4: Vapour Pressure

Correct answer: A. Lower

Explanation: Boiling begins when a liquid's vapour pressure equals the external pressure. At the same temperature, the liquid with higher vapour pressure is closer to this condition and reaches it after less heating. Therefore its boiling point is generally lower. The comparison assumes the same external pressure; boiling point is not necessarily zero or equal for both liquids.

Correct answer: A. It decreases

Explanation: The solvent mole fraction is n_solvent/(n_solvent + n_solute). Adding a nonzero amount of solute increases the denominator while the moles of solvent remain unchanged, so the fraction becomes less than its previous value. For pure solvent it is one, but after adding solute it is no longer one and can never become infinite.

Correct answer: A. Half

Explanation: Raoult’s law states P_solution = X_solvent P°_solvent. Therefore, when X_solvent = 0.5, P_solution = 0.5P°_solvent, which means the solution vapour pressure is one-half of the pure solvent vapour pressure. It is not double or four times; nor is it zero, because the solvent still has a non-zero mole fraction and can still evaporate.

Correct answer: A. It is almost none

Explanation: A non-volatile solute has an extremely low tendency to enter the vapour phase under the conditions considered. Hence its own partial vapour pressure is taken as negligible, and the measured vapour pressure is essentially due to the solvent. Its presence still lowers solvent vapour pressure by lowering the solvent mole fraction.

Correct answer: A. Total pressure is the sum of partial pressures

Explanation: Dalton's law states that, for a mixture of non-reacting gases behaving approximately ideally, the total pressure equals the arithmetic sum of the partial pressures exerted by all components. In a vapour mixture, each volatile liquid contributes one partial pressure. The law does not say that total pressure is always zero or always decreasing.

Correct answer: A. It increases

Explanation: For each volatile component in an ideal solution, Raoult's law gives P_i = X_i P°_i. At a fixed temperature, the pure-component pressure P°_i is constant, so increasing X_i increases that component's partial vapour pressure in direct proportion. The pressure does not become zero, and colour has no role in this quantitative relation.

Correct answer: A. Positive deviation

Explanation: Raoult's law gives the ideal reference vapour pressure. If experimental or actual pressure lies above that reference, the difference is called positive deviation. It indicates that molecules escape more readily than in the ideal model, often because unlike intermolecular attractions are weaker. A lower actual value would indicate negative deviation.

Correct answer: A. Negative deviation

Explanation: Deviation is judged by comparing the observed vapour pressure with the Raoult-law value. If the observed pressure is lower, the mixture shows negative deviation. This usually indicates stronger unlike-molecule attractions, which hold molecules in the liquid more effectively. Positive deviation would mean a pressure higher than the ideal prediction.

Correct answer: A. No, it mainly depends on temperature and nature

Explanation: For a pure liquid at fixed temperature, if some liquid remains and liquid-vapour equilibrium can be established, the equilibrium vapour pressure is an intensive property. Changing the amount changes the time or amount of vapour needed for equilibrium, not the final pressure. If all liquid evaporates, this conclusion no longer applies.

Correct answer: A. A slightly larger or smaller amount of liquid

Explanation: When some liquid remains present in a closed system, equilibrium vapour pressure is fixed by temperature and the nature of the liquid. Changing the amount slightly changes how long equilibrium takes to establish, but not the final pressure. If all liquid evaporates, the condition changes; the option is understood to mean a change while liquid remains present.

Correct answer: A. Vapour escapes out

Explanation: A measurable equilibrium vapour pressure requires vapour to accumulate above the liquid until evaporation and condensation rates become equal. In an open container, vapour continuously diffuses away, so the required closed-system equilibrium is not established. The liquid still contains particles and can evaporate; its temperature is not forced to zero.

Correct answer: A. It decreases

Explanation: At a specified temperature, vapour pressure reflects the escaping tendency of molecules from a liquid into the gas phase. A lower equilibrium vapour pressure means that fewer molecules can escape sufficiently to maintain the vapour phase. Therefore the liquid is less volatile and has a lower tendency to form vapour. Colour does not determine this property.

Correct answer: A. A non-volatile solute is dissolved

Explanation: At the same temperature, adding a non-volatile solute lowers the solvent mole fraction. Raoult's law then gives a lower solvent vapour pressure than for the pure solvent. Changing a name or container colour has no thermodynamic effect, and every liquid has a temperature; the decrease is not explained by those options.

Correct answer: A. More particles become vapour as temperature increases

Explanation: Increasing temperature raises the kinetic-energy distribution of liquid molecules. A greater fraction then has enough energy to escape the surface, so the equilibrium vapour pressure rises. Cooling has the opposite general effect. Heating does not stop molecular motion, and vapour pressure is neither independent of temperature nor infinite on cooling.

Correct answer: A. A non-volatile solute lowers vapour pressure

Explanation: The central idea is that a non-volatile solute lowers the solvent mole fraction and therefore lowers the solvent vapour pressure: p = x_solvent p°. This result leads to elevation of boiling point and depression of freezing point. Temperature still affects vapour pressure, and Raoult's law is a mole-fraction relation, not a colour relation.

Correct answer: A. Pressure exerted by vapour at equilibrium

Explanation: When a liquid is placed in a closed vessel, molecules evaporate while vapour molecules condense back. At dynamic equilibrium, the rate of evaporation equals the rate of condensation. The pressure exerted by the vapour at that equilibrium, at a specified temperature, is called vapour pressure. It is not the liquid's weight, volume or wall pressure.

Correct answer: A. It decreases

Explanation: A non-volatile solute does not contribute appreciably to the vapour phase. Its presence lowers the mole fraction and escaping tendency of the solvent. For an ideal solution, Raoult’s law gives psolvent = xsolvent p°solvent; since xsolvent is less than one, the solvent vapour pressure is lower than that of the pure solvent.

Correct answer: A. Lower

Explanation: A non-volatile solute does not contribute appreciably to the vapour above the solution. Its presence reduces the mole fraction and the escaping tendency of solvent molecules at the liquid surface. For an ideal solution, Raoult’s law gives p = x_solvent p°_solvent, and because x_solvent is less than 1, p is lower than the vapour pressure of the pure solvent. Therefore A is correct.

Correct answer: A. Volatile components

Explanation: A component contributes to vapour pressure only if its molecules can enter the vapour phase in appreciable amount. Each volatile component exerts a partial vapour pressure, and the total pressure is the sum of these partial pressures. A genuinely non-volatile solute contributes no significant direct vapour pressure; the container and colour do not contribute.

Correct answer: A. Solvent molecules

Explanation: The volatile solvent can escape from the liquid and establish a vapour phase. A non-volatile solute has negligible tendency to enter that phase under the stated conditions. Therefore, the vapour consists mainly of solvent molecules. The two components cannot be assumed equal, and vapour is not completely absent while solvent remains present.

Correct answer: A. Vapour pressure of the pure solvent

Explanation: In Raoult’s law for a solution containing a volatile solvent, the relation is p_solvent = x_solvent p°_solvent. Here p_solvent is the partial vapour pressure of the solvent in the solution, while p°_solvent is the vapour pressure of the pure solvent at the same temperature. Thus p° is a pressure reference, not the solute mass, solution colour, or pressure of the container. Option A is correct.

Correct answer: A. Mole fraction

Explanation: In Raoult’s law, x represents the mole fraction of the component whose partial vapour pressure is being calculated. For a solvent, p = xsolvent p°solvent. Mole fraction is the component’s moles divided by the total moles of all components, so it is dimensionless; mass, density and time have different meanings and units.

Correct answer: A. Pure water

Explanation: At the same temperature, dissolved sugar is treated as a non-volatile solute. It lowers the mole fraction and escaping tendency of water, so the water vapour pressure in the sugar solution is below the vapour pressure of pure water. Therefore pure water has the higher value; the result assumes the solution is compared at the same temperature.

Correct answer: A. Raoult's law

Explanation: For an ideal solution, each component obeys Raoult's law: its partial vapour pressure equals its mole fraction multiplied by the vapour pressure of the pure component. Ideal solutions have similar intermolecular interactions and show no significant heat or volume change on mixing. Hess's law and Newton's laws address different subjects.

Correct answer: A. Add a non-volatile solute

Explanation: Adding a non-volatile solute reduces the solvent mole fraction and therefore lowers its partial vapour pressure according to Raoult's law. Increasing temperature generally raises vapour pressure, while keeping the solvent pure does not produce lowering. Changing container colour has no thermodynamic effect on the equilibrium vapour pressure.