10 min readConcept sheet1-mark MCQ; 2-mark statement of Raoult's law; 3-mark numerical on total vapour pressure and vapour composition; also a NEET favourite

Vapour pressure of solutions and Raoult's law

For volatile liquids, each component's partial vapour pressure = mole fraction × pure vapour pressure. A non-volatile solute lowers the solvent's vapour pressure.

Raoult's lawp₁ = p₁° x₁: the partial vapour pressure of each volatile component is proportional to its mole fraction

Total vapour pressurep_total = p₁° x₁ + p₂° x₂ = p₁° + (p₂° − p₁°) x₂

Vapour compositiony₁ = p₁ ÷ p_total; the vapour is richer in the more volatile component

Link to Henry's lawRaoult's law is a special case of Henry's law in which K_H = p₁°

Vapour pressure

In a closed vessel, a liquid evaporates until its vapour is in equilibrium with the liquid. The pressure the vapour then exerts is the vapour pressure. It depends on the liquid and on the temperature. In a mixture of two volatile liquids, each one gives its own partial vapour pressure, and Dalton's law of partial pressures gives the total: p_total = p₁ + p₂.

Raoult's law for two volatile liquids

1
Component 1: p₁ = p₁° x₁, where p₁° is the vapour pressure of pure liquid 1 and x₁ its mole fraction in the liquid.
2
Component 2: p₂ = p₂° x₂.
3
Dalton's law: p_total = p₁ + p₂ = p₁° x₁ + p₂° x₂.
4
Put x₁ = 1 − x₂: p_total = p₁°(1 − x₂) + p₂° x₂.
5★
So, p_total = p₁° + (p₂° − p₁°) x₂: the total vapour pressure changes linearly with x₂, between p₁° and p₂°.
6
Mole fractions in the vapour: y₁ = p₁ ÷ p_total and y₂ = p₂ ÷ p_total.

★ key step

How to draw the vapour pressure graph of an ideal solution

Draw the x-axis for composition: x₁ = 1 (and x₂ = 0) at the left end, x₁ = 0 (and x₂ = 1) at the right end. Draw the y-axis for vapour pressure, and a second vertical axis at the right end.
Mark p₁° on the left axis and p₂° on the right axis.
Draw a straight line from p₁° on the left down to zero on the right. Label it p₁.
Draw a straight line from zero on the left up to p₂° on the right. Label it p₂.
Join p₁° to p₂° with a straight line. Label it p_total. Label both axes.

Solutions of solids in liquids

When a non-volatile solute such as glucose, urea or common salt dissolves in water, its particles take up part of the liquid surface. Fewer solvent molecules can escape, so the vapour pressure falls. Raoult's law still holds for the solvent: p₁ = p₁° x₁. Since x₁ is less than 1, p₁ is less than p₁°. The fall depends on the amount of solute, not on what it is.

Worked example: benzene and toluene

Problem

At a certain temperature, pure benzene has a vapour pressure of 90 mm Hg and pure toluene 30 mm Hg. A solution is made from 1 mol of benzene and 2 mol of toluene. Assuming ideal behaviour, find (a) the partial pressures, (b) the total vapour pressure and (c) the mole fraction of benzene in the vapour.

Working

x(benzene) = 1 ÷ 3 = 0.333; x(toluene) = 2 ÷ 3 = 0.667.
(a) p(benzene) = 90 mm Hg × 1/3 = 30 mm Hg; p(toluene) = 30 mm Hg × 2/3 = 20 mm Hg.
(b) p_total = 30 + 20 = 50 mm Hg.
(c) y(benzene) = 30 ÷ 50 = 0.60; y(toluene) = 20 ÷ 50 = 0.40.
The vapour (0.60) is richer in benzene than the liquid (0.333), because benzene is the more volatile liquid.

Answer

(a) 30 mm Hg and 20 mm Hg; (b) 50 mm Hg; (c) y(benzene) = 0.60

Raoult's law vs Henry's law

Raoult's lawFor a volatile liquid in a solution: p₁ = p₁° x₁. The constant is the vapour pressure of the pure liquid.

Henry's lawFor a gas dissolved in a liquid: p = K_H x. The constant K_H depends on the gas, the solvent and the temperature.

Both say that vapour pressure is proportional to mole fraction. Raoult's law is the special case where K_H = p₁°.

In real life

In fractional distillation of crude oil at a refinery, the vapour is always richer in the more volatile parts, just as Raoult's law predicts. That is how petrol, kerosene and diesel are separated.

Exam angle

2 marks: “State Raoult's law for a solution of two volatile liquids and write the expression for its total vapour pressure.” The partial vapour pressure of each component is directly proportional to its mole fraction: p₁ = p₁° x₁ and p₂ = p₂° x₂. p_total = p₁° + (p₂° − p₁°) x₂.

3 marks: “Pure liquids A and B have vapour pressures of 400 mm Hg and 600 mm Hg at a certain temperature. For an ideal solution with x(A) = 0.5, find the total vapour pressure and the mole fraction of A in the vapour.” p(A) = 400 × 0.5 = 200 mm Hg; p(B) = 600 × 0.5 = 300 mm Hg; p_total = 500 mm Hg; y(A) = 200 ÷ 500 = 0.40.

2 marks: “Why is the vapour pressure of a glucose solution lower than that of pure water at the same temperature?” Glucose is non-volatile. Its molecules take up part of the surface, so fewer water molecules escape. By Raoult's law, p₁ = p₁° x₁ with x₁ less than 1.

Remember it

Raoult: Pressure = Pure pressure × Portion (mole fraction). The vapour always favours the more volatile liquid.