4: Vapour Pressure

For a vapour mixture, the mole fraction of a component equals its partial pressure divided by the total pressure: yA = pA/P. Substituting the given values gives yA = 24/60 = 0.40. The value 0.24 is just the numerical partial pressure and 0.60 is the total pressure, not a mole fraction. Therefore, option A is correct.

A non-volatile solute contributes essentially no vapour to the gas phase. When its mole fraction increases, the solvent mole fraction decreases. By Raoult’s law, the solvent vapour pressure is P = xsolventP°solvent, so the solution pressure decreases. The temperature dependence does not disappear; the solute simply lowers the solvent’s escaping tendency.

Benzene and toluene are non-polar aromatic liquids with similar molecular size, shape and intermolecular forces. Consequently, benzene–toluene attractions are close to benzene–benzene and toluene–toluene attractions. Mixing then causes little enthalpy or volume change and the vapour pressures approximately follow Raoult’s law. “Nearly” is appropriate because real solutions are not perfectly ideal.

For an ideal solution, Raoult’s law gives Pi = xiP°i. At a fixed temperature, P°i is constant, so partial pressure varies directly and linearly with the liquid-phase mole fraction xi. The graph is a straight line passing through the origin for the partial pressure of a component. A circular, unrelated or always-constant relation does not follow the law.

A binary solution contains only solvent and solute, so their mole fractions must add to one: xsolvent + xsolute = 1. Therefore, xsolute = 1 − xsolvent = 1 − 0.60 = 0.40. A mole fraction cannot exceed one, so 1.60 is impossible, while 0.60 would incorrectly repeat the solvent fraction. Thus, option A is correct.

Because the solute is non-volatile, only the solvent contributes to vapour pressure. Raoult’s law gives Psolution = xsolventP°solvent = 0.60 × 120 = 72 kPa. Multiplying by the solute fraction would give the pressure lowering, not the solution pressure. The pure-solvent value would apply only when the solvent fraction is one.

A colligative property depends primarily on how many solute particles are present relative to the solvent, not on the chemical identity of those particles. For a dilute non-electrolyte solution, relative lowering equals the solute mole fraction. Thus equal numbers of particles produce comparable effects, whereas colour, smell and vessel shape do not determine this property.

The governing concept is Raoult’s law for a solution containing a non-volatile solute. The relative lowering of vapour pressure equals the solute mole fraction: xsolute = (p° − p)/p°. An 8% lowering gives (p° − p)/p° = 8/100 = 0.08. Therefore xsolute = 0.08. The value 0.92 is the solvent mole fraction, while 8 and 1.08 cannot be valid mole fractions. Hence, option A is correct.

A liquid boils when its vapour pressure becomes equal to the pressure acting on its surface. If external pressure is reduced, the vapour pressure needs to reach a smaller value, which happens at a lower temperature. Therefore, the boiling point decreases. Molecular mass does not suddenly change, vapour pressure does not become zero, and boiling is not caused by solidification.

A non-volatile solute lowers the solvent’s vapour pressure at every given temperature. Boiling requires the solution vapour pressure to equal the external pressure. Therefore the solution must be heated to a higher temperature before that equality is reached, producing elevation of boiling point. The solute does not need to vaporise, and colour is irrelevant.

Equilibrium vapour pressure is defined when evaporation and condensation occur in a closed space at equal rates. In an open vessel, vapour continuously disperses into the surroundings, so a stable vapour-liquid equilibrium above the liquid is not maintained. The difficulty is therefore the absence of a closed equilibrium system, not colour, solute, or inability to measure temperature.

For a non-volatile solute, Raoult’s law gives p = xsolvent p°. Since xsolvent = 1 − xsolute, the lowering is Δp = p° − p = p°xsolute. Dividing by p° gives Δp/p° = xsolute. Thus, relative lowering equals the solute mole fraction. The other statements contradict this direct derivation, so option A is correct.

A low vapour pressure at a given temperature means that molecules escape from the liquid only with difficulty, usually because cohesive forces are relatively strong. To boil at normal external pressure, the liquid must reach a temperature at which its vapour pressure equals that pressure. Therefore its normal boiling point may be comparatively high, though the statement is a general inference.

The composition and nature of both solutions are identical, so the only changing factor is temperature. Higher temperature increases the average kinetic energy of molecules and allows more molecules to escape into the vapour phase. Consequently, the equilibrium vapour pressure is higher. It is not necessarily double or any fixed multiple; the question asks only for the direction of change.

Vapour pressure is a pressure, so it is measured in pressure units such as pascal, kilopascal, bar or atmosphere. Kilopascal is therefore correct. Mole fraction is dimensionless, mole per kilogram is a molality-related unit, and gram per mole is a molar-mass unit. None of those three represents pressure.

The defining condition for boiling is equality between the liquid’s vapour pressure and the external pressure. If vapour pressure is still lower, bubbles cannot be sustained throughout the liquid against the external pressure, so normal boiling has not begun at that temperature. Evaporation at the surface can still occur; option C incorrectly confuses no boiling with no vapour formation.

Equilibrium vapour pressure requires a dynamic balance: molecules must leave the liquid by vaporisation and return by condensation at equal rates. A closed vessel allows vapour to accumulate and establish this balance above the liquid at a fixed temperature. In open air vapour escapes continuously, and a vessel containing only solid does not describe the stated liquid-vapour equilibrium.

Raoult’s law states that the partial pressure of a volatile component in an ideal solution is pA = xA pA°, where xA is its liquid-phase mole fraction. Substitution gives pA = 0.30 × 200 = 60 kPa. The pure pressure 200 kPa would apply only when xA is one; 170 kPa is not obtained by this law. Therefore, option A is correct.

The governing concept is Dalton’s law for the vapour mixture. The total vapour pressure is the sum of the partial pressures: Ptotal = 45 + 15 = 60 kPa. The vapour-phase mole fraction of B is yB = pB/Ptotal = 15/60 = 0.25. Option C, 0.75, is the vapour mole fraction of A, whereas 0.15 is only B’s partial pressure in kPa. Thus, option A is correct.

An ideal solution obeys Raoult’s law throughout the composition range, not merely at infinite dilution. Its unlike and like intermolecular attractions are approximately equal, so mixing causes nearly zero heat and volume change. Positive or negative deviation identifies non-ideal behaviour. A real solution may obey the law approximately when dilute, but that alone does not make it ideal at all concentrations.

For a non-volatile solute, only the solvent contributes to vapour pressure. The solvent mole fraction is xsolvent = 1 − 0.10 = 0.90. Applying Raoult’s law, p = xsolvent p° = 0.90 × 50 = 45 kPa. The value 5 kPa is the lowering of pressure, while 50 kPa would be the pure-solvent pressure. Hence, option A is correct.

Vapour pressure depends on how easily molecules escape from the liquid surface. Strong intermolecular attraction holds the molecules more firmly in the liquid, so fewer molecules enter the vapour phase at a given temperature. Consequently, the equilibrium vapour pressure is lower. It is not necessarily zero, and it need not always exceed external pressure; exceeding external pressure is related to boiling.

Increasing surface area can increase the rate at which molecules initially evaporate, so equilibrium may be reached faster. However, at a fixed temperature the equilibrium vapour pressure is determined by the nature and composition of the liquid. Once evaporation and condensation balance, the final pressure is unchanged, provided liquid remains present. Rate of approach and equilibrium value must be distinguished.

For a binary vapour mixture, Dalton’s law gives Ptotal = PA + PB. Rearranging, PB = Ptotal − PA. Substituting the values gives PB = 100 − 35 = 65 kPa. The two partial pressures must add to the total; 35 kPa repeats A’s pressure, 100 kPa is the total, and 135 kPa would incorrectly add the known total and partial pressure.

At a fixed temperature, the equilibrium vapour pressure of a pure liquid is characteristic of the liquid’s nature, because molecular attractions determine how readily molecules escape. For a solution, composition also matters through mole fractions and Raoult’s law. The amount of liquid does not change the equilibrium pressure as long as some liquid remains, and vapour pressure is not a molar-mass value.