressure to equal external pressure, the solution must be heated to a higher temperature before boiling begins. This related colligative property is called elevation of boiling point. Colour, smell and container width are not consequences of vapour-pressure lowering.

Correct answer: A. Colligative property

Explanation: Lowering of vapour pressure is a colligative property because, for dilute solutions with a non-volatile solute, its value depends mainly on the number of dissolved solute particles relative to solvent particles. It does not primarily depend on the chemical identity of those particles. Colour, smell and shape are not the classification principle here.

Correct answer: A. Solution with non-volatile solute and volatile solvent

Explanation: A volatile solvent can establish a measurable vapour phase, while a non-volatile solute does not add its own appreciable vapour pressure. The solute then lowers the solvent's vapour pressure, making the effect easy to identify and relate to Raoult's law. A solid mixture or empty container does not provide this liquid-solution situation.

Correct answer: A. Solute particles are reducing evaporation of solvent

Explanation: When a non-volatile solute is dissolved, it lowers the solvent mole fraction and reduces the number or escaping tendency of solvent molecules at the surface. Consequently, fewer solvent molecules enter the vapour phase at equilibrium, so the solvent vapour pressure falls. The solvent is not destroyed or converted into solute, and the solution still has a definite temperature.

Correct answer: A. Because it depends on number of solute particles

Explanation: A colligative property depends mainly on the number of dissolved particles relative to the amount of solvent, not on the chemical identity of those particles, for dilute ideal solutions. Lowering of vapour pressure follows this rule: more particles produce a larger lowering. Colour, container design, and the solvent's name are irrelevant to the definition.

Correct answer: A. Lowering of vapour pressure

Explanation: Dissolved salt behaves approximately as a non-volatile solute in water. It lowers the mole fraction and escaping tendency of water, so the water vapour pressure decreases. This direct observation is called lowering of vapour pressure and is a colligative property; it also contributes to boiling-point elevation. The other choices describe unrelated physical phenomena.

Correct answer: A. 0.25

Explanation: For a solution with a non-volatile solute, Raoult’s law gives the relative lowering as (P° − P)/P° = X_solute. The stated relative lowering is 0.25, so X_solute = 0.25 directly. In a binary solution, 0.75 would be the corresponding solvent mole fraction, since the two fractions sum to one. Values such as 1.25 or 25 cannot be mole fractions.

Correct answer: A. Increase in boiling point of salt water

Explanation: Salt is treated as a non-volatile solute when dissolved in water. It lowers the mole fraction and vapour pressure of water. Since boiling requires the vapour pressure to equal external pressure, the salt solution must be heated to a higher temperature; hence its boiling point rises. The other observations arise from optical, magnetic, or acoustic phenomena.

Correct answer: B. 0.20

Explanation: For a solution containing a non-volatile solute, the relative lowering of vapour pressure is (p° − p)/p°. Under the ideal dilute-solution relation, this quantity equals the mole fraction of the solute, xsolute. Since xsolute = 0.20, the relative lowering is 0.20. It is dimensionless; 20 kPa would represent an absolute lowering only if the pressure were calculated separately.

Correct answer: C. 0.25

Explanation: First calculate the absolute lowering: Δp = p° − p = 120 − 90 = 30 kPa. Relative lowering is the absolute lowering divided by the vapour pressure of the pure solvent: Δp/p° = 30/120 = 0.25. Hence option C is correct. The value 30 kPa is the absolute decrease, whereas 0.25 is the required dimensionless relative decrease.

Correct answer: A. mole fraction of solute

Explanation: In a dilute solution with a non-volatile solute, the solute reduces the fraction of solvent molecules present at the surface and therefore lowers the solvent’s vapour pressure. Raoult’s law gives Δp/p° ≈ xsolute, so the lowering is approximately proportional to the solute mole fraction. Colour, container height, and taste are not the governing variables in this relation.

Correct answer: A. solution with more non-volatile solute particles in the same solvent

Explanation: Lowering of vapour pressure is a colligative effect and, for comparable solutions of the same solvent, increases with the solute mole fraction or number of dissolved particles. More non-volatile particles reduce the solvent's escaping tendency more strongly. A pure solvent and a liquid without solute have no solute-caused lowering, while fewer particles cause a smaller effect.

Correct answer: A. it lowers the solvent vapour pressure

Explanation: A non-volatile solute has negligible vapour pressure of its own, so it dilutes the solvent in the liquid phase. The solvent mole fraction becomes less than one, and Raoult's law gives a lower solvent vapour pressure. It does not make the solvent more volatile or remove the normal dependence on temperature and equilibrium conditions.

Correct answer: A. solution contains appreciable non-volatile solute

Explanation: At the same temperature, pure solvent has the maximum solvent mole fraction and therefore a higher vapour pressure than a solution containing non-volatile solute. A large decrease indicates a significant reduction in solvent mole fraction, usually due to an appreciable amount of solute. It does not prove complete evaporation or zero temperature.

Correct answer: A. boiling point increases

Explanation: Boiling requires the liquid vapour pressure to equal the external pressure. Adding a non-volatile solute lowers vapour pressure at every given temperature, so the solution must be heated to a higher temperature before equality is reached. This is boiling-point elevation. The liquid can still boil; its boiling temperature is simply higher.

Correct answer: B. 10 kPa

Explanation: For a non-volatile solute, the relative lowering of vapour pressure is equal to the solute mole fraction: Δp/p° = xsolute = 0.05. Therefore the actual lowering is Δp = 0.05 × 200 kPa = 10 kPa. Option B is correct. The value 0.05 is the relative lowering, while 10 kPa is the required pressure decrease; 190 kPa would be the solution pressure, not the lowering.

Correct answer: A. because it depends on number of solute particles

Explanation: A colligative property depends primarily on the number of dissolved particles relative to solvent particles, not on the chemical identity of those particles. For a non-volatile solute, relative lowering equals the solute mole fraction, Δp/p° = xsolute. Thus equal particle numbers produce comparable effects under the same conditions, apart from association or dissociation.

Correct answer: A. it becomes approximately double

Explanation: For a non-volatile solute in a dilute solution, relative lowering is approximately equal to the solute mole fraction: Δp/p° ≈ xsolute. Therefore, doubling xsolute doubles the relative lowering approximately, as long as the dilute-solution assumption remains valid. It does not halve or disappear; deviations may arise at higher concentrations.

Correct answer: A. 0.12

Explanation: A 12 percent lowering means that the relative lowering of vapour pressure is 12/100 = 0.12. For a solution containing a non-volatile solute, Raoult’s law gives relative lowering = xsolute. Therefore the solute mole fraction is 0.12, so option A is correct. The value 0.88 is the corresponding solvent mole fraction, while 12 and 1.12 are not valid mole fractions in this context.

Correct answer: A. lowering depends on number of solute particles

Explanation: A colligative property depends on the number of dissolved particles relative to the solvent, not on the chemical identity alone. For a non-volatile solute, relative vapour-pressure lowering equals the solute mole fraction, which reflects the effective particle count. Electrolyte dissociation and solute association change the effect because they change that count. Colour, smell and vessel name have no role in the law.

Correct answer: B. 0.85

Explanation: For a solution containing a non-volatile solute, Raoult’s law gives relative lowering of vapour pressure as Δp/p° = x_solute. Therefore x_solute = 0.15. Mole fractions of solvent and solute add to one, so x_solvent = 1 − 0.15 = 0.85. Option A is the solute fraction, not the solvent fraction; values above one cannot be mole fractions.

Correct answer: B. Because it depends on number of solute particles

Explanation: A colligative property depends primarily on the number of dissolved particles relative to solvent particles, not on the chemical identity of those particles. Relative lowering equals the solute mole fraction for a non-volatile solute, so it reflects particle number. Colour, container shape, and the name of the solvent are not its defining basis.

Correct answer: B. Because solution vapour pressure decreases and higher temperature is needed to reach external pressure

Explanation: Adding a non-volatile solute lowers the solvent’s vapour pressure at every given temperature. Boiling requires vapour pressure to equal external pressure. The solution must therefore be heated to a temperature higher than the pure solvent’s boiling point before equality is reached. The external pressure is not increased by the solute, and vapour can still form.

Correct answer: B. 0.90

Explanation: A 10% lowering means Δp/p° = 10/100 = 0.10. For a solution with a non-volatile solute, this relative lowering equals the solute mole fraction, so x_solute = 0.10. Since x_solvent + x_solute = 1, x_solvent = 1 − 0.10 = 0.90. Option A is the solute fraction, not the solvent fraction, and mole fractions cannot exceed one.

Correct answer: B. 60 kPa

Explanation: The solute mole fraction is x_solute = 1 − x_solvent = 1 − 0.6 = 0.4. For a non-volatile solute, the lowering is Δp = x_solute p° = 0.4 × 150 = 60 kPa. Equivalently, the solution pressure is 0.6 × 150 = 90 kPa, and the lowering is 150 − 90 = 60 kPa. Thus 90 kPa is the solution pressure, not the lowering.