4: Vapour Pressure

Correct answer: A. Mole fraction of the solute

Explanation: For a solution containing a non-volatile solute, Raoult's law leads to (P° − P)/P° = X_solute. Thus the relative lowering of vapour pressure equals the solute mole fraction in the simple ideal case. It depends on the number of solute particles relative to total particles, not on colour, container height, or smell.

Correct answer: B. It decreases because solvent mole fraction decreases.

Explanation: A non-volatile solute does not add appreciable vapour, but it increases total moles and therefore lowers the solvent mole fraction. At constant temperature, Raoult's law gives p = x_solvent p°, so the solvent vapour pressure decreases. More solution particles do not mean more solvent vapour, and the pressure is not zero at every concentration.

Correct answer: A. 80 units

Explanation: For a solution with a non-volatile solute, Raoult’s law gives P_solution = X_solvent P°_solvent. Substituting the given values, P_solution = 0.8 × 100 = 80 units. The value cannot remain 100 because the solvent mole fraction is less than one. Dividing by 0.8 gives 125, which is not the Raoult-law expression, while 20 is only the lowering in pressure.

Correct answer: A. Contributions of both solvent and solute

Explanation: A volatile solute can enter the vapour phase along with the solvent. Consequently, the total pressure above the solution is the sum of the partial pressures of both components, provided the vapour behaves approximately ideally. This differs from a non-volatile-solute case, where the solute's own vapour contribution is neglected.

Correct answer: A. The sum of the two partial vapour pressures

Explanation: In an ideal solution of two volatile liquids, each component contributes a partial vapour pressure according to Raoult’s law: p_A = X_A P°_A and p_B = X_B P°_B. Dalton’s law then gives the total pressure as P_total = p_A + p_B. Thus both volatile components contribute to the total; it is not determined simply by the lower pressure, mass, or colour of the liquids.

Correct answer: A. Nearly similar in nature

Explanation: An ideal solution is one in which interactions between unlike molecules are approximately comparable to interactions between like molecules: A–B is similar to A–A and B–B. Consequently, mixing causes little enthalpy change and little volume change, and Raoult's law is obeyed. The attractions are not zero and need not be purely ionic.

Correct answer: A. When new attractions are weaker

Explanation: In positive deviation, the observed total vapour pressure is higher than Raoult's-law prediction. This occurs when unlike A–B attractions are weaker than the original A–A and B–B attractions. Molecules are then held less strongly in the liquid and escape more readily. Stronger unlike attraction would instead favour negative deviation.

Correct answer: A. When new attractions are stronger

Explanation: Negative deviation means that the actual vapour pressure is lower than the ideal Raoult-law value. It can occur when unlike A–B attractions are stronger than the original attractions, holding molecules more firmly in the liquid. Fewer molecules escape, so pressure decreases. Weaker unlike attractions produce positive deviation instead.

Correct answer: A. Mole fraction of the solvent

Explanation: For an ideal solution, Raoult’s law states that the partial vapour pressure of a volatile component equals its mole fraction multiplied by its vapour pressure in the pure state: pi = xi pi°. Therefore, the solvent’s partial pressure is directly proportional to its mole fraction. Colour, container thickness and particle size are not the governing variables.

Correct answer: A. It decreases

Explanation: Raoult's law gives p_solvent = x_solvent p°_solvent at fixed temperature. Since p° is constant for the chosen solvent and temperature, reducing x_solvent directly reduces p_solvent in the same proportion. No particular doubling can be inferred, and the pressure does not remain unchanged unless the mole fraction remains unchanged.

Correct answer: A. Decrease in number of solvent molecules at the surface

Explanation: When a non-volatile solute is added, the proportion of solvent molecules among all particles decreases. Consequently, fewer solvent molecules are available at the surface and fewer escape into the vapour phase at a given temperature. This lowers the solvent's equilibrium vapour pressure. Colour, name, and brightness have no role in this explanation.

Correct answer: A. 80 kilopascal

Explanation: For an ideal solution, Raoult’s law gives psolvent = xsolvent p°solvent. Substituting the given values, psolvent = 0.8 × 100 kPa = 80 kPa. The answer is lower than 100 kPa because the solvent mole fraction is less than one. The value 20 kPa is the lowering, not the partial vapour pressure itself.

Correct answer: A. 30 kilopascal

Explanation: Use Raoult’s law for the solvent: p = x p°. Here x = 0.6 and p° = 50 kPa, so p = 0.6 × 50 = 30 kPa. Since the solvent mole fraction is below one, its partial vapour pressure must be below the pure-solvent value of 50 kPa. The other choices do not follow the required multiplication.

Correct answer: A. Sum of partial vapour pressures of all volatile components

Explanation: Dalton’s law states that the total pressure of a mixture of gases or vapours is the sum of the partial pressures contributed by its components. Thus, for an ideal solution containing volatile components, Ptotal = pA + pB + …, with each partial pressure obtained from Raoult’s law. Container weight, colour and the solute’s name do not determine the total vapour pressure.

Correct answer: A. 0.75 times the vapour pressure of pure solvent

Explanation: For an ideal solution, Raoult's law gives p_solvent = x_solvent p°_solvent. With x_solvent = 0.75, p_solvent = 0.75 p°_solvent. Thus the solution vapour pressure is three-fourths of the pure-solvent value, not equal to it, twice it, or zero. This comparison assumes the same temperature.

Correct answer: A. P = X_solvent P°

Explanation: When the solute is non-volatile, essentially only the solvent contributes to the vapour pressure. Raoult’s law states that the partial vapour pressure of the solvent is P = X_solvent P°, where X_solvent is its mole fraction and P° is the vapour pressure of the pure solvent at the same temperature. The other options use an incorrect sum, inverse relation, or subtraction without consistent units.

Correct answer: A. Benzene and toluene

Explanation: Benzene and toluene are chemically similar, and the attractions between unlike molecules are close to those in the two pure liquids. Their mixture therefore shows nearly ideal behaviour and approximately follows Raoult's law. Acetone–chloroform has strong specific attraction, while ethanol–water and HCl–water show significant non-ideal effects.

Correct answer: A. Because salt lowers the vapour pressure of water

Explanation: For a sufficiently dilute solution, dissolved salt behaves as a non-volatile solute and lowers the mole fraction and vapour pressure of water. Boiling requires vapour pressure to equal external pressure, so the salt solution must be heated to a higher temperature. This is boiling-point elevation, not conversion of water to gas or removal of external pressure.

Correct answer: A. When rates of evaporation and condensation become equal

Explanation: In a closed container, evaporation transfers molecules from liquid to vapour while condensation returns molecules to the liquid. Initially evaporation may dominate, but vapour accumulation increases condensation until the two rates become equal. This dynamic equilibrium gives a constant vapour pressure at fixed temperature, even though both processes continue microscopically.

Correct answer: A. Sum of both partial vapour pressures

Explanation: In an ideal solution, each volatile component obeys Raoult's law: p_A = x_A P°_A and p_B = x_B P°_B. Dalton's law then gives P_total = p_A + p_B. Both components contribute, even if one is more volatile than the other; the more volatile component does not replace the contribution of the less volatile one.

Correct answer: A. It applies over the entire concentration range

Explanation: An ideal solution has similar intermolecular interactions between unlike molecules and between like molecules. Consequently, each component's partial vapour pressure remains proportional to its mole fraction throughout the composition range, not merely at infinite dilution. Thus Raoult's law applies over the entire concentration range for an ideal solution at a fixed temperature.

Correct answer: A. Nearly similar

Explanation: An ideal solution is one in which interactions between unlike molecules, A–B, are approximately equal to interactions between like molecules, A–A and B–B. Because mixing does not cause a major change in intermolecular forces, each component follows Raoult’s law over the relevant composition range. Attractions are not zero and need not be ionic; they are simply nearly similar in strength.

Correct answer: A. Greater than expected value

Explanation: Positive deviation means that the observed partial or total vapour pressure is greater than the value predicted by Raoult's law. This commonly occurs when unlike-molecule attractions are weaker than the average like-molecule attractions, allowing particles to escape more easily. It does not mean pressure is independent of composition or necessarily infinite.

Correct answer: A. Less than expected value

Explanation: Negative deviation means that the measured vapour pressure is lower than the ideal value calculated using Raoult's law. Stronger unlike-molecule attractions hold particles more firmly in the liquid, reducing their escape tendency. The pressure is not necessarily zero or equal to external pressure; it is simply below the corresponding ideal prediction at that composition and temperature.

Correct answer: A. Positive deviation

Explanation: If the new unlike-molecule attractions are weaker than the original like-molecule attractions, particles are held less strongly after mixing. Their escaping tendency and vapour pressure become higher than Raoult's-law predictions, giving positive deviation. The mixture still has vapour; weaker attraction does not imply no vapour, and it does not guarantee ideal behaviour.