5: Colligative Properties

Correct answer: A. Freezing point is depressed

Explanation: For a solution containing a non-volatile solute, the liquid solvent has a lower chemical potential and lower vapour pressure than the pure solvent. The solution must be cooled to a lower temperature before its solid and liquid phases are in equilibrium. Thus the freezing point is depressed. This is another colligative effect, not an infinite or meaningless value.

Correct answer: A. Mole fraction of solute

Explanation: For a solution containing a non-volatile solute, relative lowering is (p° − p)/p°. By Raoult’s law this equals the mole fraction of solute, xsolute, for an ideal dilute solution. Thus it depends on the relative number of solute particles, not on container colour, smell or elapsed time under fixed conditions.

Correct answer: A. In a solution with more non-volatile solute particles

Explanation: For comparable amounts of solvent at the same temperature, a greater number of non-volatile solute particles gives a larger solute mole fraction. Raoult’s law then predicts a larger relative lowering of vapour pressure. Pure solvent has no solute and hence no solute-caused lowering; colour and container temperature alone do not determine this comparison.

Correct answer: A. It increases

Explanation: For a non-volatile solute, the relative lowering of vapour pressure is related to the solute mole fraction. Increasing x_solute lowers x_solvent and therefore lowers the solvent vapour pressure further. The difference p°−p consequently increases, assuming the same temperature and the usual solution model. It does not become zero or remain unchanged.

Correct answer: A. Colligative property

Explanation: Lowering of vapour pressure is a colligative property because, for dilute ideal solutions, its magnitude depends primarily on the number or concentration of dissolved solute particles rather than their chemical identity. A mole-fraction relation expresses this effect. Colour, odour and particle shape do not define the property.

Correct answer: A. The one with more non-volatile solute particles

Explanation: For comparable amounts of solvent and at the same temperature, a greater number of non-volatile solute particles produces a larger solute mole fraction and a smaller solvent mole fraction. Raoult's law then predicts a larger fall from the pure-solvent vapour pressure. Pure solvent has no solute-induced lowering, and container colour is irrelevant.

Correct answer: A. Mole fraction of solute

Explanation: For a solution containing a non-volatile solute, Raoult’s law gives P = X_solvent P°. The lowering is P° − P = P°(1 − X_solvent) = P°X_solute, because X_solvent + X_solute = 1. Dividing by P° gives (P° − P)/P° = X_solute. Thus relative lowering depends on solute mole fraction, not colour, container height, or odour.

Correct answer: A. Boiling point increases

Explanation: Boiling occurs when vapour pressure equals external pressure. If a solute lowers the vapour pressure at every given temperature, the solution must be heated further before it reaches that equality. Therefore its boiling point is higher than that of the pure solvent under the same pressure. This increase is called elevation of boiling point.

Correct answer: A. Freezing point decreases

Explanation: Adding a non-volatile solute lowers the chemical potential and escaping tendency of the liquid solvent. Equilibrium between the solid solvent and the solution is therefore reached at a lower temperature than for the pure solvent. This is freezing-point depression. The effect is not an infinite change and does not make the freezing point independent of pressure.

Correct answer: A. Vapour pressure decreases, so higher temperature is needed for boiling

Explanation: Adding a non-volatile solute lowers the solvent mole fraction and therefore lowers the solution vapour pressure at a given temperature. Boiling requires vapour pressure to equal external pressure, so the solution must be heated further to reach that value. This produces boiling-point elevation. The other choices give the wrong pressure change or mechanism.

Correct answer: A. Nearly same

Explanation: Lowering of vapour pressure is colligative: under comparable conditions it depends on the number of solute particles. Equal mole amounts of non-dissociating solutes produce equal numbers of particles, so their lowerings are nearly the same in the same solvent. Chemical identity or colour does not change this ideal colligative comparison.

Correct answer: A. Lowering increases

Explanation: Dissociation increases the effective number of solute particles in a given amount of solution. Since lowering of vapour pressure is colligative, the larger particle count gives a larger lowering, often represented by a van't Hoff factor greater than one. The solvent pressure remains below its pure value; it does not become higher merely because dissociation occurs.

Correct answer: A. Lowering decreases

Explanation: Association combines separate solute particles into fewer effective particles. Because the colligative lowering depends on particle number, the effective concentration is smaller than expected and the vapour-pressure lowering decreases. In van't Hoff terms, the factor is less than one. Association does not make the lowering infinite or force the solvent pressure to zero.

Correct answer: A. nearly same

Explanation: Lowering of vapour pressure is colligative: for the same solvent and temperature it depends on the effective number of dissolved particles. Therefore equal effective particle amounts give nearly equal lowering, even if the solutes are chemically different. The comparison assumes comparable solution conditions; dissociation or association must be included in the actual particle count.

Correct answer: A. because solute dissociates into ions and increases particle number

Explanation: An electrolyte such as an ionic salt can dissociate into two or more ions in a polar solvent. The number of solute particles then exceeds the number of formula units added, increasing the colligative effect and hence the lowering of solvent vapour pressure. The ions generally remain in solution; their vapour formation is not the reason.

Correct answer: A. because several particles combine to form fewer particles

Explanation: An associating solute forms larger species, such as dimers, from smaller dissolved particles. The effective number of particles becomes lower than the number calculated from the amount added, so the colligative lowering of vapour pressure is smaller than the ideal prediction. This is a particle-count effect, not solvent disappearance or solute vaporisation.

Correct answer: B. It dissociates into ions and increases effective particle number

Explanation: An electrolyte such as an ionic compound can dissociate into two or more ions in solution. The effective number of solute particles then becomes larger than the number of formula units added, increasing the colligative effect and lowering solvent vapour pressure more. The magnitude is represented by the van’t Hoff factor when dissociation is considered.

Correct answer: A. Nearly the same

Explanation: Lowering of vapour pressure is a colligative property. Under the same temperature and with the same solvent, its magnitude depends on the effective number of dissolved particles relative to solvent particles. If those effective particle conditions are the same, the lowering will be nearly the same, even if the chemical identities differ. Association or dissociation must be included in counting particles.

Correct answer: A. Lower solvent mole fraction and presence of solute particles

Explanation: For a non-volatile solute, adding particles reduces the solvent mole fraction from one and lowers the solvent’s escaping tendency. Raoult’s law expresses this as p_solution = x_solvent p°_solvent, or Δp/p° = x_solute. Thus the pressure lowering is linked to solvent composition and effective solute-particle number, not container name, colour, or glass thickness.

Correct answer: A. It will be the same

Explanation: For a non-volatile, non-dissociating solute, the lowering depends on the solute mole fraction. Equal moles placed in equal moles of the same solvent give equal mole fractions and equal numbers of dissolved particles. Thus the lowering is the same, regardless of molar mass, colour, or chemical name. The conclusion assumes comparable ideal dilute-solution behaviour.

Correct answer: A. 0.05

Explanation: Relative lowering of vapour pressure is calculated as (p⁰ − p)/p⁰, where p⁰ is the vapour pressure of the pure solvent and p is that of the solution. Here the absolute decrease is 100 − 95 = 5 units. Dividing by the pure-solvent pressure gives 5/100 = 0.05. Thus 0.05, or 5%, is correct; 5 is only the absolute decrease, not the relative value.

Correct answer: A. Convert mass into moles

Explanation: Raoult’s law and the colligative-property equations use mole fractions, not masses directly. Therefore, when masses are supplied, first divide each mass by the appropriate molar mass to obtain moles, then form the required mole fraction. The solution’s ideality must still be checked from the wording, but converting mass to moles is the essential first calculation step.

Correct answer: A. 0.15

Explanation: The relative lowering is (p° − p)/p° = (100 − 85)/100 = 15/100 = 0.15. For a solution containing a non-volatile solute, this relative lowering equals the solute mole fraction, xsolute. Hence xsolute = 0.15. The value 0.85 is the solvent mole fraction, and 1.15 cannot be a mole fraction because mole fractions range from zero to one.

Correct answer: A. Vapour pressure decreases, so higher temperature is needed to reach external pressure

Explanation: A non-volatile solute contributes essentially no vapour, but it lowers the solvent mole fraction and hence the solution vapour pressure. Boiling requires the solution vapour pressure to equal the external pressure. Since the pressure starts lower at any given temperature, the solution must be heated further to reach equality, producing boiling-point elevation.

Correct answer: A. Nearly the same

Explanation: Vapour-pressure lowering is a colligative property. Under the same solvent and temperature, its value depends primarily on the number of independent dissolved particles, not on their chemical identity. Thus equal effective particle numbers give nearly equal lowering in the ideal dilute limit. Association, dissociation, concentration, or non-ideal interactions must be included if they change the effective count or assumptions.