4: Vapour Pressure

Correct answer: A. Vapour pressure

Explanation: In a closed container, some liquid molecules escape from the surface and form vapour. After dynamic equilibrium is reached, the vapour molecules collide with the container and liquid surface, exerting a pressure at that temperature. This equilibrium pressure is called vapour pressure. Osmotic pressure concerns solvent flow through a semipermeable membrane, not vapour above a liquid.

Correct answer: A. It increases

Explanation: When temperature rises, the average kinetic energy of liquid molecules increases. Consequently, more molecules acquire enough energy to escape from the liquid surface and enter the vapour phase. This increases the number of vapour molecules above the liquid and therefore raises the equilibrium vapour pressure. It does not become zero or remain constant under ordinary conditions.

Correct answer: A. Number of solvent particles at the surface decreases

Explanation: A non-volatile solute has negligible vapour pressure and does not escape appreciably into the gas phase. Its presence lowers the mole fraction and availability of solvent molecules at the surface. Thus fewer solvent molecules evaporate, so the solvent vapour pressure becomes lower than that of the pure solvent.

Correct answer: A. Mole fraction of the solvent

Explanation: For an ideal solution containing a non-volatile solute, Raoult's law is written as P_solvent = X_solvent P°_solvent. At fixed temperature, the pure-solvent pressure P° is constant, so the solution vapour pressure is directly proportional to the solvent mole fraction. Colour, container shape, and smell do not appear in this relation.

Correct answer: A. P = X_solvent × P°

Explanation: For a solution containing a non-volatile solute, Raoult’s law states that the partial vapour pressure of the solvent is P = X_solvent P°. Here, X_solvent is the mole fraction of the solvent and P° is the vapour pressure of the pure solvent at the same temperature. Thus multiplication is required; the addition, subtraction, and division forms are incorrect.

Correct answer: A. Lower vapour pressure

Explanation: A non-volatile solute has negligible vapour pressure of its own, so it does not significantly enter the vapour phase. Its presence reduces the mole fraction of the solvent and decreases the number of solvent molecules escaping from the surface. By Raoult’s law, P_solution = X_solvent P°; because X_solvent is less than 1, the solution vapour pressure is lower than P°.

Correct answer: A. The fraction of water molecules available at the surface decreases

Explanation: Dissolved sugar is treated as a non-volatile solute under ordinary conditions. It occupies part of the solution and reduces the mole fraction of water, so fewer water molecules are available to escape per unit surface in equilibrium. Consequently, water's vapour pressure falls. Sugar is not more volatile, and attractions do not disappear.

Correct answer: A. A closed container

Explanation: Vapour pressure is the pressure exerted by vapour in equilibrium with its liquid at a specified temperature. A closed container is needed so that vapour can accumulate and condensation can balance evaporation. In an open or porous container, vapour escapes continuously, so a stable equilibrium pressure cannot normally be established or measured.

Correct answer: A. More volatile

Explanation: Vapour pressure measures the tendency of molecules to escape from a liquid and exist in the vapour phase at a given temperature. A high vapour pressure means that many molecules can escape relatively easily. Such a liquid is called more volatile and usually evaporates readily. A low vapour pressure indicates stronger resistance to escape and lower volatility, not higher volatility.

Correct answer: A. Low

Explanation: Strong intermolecular attraction holds molecules more firmly in the liquid phase. At a given temperature, fewer molecules possess enough energy to escape into the vapour phase, so the equilibrium vapour pressure is lower. The pressure is not necessarily zero, because some molecules can still escape; “low,” rather than “always zero,” is the correct conclusion.

Correct answer: A. High

Explanation: Intermolecular attractions hold liquid molecules together. When these attractions are weak, a molecule needs less energy to leave the liquid surface and enter the vapour phase. Therefore, at the same temperature, more molecules escape and the equilibrium vapour pressure is generally higher. Strong attractions would make escape more difficult and usually produce lower vapour pressure.

Correct answer: A. When the liquid vapour pressure becomes equal to external pressure

Explanation: Boiling occurs throughout the liquid when vapour bubbles can form and persist. This becomes possible when the liquid's vapour pressure equals the pressure exerted by the surroundings. Surface evaporation can occur below this condition, so merely seeing vapour or using an open container does not define boiling. Colour and mass are not the criterion.

Correct answer: A. Its vapour pressure is lower at the same temperature

Explanation: A liquid boils when its vapour pressure reaches the external pressure. If its vapour pressure is lower than that of another liquid at the same temperature, it must be heated more to reach the same external pressure, giving a higher boiling point. The liquid still forms vapour; infinite pressure and absence of particles are impossible claims.

Correct answer: A. Because its vapour pressure decreases

Explanation: A non-volatile solute lowers the solvent mole fraction and therefore lowers the solution vapour pressure at a given temperature. The solution must then be heated to a higher temperature before its vapour pressure equals the external pressure. That higher temperature is the elevated boiling point. The solute is not the component escaping as vapour.

Correct answer: A. Lower

Explanation: Dissolved salt is treated as a non-volatile solute in this context. It contributes particles to the liquid phase but not appreciable vapour, reducing the mole fraction and escaping tendency of water. Thus, at the same temperature, salt water has a lower vapour pressure than pure water. The amount of lowering depends on solution concentration.

Correct answer: A. The vapour pressure of the pure solvent

Explanation: In Raoult’s law, the symbol P° denotes the vapour pressure of the pure solvent at the specified temperature. The vapour pressure of the solvent in the solution is written as P = X_solvent P°. Since the solvent mole fraction is normally below one after adding solute, the solution pressure is usually lower than P°. P° is not a mass, colour, or container-volume quantity.

Correct answer: A. Mole fraction of the solvent

Explanation: The symbol X_solvent denotes the mole fraction of the solvent, calculated as moles of solvent divided by total moles of all components in the solution. In Raoult's law, decreasing this fraction decreases the solvent's partial vapour pressure at fixed temperature. It is not a temperature, pressure-only, or colour symbol.

Correct answer: A. It decreases

Explanation: Raoult's law gives P_solvent = X_solvent P°_solvent. At constant temperature, P°_solvent is fixed, so the solvent pressure changes in the same direction as X_solvent. A decrease in mole fraction therefore causes a proportional decrease in vapour pressure. The result is not automatically zero or double; the amount of change depends on the numerical change in X.

Correct answer: A. Colligative property

Explanation: A colligative property depends primarily on the number of dissolved solute particles, not on their chemical identity, under the stated ideal conditions. Lowering of vapour pressure is one such property. For electrolytes, dissociation changes the effective particle number, but colour, shape and odour do not define the effect.

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

Explanation: For comparable solutions at the same temperature, lowering of vapour pressure is governed by the number of non-volatile solute particles. More such particles produce a larger solute mole fraction and a larger relative lowering. Pure solvent and a solute-free sample show no solute-caused lowering; container size alone is irrelevant.

Correct answer: A. Vapour pressure decreases further

Explanation: Adding more non-volatile solute increases the solute mole fraction and decreases the solvent mole fraction. From p = x_solvent p° at constant temperature, the solvent vapour pressure therefore falls further. The effect is not an increase or infinity; the option about external pressure is also unrelated to this concentration change.

Correct answer: A. Evaporation

Explanation: Evaporation is the process in which molecules escape from the surface of a liquid and enter the vapour phase. It can occur below the boiling point because some surface molecules possess enough kinetic energy to escape. Freezing changes a liquid into a solid, condensation changes vapour into liquid, and solution formation refers to mixing substances rather than surface vapour formation.

Correct answer: A. Rates of evaporation and condensation

Explanation: In a closed vessel, molecules continuously leave the liquid by evaporation and return from vapour by condensation. At dynamic equilibrium, these opposite rates become equal, so the amounts of liquid and vapour and the pressure remain constant. The processes in the other options are not the paired phase changes defining vapour equilibrium.

Correct answer: A. Nature of liquid and temperature

Explanation: At equilibrium, vapour pressure is determined mainly by temperature and the nature of the liquid, which includes the strength of intermolecular attractions. Heating changes molecular energies, while different liquids have different escape tendencies. Container colour, word length and the observer do not determine the thermodynamic vapour pressure.

Correct answer: A. Because its particles easily become vapour

Explanation: At a fixed temperature, higher vapour pressure indicates a greater equilibrium tendency for molecules to occupy the vapour phase. This usually reflects weaker effective intermolecular attraction or greater volatility, so surface molecules escape more readily and evaporation is faster. The other statements contradict the molecular picture of a liquid.