Ch. 2 | Solutions0/13

Raoult's Law and Vapour Pressure of Solutions

How dissolved substances reduce vapour pressure — and Dalton's law of composition

Quantitative ReasoningLevel 2 · Application

You mix equal moles of benzene (vapour pressure = 95 mmHg) and toluene (vapour pressure = 29 mmHg) at 25°C. A classmate says: "The total vapour pressure must be close to 95 mmHg since benzene dominates with its high vapour pressure." Another says it must be exactly 62 mmHg — the arithmetic mean. Who is right, and how would you calculate the true answer?

Real Life Hook

François-Marie Raoult spent twelve years (1878–1890) patiently measuring the vapour pressures of hundreds of solutions. He discovered the same law applied to all dilute solutions regardless of the chemical identity of the solute — as long as it was non-volatile. His insight was that only the solvent evaporates, and its ability to evaporate is reduced in proportion to how much solute is blocking the surface. A simple geometric idea that now underpins everything from industrial distillation to osmotic drug delivery.

Raoult's Law — For Volatile Solutes

For a solution of two volatile components (A and B), each component obeys Raoult's law independently:

pA=pAxApB=pBxBp_A = p_A^\circ \cdot x_A \qquad p_B = p_B^\circ \cdot x_B

By Dalton's law of partial pressures, the total vapour pressure:
ptotal=pA+pB=pAxA+pBxBp_{\text{total}} = p_A + p_B = p_A^\circ x_A + p_B^\circ x_B

where pAp_A^\circ and pBp_B^\circ are the vapour pressures of pure A and pure B, and xAx_A, xBx_B are their mole fractions in solution.

Key result: ptotalp_{\text{total}} lies between pAp_A^\circ and pBp_B^\circ — the total pressure is always intermediate.

Raoult's Law — For Non-Volatile Solutes

When the solute is non-volatile (solid dissolved in liquid — e.g. NaCl in water), it contributes zero vapour pressure. Only the solvent evaporates:

psolution=p1x1p_{\text{solution}} = p_1^\circ \cdot x_1

Since x1=1x2x_1 = 1 - x_2 (mole fraction of solvent = 1 − mole fraction of solute):
psolution=p1(1x2)p_{\text{solution}} = p_1^\circ (1 - x_2)

Relative lowering of vapour pressure (RLVP):
p1psolutionp1=x2\frac{p_1^\circ - p_{\text{solution}}}{p_1^\circ} = x_2

RLVP equals the mole fraction of the solute — a colligative property that depends only on how many solute particles are present, not what they are.

🖼 Image PendingGraph showing total vapour pressure vs mole fraction for benzene-toluene system obeying Raoult's law

AI Generation Prompt

Vapour pressure vs mole fraction graph for benzene-toluene ideal solution. X-axis: mole fraction of benzene (0 to 1). Y-axis: vapour pressure in mmHg (0 to 120). Three lines: (1) p_toluene: linear from 29 mmHg at x=0 to 0 at x=1, in blue. (2) p_benzene: linear from 0 at x=0 to 95 mmHg at x=1, in orange. (3) p_total: linear from 29 mmHg at x=0 to 95 mmHg at x=1, in white/amber, labelled p_total = p°_toluene × x_toluene + p°_benzene × x_benzene. Mark the midpoint at x_benzene = 0.5 showing p_total = 62 mmHg. Dark background, orange accent labels, clean technical illustration style.

📸 Benzene-toluene vapour pressure diagram: linear variation of p_A, p_B, and p_total with mole fraction — a textbook ideal solution

Remember

Raoult's law applies to the solvent, not the solute. The solvent's partial pressure = psolvent×xsolventp^\circ_{\text{solvent}} \times x_{\text{solvent}}.

RLVP = mole fraction of solute: ppsp=x2\frac{p^\circ - p_s}{p^\circ} = x_2 — valid for dilute solutions only.

Composition of vapour ≠ composition of liquid. The vapour is richer in the more volatile component. This is the basis of distillation.

📖NCERT 2.6NCERT Intext

Problem

The vapour pressure of pure benzene at 25°C is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g is dissolved in 39.0 g of benzene (molar mass = 78 g/mol). The vapour pressure of the solution is 0.845 bar. Find the molar mass of the solid.

JEE / NEET Exam InsightJEE / NEET
RLVP for dilute solutions: ppspn2n1\frac{p^\circ - p_s}{p^\circ} \approx \frac{n_2}{n_1} (approximation when n2n1n_2 \ll n_1). The exact formula is n2n1+n2\frac{n_2}{n_1 + n_2}.
Vapour composition: Mole fraction of A in vapour: yA=pAptotal=pAxAptotaly_A = \frac{p_A}{p_{\text{total}}} = \frac{p_A^\circ x_A}{p_{\text{total}}}. The vapour is always richer in the more volatile component.
Colligative property check: RLVP depends only on x2x_2 (number of solute particles), not on the nature of the solute. 1 mol glucose and 1 mol urea in the same solvent give the same RLVP.
Quick Check

Q1.According to Raoult's law, the partial vapour pressure of a component in a liquid mixture is: