4: Vapour Pressure

Correct answer: B. It increases

Explanation: For an ideal solution, Raoult’s law gives P_i = X_i P_i°, where P_i° remains fixed when temperature and the pure-component identity are unchanged. Thus the partial pressure is directly proportional to the component’s mole fraction. If X_i increases, P_i increases in the same ratio. It cannot become zero or infinite merely because the mole fraction rises, and option A gives the opposite trend.

Correct answer: B. Positive deviation

Explanation: Deviation is identified by comparing the observed pressure with the pressure predicted by Raoult’s law. If the observed total or partial pressure is higher than the ideal value, the solution shows positive deviation. This generally reflects weaker unlike-molecule attractions, allowing molecules to escape more easily. Lower-than-expected pressure would instead indicate negative deviation.

Correct answer: B. New attractions being stronger

Explanation: Negative deviation means that the actual vapour pressure is lower than the value predicted for an ideal solution. This occurs when unlike molecules attract one another more strongly than the like molecules did. The stronger interactions hold particles in the liquid and reduce escape into vapour. Weak unlike attractions would generally produce positive deviation.

Correct answer: A. Because attraction between components is comparatively weaker

Explanation: In positive deviation, unlike-molecule attractions are weaker than the attractions assumed for an ideal solution. Particles are therefore held less strongly in the liquid and can escape into the vapour phase more readily. This raises the observed vapour pressure above the Raoult-law value. Stronger attractions would restrict escape and cause negative deviation.

Correct answer: A. It increases

Explanation: A liquid boils when its vapour pressure becomes equal to the external pressure. Adding a non-volatile solute lowers the vapour pressure at every given temperature. Consequently, the solution must be heated to a higher temperature before its vapour pressure reaches the external pressure. Therefore its boiling point rises, producing elevation in boiling point. Lowering the boiling point or claiming no relation contradicts this condition for boiling.

Correct answer: A. External pressure

Explanation: Boiling begins when vapour bubbles can form throughout the liquid and expand rather than collapse. This occurs when the liquid’s vapour pressure equals the pressure acting on its surface, called external pressure. The condition is independent of colour, container height, or solute mass as direct equality terms. Changing external pressure can, however, change the boiling temperature.

Correct answer: A. It decreases

Explanation: Boiling begins when the vapour pressure of a liquid equals the pressure exerted on its surface. If the external pressure is reduced, this equality is reached at a lower temperature because the liquid does not need to develop as high a vapour pressure. Therefore the boiling point decreases, which explains why liquids boil at lower temperatures at high altitude. The other choices do not follow the pressure-equality condition.

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

Explanation: A colligative property depends primarily on the number of dissolved solute particles relative to the solvent particles, not on the chemical identity of those particles. For a non-volatile solute, relative lowering is Xsolute, so increasing the effective particle count increases the lowering. Colour, container design, and the name of the solvent do not define this property.

Correct answer: A. Liquid A is more volatile

Explanation: At a fixed temperature, higher vapour pressure means that a liquid has a greater tendency for molecules to escape into the vapour phase. Therefore, liquid A is more volatile than liquid B. Usually, the more volatile liquid also has a lower normal boiling point, so equal boiling points are not required. A higher-pressure liquid certainly can form vapour.

Correct answer: B. Low

Explanation: Strong intermolecular attractions hold liquid molecules more firmly and make it harder for them to escape from the surface. Consequently, fewer molecules enter the vapour phase at a given temperature, so the equilibrium vapour pressure is lower. It is not necessarily zero, because some molecules can still escape. Mole fraction alone is not a pressure value.

Correct answer: A. More particles get enough energy to leave the surface

Explanation: Increasing temperature raises the average kinetic energy of liquid molecules and broadens the energy distribution. As a result, a larger fraction of molecules has enough energy to overcome intermolecular attractions and escape from the surface. More molecules in the vapour phase produce a higher equilibrium vapour pressure. Temperature does not stop motion or remove particle mass.

Correct answer: A. Rates of evaporation and condensation

Explanation: In a closed container, liquid molecules continuously evaporate and vapour molecules continuously condense back into the liquid. At dynamic equilibrium, the rate of evaporation equals the rate of condensation, so the macroscopic vapour pressure remains constant. The processes do not stop; only their opposing rates become equal. The other pairs are unrelated phase processes.

Correct answer: A. Vapour keeps escaping out

Explanation: Equilibrium vapour pressure requires a closed space in which vapour remains above the liquid and can condense back. In an open container, escaping vapour is continually removed, so the condensation rate cannot balance evaporation in the same way. The liquid still contains particles and can evaporate, and vapour pressure remains temperature-dependent; the problem is the open system.

Correct answer: A. Vapour pressure increases as temperature increases

Explanation: In the stated graph, temperature is the independent variable and vapour pressure is the measured quantity. An upward trend means that larger temperature values correspond to larger vapour-pressure values. This agrees with the physical idea that heating gives more liquid molecules enough energy to escape into the vapour phase. A downward line would indicate decrease, whereas a horizontal line would indicate no change in the plotted range.

Correct answer: A. Temperature

Explanation: Vapour pressure is the equilibrium pressure of vapour above a liquid at a specified temperature. It changes significantly with temperature, so the vapour pressure of a solution can be compared fairly with that of the pure solvent only when both are measured at the same temperature. Colour, container name, and odour are not the controlling thermodynamic condition. The composition may differ because that difference is the subject of the comparison.

Correct answer: A. 120

Explanation: For a solution containing a non-volatile solute, Raoult’s law states P_solution = X_solvent P°_solvent. Substituting the given values gives P_solution = 0.6 × 200 = 120 units. Hence option A is correct. The value 200 is the pressure of the pure solvent before adding solute, while 140 and 60 do not result from the law. The solution pressure is lower because the solvent mole fraction is less than one.

Correct answer: A. Solvent

Explanation: A non-volatile solute has negligible vapour pressure under the conditions considered, so it contributes essentially no vapour particles. The solvent is the component that evaporates and supplies the vapour above the solution. Thus the solution pressure is the solvent’s partial pressure, reduced from P° according to its mole fraction. The container and outside air are not vapour-pressure components.

Correct answer: A. Both components can contribute partial pressures

Explanation: A volatile component can escape from the liquid phase and contribute molecules to the vapour phase. Therefore, when both components A and B are volatile, each has a partial vapour pressure, and the total pressure is their sum: P_total = P_A + P_B. Calling one component only a solvent does not prevent it or the other component from vaporising. The alternatives incorrectly treat the second component as non-volatile or make the total pressure zero.

Correct answer: A. The effective number of solute particles is greater

Explanation: Lowering of vapour pressure is a colligative property, so under comparable conditions it depends primarily on the effective number of dissolved particles. A greater lowering generally indicates a greater effective solute-particle concentration, assuming the solvent, temperature, and amount basis are comparable. Particle colour and location only in the vapour do not determine this effect, and zero particles would produce no solute-induced lowering.

Correct answer: A. Salt lowers vapour pressure of water

Explanation: Dissolved salt is treated as a non-volatile solute in the usual idealized explanation. It lowers the mole fraction and vapour pressure of water. Since boiling requires vapour pressure to equal external pressure, the salt solution must be heated to a higher temperature to reach that condition. This is boiling-point elevation; salt does not increase vapour pressure or make external pressure zero.

Correct answer: A. It may be higher

Explanation: Boiling occurs when vapour pressure reaches the external pressure. A liquid with lower vapour pressure at a given temperature must be heated more to reach that same pressure. Thus it generally has a higher boiling point under the same external pressure. The exact comparison requires comparable conditions, so “may be higher” is the suitably cautious option.

Correct answer: A. Higher temperature must be supplied

Explanation: The boiling condition is P vapour = P external. Increasing the external pressure means that the liquid must develop a higher vapour pressure before boiling can begin. Since vapour pressure increases with temperature, a higher temperature is required. Consequently, the boiling point rises. Neither making vapour pressure zero nor removing mole fraction represents the physical condition for boiling.

Correct answer: A. Whether the solute is volatile or non-volatile

Explanation: The first conceptual decision is whether each component can enter the vapour phase. A non-volatile solute contributes negligible vapour and lowers the solvent pressure, whereas a volatile component contributes its own partial pressure. This determines whether to use the single-solvent form of Raoult’s law or add partial pressures for volatile components. Colour and naming details are irrelevant.

Correct answer: A. When effective number of solute particles is maximum

Explanation: For a non-volatile solute, relative lowering of vapour pressure equals the solute mole fraction in the ideal dilute treatment, and more generally it depends on the effective particle concentration. The greatest effective number of solute particles produces the greatest reduction in solvent mole fraction and hence the maximum lowering. Pure solvent gives zero lowering, while association reduces the particle count.

Correct answer: A. The tendency is low

Explanation: Vapour pressure is the equilibrium pressure produced by molecules that escape from a liquid into the vapour phase. At the same temperature, a lower vapour pressure means fewer molecules escape or the escaping tendency is weaker. Thus the liquid is less volatile and has a lower tendency to form vapour. High vapour pressure would indicate greater volatility; colour does not provide the relevant thermodynamic measure.