4: Vapour Pressure

Correct answer: A. It will increase

Explanation: For component A in an ideal solution, Raoult’s law gives pA = xA pA°, where pA° remains fixed if temperature is fixed. Thus increasing xA increases pA in direct proportion. The pressure does not remain zero or become independent of composition. In a binary solution, increasing xA generally also decreases xB, but that does not alter the direct conclusion for A.

Correct answer: A. It has high vapour pressure

Explanation: At the same temperature, a liquid with higher equilibrium vapour pressure has a greater tendency to enter the gas phase and is therefore more volatile. It usually has weaker effective intermolecular attraction and a lower boiling point than a less volatile liquid. Zero vapour pressure and inability to form vapour describe the opposite behaviour, not volatility.

Correct answer: A. It will decrease

Explanation: At constant temperature, Raoult’s law gives psolvent = xsolvent p°solvent. Since the pure-solvent pressure is fixed, the partial pressure changes directly with the solvent mole fraction. Reducing xsolvent therefore reduces psolvent. A physical vapour pressure cannot become negative; it also does not remain unchanged unless the mole fraction remains unchanged.

Correct answer: A. Negative deviation

Explanation: The ideal value is the pressure predicted by Raoult’s law for the same composition and temperature. If the measured pressure is lower, the mixture shows negative deviation. This generally indicates stronger unlike interactions, which hold molecules in the liquid more effectively. Positive deviation would mean a pressure higher than the predicted value, while no deviation means equality.

Correct answer: A. Positive deviation

Explanation: Compare the observed total or partial pressure with the Raoult-law prediction at the same composition and temperature. If the observed pressure is higher, the deviation is positive. Weaker unlike attractions allow more molecules to escape into vapour. Negative deviation would give a lower pressure, while complete condensation and absence of mole fraction are not descriptions of this comparison.

Correct answer: A. Because it decides how easily molecules escape into vapour

Explanation: Vapour pressure is produced by molecules escaping from the liquid surface. If unlike-molecule attraction is stronger than expected, escape is harder and the pressure is lower, giving negative deviation. If it is weaker, escape is easier and the pressure is higher, giving positive deviation. Ideal behaviour occurs when unlike and like attractions are approximately equal.

Correct answer: A. Vapour pressure will decrease

Explanation: A strong new solvent–solute or unlike-molecule attraction stabilises the liquid phase. Molecules then require more energy to escape, so fewer enter the vapour phase at the same temperature. The observed vapour pressure falls below the Raoult-law prediction, giving negative deviation. It cannot become infinite or independent of temperature merely because the attraction is strong.

Correct answer: A. It becomes constant when rates of vaporisation and condensation are equal in a closed container

Explanation: Equilibrium vapour pressure is established in a closed system when evaporation and condensation continue at equal rates. Because the opposing rates are equal, the amount of vapour and the pressure remain constant even though molecules are still moving dynamically. An open container cannot generally maintain this equilibrium, and colour or solid-state behaviour does not define vapour pressure.

Correct answer: A. At the surface of the liquid

Explanation: Evaporation occurs from the liquid surface. Solute particles occupy part of the surface and, more fundamentally, reduce the mole fraction and probability of solvent molecules being present at the surface. Fewer solvent molecules can escape per unit time, so the equilibrium vapour pressure falls. The outer wall or an imagined separate layer is not the relevant location.

Correct answer: A. Vapour pressure will be higher than ideal value

Explanation: Weak unlike attraction means that solvent and solute molecules are held together less strongly after mixing than the ideal model assumes. Molecules therefore escape into the vapour phase more readily, giving a vapour pressure above the Raoult-law value. This is positive deviation. The pressure is still a meaningful, temperature-dependent physical quantity and does not become zero.

Correct answer: A. Vapour pressure will be lower than ideal value

Explanation: Strong solvent–solute attraction stabilises the mixed liquid and makes escape into the vapour phase more difficult. Consequently, fewer molecules contribute to vapour at a given temperature, and the observed pressure is lower than the Raoult-law prediction. This is negative deviation and may be associated with a maximum-boiling azeotrope. Colour is unrelated, and the pressure does not necessarily increase.

Correct answer: A. The more volatile component

Explanation: The more volatile component has the higher pure-component vapour pressure and therefore contributes a larger partial pressure at a given liquid composition. Since vapour-phase mole fraction is proportional to partial pressure, that component is enriched in the vapour compared with the liquid. This principle underlies fractional distillation. The less volatile component is relatively enriched in the liquid residue.

Correct answer: A. Vapour pressure of a component in solution changes with its mole fraction

Explanation: Raoult’s law states that, for an ideal solution, the partial vapour pressure of a component equals its mole fraction multiplied by its pure vapour pressure: pi = xi pi°. Thus composition directly controls the component’s contribution to the vapour phase at fixed temperature. The law is not based on colour, does not make every pressure zero, and says nothing about all liquids freezing at the same temperature.

Correct answer: B. 66 kPa

Explanation: For a solution containing a non-volatile solute, Raoult’s law gives p = x_solvent p°_solvent. The solvent mole fraction is 1 − 0.12 = 0.88. Therefore, p = 0.88 × 75 = 66 kPa. The pressure is lower than 75 kPa because the solute reduces the fraction of solvent molecules escaping into the vapour. Thus, option B is correct; 9 kPa is only the lowering, not the final pressure.

Correct answer: B. Strong attraction between unlike molecules

Explanation: A vapour pressure lower than the Raoult-law prediction represents a negative deviation. It occurs when unlike molecules, such as A–B pairs, attract one another more strongly than the A–A and B–B interactions in the pure liquids. These molecules escape less readily from the liquid, so the vapour pressure falls. Weak unlike attraction would instead favour positive deviation. Hence option B is correct.

Correct answer: A. An increase in volume is possible

Explanation: Positive deviation means that unlike-molecule attractions are weaker than the average attractions in the pure components. On mixing, the molecules may therefore pack less efficiently and remain slightly farther apart. This can produce a positive volume change, or expansion, although it is not an unavoidable result for every positive-deviation system. Thus “an increase in volume is possible” is the precise statement. Options B and C use unjustified absolute claims.

Correct answer: B. Heat may be released

Explanation: Negative deviation generally arises when unlike molecules attract more strongly than like molecules. Formation of these stronger interactions is energetically favorable and commonly releases heat, so mixing is exothermic. Zero heat change is characteristic of an ideal solution, whereas heat absorption is associated more often with positive deviation.

Correct answer: A. 0.12

Explanation: For a solution with a non-volatile solute, the relative lowering of vapour pressure equals the solute mole fraction: (p° − p)/p° = x_solute. A lowering of 12% is 12/100 = 0.12. Therefore x_solute = 0.12. The value 0.88 is the solvent mole fraction, not the solute mole fraction, while 12 and 1.12 cannot be mole fractions because a mole fraction is dimensionless and lies between 0 and 1.

Correct answer: A. As long as some liquid remains

Explanation: At fixed temperature, a liquid and its vapour establish a dynamic equilibrium. If some liquid remains, evaporation and condensation adjust until the same saturated vapour pressure is restored, even after a small amount is removed. Once all liquid is gone, the vapour is no longer saturated and pressure can depend on amount and volume.

Correct answer: A. Its pure vapour pressure and its liquid-phase mole fraction

Explanation: Raoult’s law for component i is p_i = x_i p_i°. Here x_i is the mole fraction of that component in the liquid phase, and p_i° is its vapour pressure in the pure state at the same temperature. Thus the partial pressure is the product of pure vapour pressure and liquid mole fraction. Option B resembles Dalton’s relation for vapour composition, but it is not the direct Raoult-law expression.

Correct answer: A. 54 kPa

Explanation: For an ideal solution, Raoult’s law gives the partial pressure of A as p_A = x_A p_A°. Substituting the given values gives p_A = 0.6 × 90 kPa = 54 kPa. The result is below the pure vapour pressure because only 60% of the liquid particles are A. The value 36 kPa is obtained from the complementary fraction 0.4, while 90 kPa would require x_A = 1. Therefore, option A is correct.

Correct answer: A. Unlike molecular attraction is comparatively weak

Explanation: A point above the ideal pressure-composition line represents positive deviation from Raoult’s law. In such a mixture, unlike molecules attract one another less strongly than the original like molecules. Their escape tendency therefore increases, raising the total vapour pressure. Strong unlike attraction would place the curve below the ideal line.

Correct answer: B. Unlike molecular attraction is strong

Explanation: A pressure below the ideal line is negative deviation. The unlike molecules then attract each other more strongly than the like molecules they replace. This stronger bonding lowers the escaping tendency and hence lowers vapour pressure. The other numerical or volatility statements do not explain the graph’s negative deviation.

Correct answer: B. 0.82

Explanation: For a solution containing a non-volatile solute, the relative lowering of vapour pressure equals the solute mole fraction. An 18% lowering therefore gives x_solute = 0.18. Since the mole fractions of solvent and solute add to one, x_solvent = 1 − 0.18 = 0.82. The value 0.18 belongs to the solute, not the solvent; 1.18 and 18 are impossible mole fractions. Thus option B is correct.

Correct answer: A. Boiling condition

Explanation: Boiling begins when the vapour pressure of a liquid becomes equal to the pressure exerted on its surface by the surroundings. At this point, vapour bubbles can form throughout the liquid rather than only evaporating at the surface. Freezing concerns the solid–liquid transition, precipitation concerns formation of a solid from a solution, and ideality concerns intermolecular interactions. Therefore, pressure equality identifies the boiling condition, option A.