4: Vapour Pressure

Because the solute is non-volatile, it contributes no vapour pressure; only the solvent contributes. Raoult’s law therefore gives psolution = xsolvent p°solvent = 0.90 × 50 kPa = 45 kPa. Option B is correct. The pure-solvent value, 50 kPa, would apply only when the solvent mole fraction is 1. The value 5 kPa incorrectly uses the complement as a multiplier.

Boiling is a bulk phenomenon that begins when the vapour pressure of a liquid becomes equal to the pressure acting on its surface. At that point, vapour bubbles can form throughout the liquid rather than collapsing under the external pressure. Thus, changing external pressure changes the boiling point. Zero vapour pressure is not required, and liquids can boil in open containers as well.

Atmospheric pressure decreases with increasing altitude. Water boils when its vapour pressure equals the surrounding atmospheric pressure, so at a mountain location it reaches the required lower pressure at a lower temperature. Its molar mass has not changed, salt is not automatically added, and its vapour pressure does not become zero. Therefore the lower external pressure explains the lower boiling point.

In an ideal solution, the attractions between unlike molecules are approximately equal to those between like molecules. Replacing like contacts with unlike contacts therefore causes almost no net enthalpy change, so heat of mixing is approximately zero. The molecules also pack with nearly no volume change, giving ΔVmix approximately zero.

At a given temperature, higher vapour pressure means a greater tendency of molecules to leave the liquid and enter the vapour phase, so the liquid is more volatile. For comparable external pressure, a more volatile liquid generally reaches that pressure at a lower temperature and has a lower boiling point, not a higher one. The other statements contradict molecular motion.

Vapour pressure is defined as the pressure of vapour in equilibrium with its liquid at a specified temperature. In an open vessel, vapour continuously escapes, so a stable evaporation-condensation equilibrium and measurable equilibrium pressure cannot be established. A closed vessel retains vapour until the forward and reverse rates become equal.

For an ideal solution, Raoult's law gives pi = xi p°i at fixed temperature. Since the pure-component pressure p°i is unchanged, reducing the liquid mole fraction xi directly reduces the component's partial vapour pressure. The pressure does not increase independently of composition; it would become zero only if the component were absent.

The total pressure of the vapour is the sum of the partial pressures: P = 30 + 70 = 100 kPa. By Dalton’s law, the vapour-phase mole fraction of A equals its partial pressure divided by total pressure: yA = pA/P = 30/100 = 0.30. Therefore option A is correct. The liquid-phase mole fraction cannot be read directly from these pressures without additional pure-component data.

The more volatile component has a higher pure-component vapour pressure at the same temperature. In an ideal or approximately ideal mixture, its partial pressure is xi p°i, so it contributes a larger share to the vapour than its liquid fraction alone might suggest. Greater volatility, not mass, colour or failure to dissolve, causes vapour enrichment.

Very low vapour pressure means that only a small fraction of molecules can escape into the vapour at the given temperature. A reasonable molecular explanation is strong intermolecular attraction, which holds the molecules in the liquid. This does not prove that the particles are ions, and a substance with low vapour pressure is still a liquid under the stated conditions.

Since the solute is non-volatile, the vapour pressure of the solution is due only to the solvent. Applying Raoult’s law, psolution = xsolvent × p°solvent = 0.88 × 75 kPa = 66 kPa. Hence option B is correct. The pressure must be below 75 kPa because adding solute reduces the solvent mole fraction; 85.2 kPa would incorrectly increase the pure-solvent pressure.

Vapour pressure is a pressure, so it is measured in pressure units such as pascal (Pa), kilopascal (kPa), bar, or atmosphere. Among the given choices, pascal is the correct SI unit. Mole measures amount of substance, kilogram measures mass, and mole fraction is a dimensionless ratio. These quantities may occur in solution calculations but are not units of pressure.

A total pressure above the Raoult-law value is positive deviation. It indicates that unlike molecules are held less strongly than the corresponding like molecules, so mixing increases the escaping tendency of one or both components. Their partial pressures become higher than ideal predictions. Strong unlike attraction would lower pressure and produce negative deviation.

For an ideal solution containing a non-volatile solute, Raoult’s law gives the solution vapour pressure as p = xsolvent p°solvent. Substitution gives p = 0.70 × 90 kPa = 63 kPa. Thus option A is correct. The value 27 kPa is the lowering, since 90 − 63 = 27 kPa; 90 kPa would be the pressure of pure solvent, not the solution.

The lowering of vapour pressure is Δp = p° − p = 100 − 82 = 18 kPa. Relative lowering is the lowering divided by the vapour pressure of the pure solvent: Δp/p° = 18/100 = 0.18. It has no unit because it is a ratio. Thus, 18 kPa is the absolute lowering, not the relative lowering, so option A is correct.

A vapour pressure greater than the value predicted by Raoult’s law represents positive deviation. This occurs when unlike molecules attract one another less strongly than the like molecules in the pure liquids. Their escape into the vapour phase is therefore easier, increasing vapour pressure. Strong unlike-molecule attraction would instead produce negative deviation.

Ideal pressure is predicted by Raoult’s law. If the observed total pressure is lower than that prediction, the mixture shows negative deviation. Stronger unlike-molecule attraction holds molecules more firmly in the liquid and reduces their escape into vapour. It does not mean that vapour formation stops completely.

Dalton’s law states that the total pressure of a mixture of gases or vapours equals the sum of the partial pressures contributed by all components. Therefore, for a binary solution containing volatile A and B, Ptotal = pA + pB. Neither component can be ignored, and subtraction has no basis here. Hence, adding both partial vapour pressures gives the correct result.

A non-volatile solute contributes essentially no vapour of its own. Its presence lowers the mole fraction of solvent in the liquid and reduces the number or tendency of solvent molecules reaching the surface and escaping. Thus solvent partial pressure falls. No change in solvent formula or compulsory cooling is involved.

At a fixed temperature, higher vapour pressure means that a greater tendency exists for molecules to escape from the liquid surface. Such a liquid is called more volatile. Greater volatility is generally associated with weaker intermolecular attraction and a lower, not higher, boiling point. The other choices contradict this relationship or use impossible physical descriptions.

Raoult’s law states that, for an ideal solution, the partial vapour pressure of component i is Pi = xiP°i. Thus, at a fixed temperature, its partial pressure is directly proportional to its mole fraction in the liquid phase and its pure-component vapour pressure. Vessel size alone does not determine this relation, and vapour pressure is not generally zero.

For a non-volatile solute, Psolution = xsolvent P°solvent. Equality with the pure-solvent pressure requires xsolvent = 1, which means xsolute = 0. Adding any positive amount of non-volatile solute makes the solvent mole fraction less than one and lowers the pressure. Colour has no role in this equality.

Heating increases the average kinetic energy of liquid molecules. Consequently, more molecules acquire enough energy to overcome cohesive forces at the surface and enter the vapour phase. The equilibrium vapour pressure therefore rises. Molecular mass and the liquid’s name do not change merely because temperature rises, and external pressure need not decrease.

In a closed vessel, evaporation and condensation continue until their rates become equal. At a given temperature, this dynamic equilibrium has a definite vapour pressure determined mainly by the liquid’s nature. Changing the amount changes the time or volume occupied, but not the equilibrium pressure, provided some liquid remains to maintain contact with vapour.

At a fixed temperature, the more volatile component has the higher pure vapour pressure and a greater tendency to escape from the liquid. By Raoult’s law its partial pressure is xiP°i, and this larger contribution gives it a larger fraction in the vapour than in the liquid. It need not be heavier or have unit mole fraction.