4: Vapour Pressure

Correct answer: A. 6 kPa

Explanation: The solute mole fraction is x solute = 1 − 0.92 = 0.08. For a non-volatile solute, the lowering is Δp = x solute p° solvent. Hence Δp = 0.08 × 75 = 6 kPa. Equivalently, the solution pressure is 0.92 × 75 = 69 kPa and the lowering is 75 − 69 = 6 kPa. Thus 69 kPa is the solution pressure, not the lowering.

Correct answer: A. straight line

Explanation: For an ideal binary solution, p A = x A p° A and p B = (1 − x A)p° B. Each partial pressure therefore changes linearly with liquid composition. Their sum, P = p A + p B, is also a linear function of x A, joining the pure-B and pure-A vapour-pressure values. Non-ideal solutions may show curved positive or negative deviations instead.

Correct answer: A. more molecules get enough energy to leave the surface

Explanation: When temperature rises, the average kinetic energy of liquid molecules increases and the energy distribution broadens. A larger fraction of molecules can overcome intermolecular attractions and escape from the surface. In a closed vessel, more escaping molecules produce a higher equilibrium vapour pressure. Molecular mass does not continuously change, and solvent mole fraction is irrelevant for a pure liquid unless a solution is being considered.

Correct answer: A. the liquid is at boiling condition

Explanation: The defining condition for boiling is that the liquid’s equilibrium vapour pressure equals the pressure acting on its surface. At this point vapour bubbles can form throughout the liquid and grow instead of collapsing. Equality alone does not imply freezing, zero temperature or absence of vapour; the actual boiling temperature depends on the liquid and the external pressure.

Correct answer: A. 0.40

Explanation: For a gas mixture, Dalton’s law relates vapour-phase mole fraction to partial pressure: y A = p A/P total. Substitution gives y A = 36/90 = 0.40. The number 0.36 is the partial pressure expressed without its unit, not the mole fraction, and 2.50 is impossible because a mole fraction must lie between zero and one. The liquid mole fraction cannot be inferred without more data.

Correct answer: A. nearly ideal solution

Explanation: An ideal solution is one in which the interactions between unlike molecules are approximately equal to those between like molecules. Replacing an A–A or B–B contact with an A–B contact then causes little change in the escaping tendency, enthalpy or volume. The vapour pressures consequently follow Raoult’s law closely. Stronger unlike attraction would give negative deviation, while weaker attraction would give positive deviation.

Correct answer: A. it may become lower than expected

Explanation: Very strong unlike-molecule attraction stabilises the mixed liquid phase and makes escape into vapour more difficult. The partial pressures therefore become lower than the values predicted by Raoult’s law, giving negative deviation. The pressure does not have to become zero, because some molecules still enter the vapour. Also, total pressure is the sum of non-negative partial pressures and cannot be less than an individual partial pressure.

Correct answer: A. because it is ratio of two similar pressures

Explanation: Relative lowering is defined as (p° − p)/p°, where both the numerator and denominator are pressures measured in the same unit. Dividing one pressure by another cancels units such as pascal or kilopascal, leaving a pure number. It may be written as a decimal or percentage. It is not itself a mass, temperature or volume, so those options do not explain its unitless nature.

Correct answer: A. 0.80

Explanation: A 20% lowering means (p° − p)/p° = 0.20. For a non-volatile solute, this relative lowering equals x solute, so x solute = 0.20. Since the solvent and solute mole fractions sum to one, x solvent = 1 − 0.20 = 0.80. A mole fraction cannot be 1.20 or 20, and 0.20 refers to the solute, not the solvent.

Correct answer: A. because surface area may affect rate but not equilibrium pressure

Explanation: Increasing surface area can increase the initial rate at which molecules leave the liquid, so equilibrium may be reached more quickly. In a closed vessel at fixed temperature, however, the final equilibrium condition requires the rates of evaporation and condensation to be equal. The pressure at that condition is determined by temperature and liquid nature, not surface area. Evaporation and condensation do not become impossible.

Correct answer: A. higher

Explanation: Volatility is the tendency of a substance to pass from liquid to vapour. At the same temperature, a more volatile liquid has molecules that escape more readily, so its equilibrium vapour pressure in the pure state is higher. Pure vapour pressure is a pressure, whereas mole fraction is a dimensionless composition quantity; they cannot generally be equal. A zero vapour pressure would indicate no appreciable vapour formation.

Correct answer: A. 100 kPa

Explanation: For a mixture of vapours, Dalton’s law states that the total pressure equals the sum of the partial pressures of all components. Therefore P = p A + p B = 55 + 45 = 100 kPa. Neither individual partial pressure is the total pressure, and 10 kPa is their difference rather than their sum. The ideal-solution condition is consistent with using the calculated partial pressures.

Correct answer: A. 0.25

Explanation: Relative lowering is defined as the lowering divided by the vapour pressure of the pure solvent: Δp/p° = 24/96. Simplifying gives 0.25, a dimensionless quantity, or 25% if expressed as a percentage. The value 0.75 is the corresponding solvent mole fraction for a non-volatile solute, while 24 is the absolute lowering and not the relative value.

Correct answer: A. 0.24

Explanation: For a binary solution containing a non-volatile solute, x solute = 1 − x solvent = 1 − 0.76 = 0.24. Raoult’s law then gives relative lowering (p° − p)/p° = x solute = 0.24. The value 0.76 is the solvent mole fraction itself, while 1.76 is impossible for a mole fraction and 0.52 has no basis in the given data.

Correct answer: A. at fixed temperature equilibrium vapour pressure depends on nature of liquid

Explanation: At a fixed temperature, the equilibrium vapour pressure of a liquid depends on the nature of the liquid, especially the strength of its intermolecular attractions. It is not equal to the amount of liquid and is not determined only by vessel height. Vapour pressure is directly related to boiling point: boiling occurs when it equals external pressure, so a lower vapour pressure generally corresponds to a higher boiling point at the same external pressure.

Correct answer: A. 60 kPa

Explanation: For an ideal solution with a non-volatile solute, Raoult’s law gives p solution = x solvent p° solvent. Substituting the values, p solution = 0.75 × 80 = 60 kPa. The value 20 kPa is the lowering, 80 kPa is the pressure of pure solvent, and 75 is the mole fraction written as a number rather than a pressure. The given mole fraction belongs to the solvent, so it is used directly.

Correct answer: C. 0.25

Explanation: For a solution containing a non-volatile solute, the relative lowering of vapour pressure equals the mole fraction of the solute: x_solute = (P° − P)/P°. Here, the lowering is 120 − 90 = 30 kPa. Therefore, x_solute = 30/120 = 0.25. Thus, option C is correct. The value 0.75 is the solvent mole fraction, not the solute mole fraction.

Correct answer: B. The escaping tendency of solvent molecules from the surface decreases

Explanation: A non-volatile solute does not itself contribute appreciably to the vapour phase. Its presence lowers the mole fraction and escaping tendency of solvent molecules, so fewer solvent molecules enter the vapour at equilibrium. This is a physical effect described by Raoult’s law, not a change in solvent mass, formula, or external pressure.

Correct answer: B. More molecules gain enough energy to enter the vapour phase

Explanation: Raising temperature increases the average kinetic energy of liquid molecules. A larger fraction can overcome intermolecular attractions and enter the vapour phase. The new equilibrium is established at a higher vapour pressure. Molecular mass, the name of the liquid, and solvent mole fraction do not provide this temperature effect.

Correct answer: B. 40 kPa

Explanation: Because the solute is non-volatile, only the solvent contributes to the vapour pressure. The solvent mole fraction is x_solvent = 1 − x_solute = 1 − 0.2 = 0.8. By Raoult’s law, P_solution = x_solvent P°_solvent = 0.8 × 50 = 40 kPa. Hence, option B is correct. The pressure is lower than 50 kPa because adding a non-volatile solute reduces the fraction of solvent molecules at the surface.

Correct answer: B. It obeys Raoult's law at all concentrations

Explanation: An ideal solution obeys Raoult’s law over the entire composition range, not merely at infinite dilution. Its A–B interactions are approximately comparable to A–A and B–B interactions, so mixing has nearly zero enthalpy and volume change. Positive deviation and heat release are non-ideal behaviours, so B is the only correct statement.

Correct answer: B. Total vapour pressure increases

Explanation: The more volatile component has the larger pure-component vapour pressure, say p_A° > p_B°. For an ideal solution, P = x_A p_A° + (1 − x_A)p_B°. Increasing x_A replaces some contribution from the lower-pressure component B with the higher-pressure contribution from A, so total pressure rises. It does not become zero or composition-independent; those choices contradict Raoult’s law.

Correct answer: B. 0.6

Explanation: For a binary vapour mixture, the partial pressure of B is p_B = P − p_A = 100 − 40 = 60 kPa. Dalton’s law relates vapour mole fraction to partial pressure: y_B = p_B/P. Hence y_B = 60/100 = 0.60. Option A is the mole fraction of A in the vapour, while 1.4 is impossible because a mole fraction must lie between zero and one.

Correct answer: B. When the solute mole fraction is nearly zero

Explanation: For a non-volatile solute, Raoult’s law gives p_solution = x_solvent p°_solvent. If the solute mole fraction is nearly zero, x_solvent is nearly one, so p_solution is nearly p°_solvent. A high solute fraction substantially lowers pressure, and zero solvent fraction leaves no solvent vapour pressure. Whether the solute is ionic alone does not determine this conclusion without concentration and dissociation information.

Correct answer: A. Component A

Explanation: At equal liquid mole fractions, Raoult’s law gives p_A = x_Ap_A° and p_B = x_Bp_B°. Since x_A = x_B but p_A° < p_B°, the partial pressure of A is lower. In the vapour phase, y_i = p_i/P, so A also has the smaller vapour-phase mole fraction. B is more volatile and therefore becomes enriched in the vapour; equal liquid fractions do not imply equal vapour fractions.