Vapour Pressure of Solutions and Raoult's Law

For a solution of two volatile liquids, Raoult's law states that the partial vapour pressure of each component is proportional to its mole fraction in the liquid: p₁ = p₁° x₁ and p₂ = p₂° x₂, where p° is the vapour pressure of the pure component. Total pressure p = p₁° x₁ + p₂° x₂.

Solutions of a non-volatile solute

When the solute is non-volatile, only the solvent contributes to vapour pressure: p = p₁° x₁ (solvent). This means the vapour pressure is lowered on adding solute. The relative lowering of vapour pressure, (p₁° − p)/p₁° = x₂ (mole fraction of solute), which is a colligative property.

Ideal and non-ideal solutions

• Ideal solutions obey Raoult's law over the whole range; ΔmixH = 0, ΔmixV = 0. Solute–solvent interactions equal solute–solute and solvent–solvent (e.g. benzene + toluene, n-hexane + n-heptane).
• Positive deviation: vapour pressure higher than predicted; A–B interactions weaker than A–A/B–B; ΔmixH > 0 (e.g. ethanol + acetone). Forms minimum-boiling azeotrope.
• Negative deviation: vapour pressure lower than predicted; A–B interactions stronger; ΔmixH < 0 (e.g. chloroform + acetone). Forms maximum-boiling azeotrope.

Azeotropes

Binary mixtures that boil at constant temperature with the same composition in liquid and vapour — they cannot be separated by fractional distillation.

Exam Tricks & Tips

• 🎯 Relative lowering of vapour pressure = mole fraction of solute (x₂) — memorise this exact statement.
• 🎯 Positive deviation → minimum-boiling azeotrope; negative deviation → maximum-boiling azeotrope.
• 🎯 Ideal solution: ΔmixH = 0 and ΔmixV = 0 — a common 1-mark definition.
• 🎯 Chloroform + acetone shows negative deviation because of intermolecular H-bonding between them (a classic example).
• 🎯 Only the SOLVENT term survives when the solute is non-volatile.
• ❌ Common mistake: using the solute's mole fraction in p = p°x for a non-volatile solute. Use the SOLVENT's mole fraction, because the non-volatile solute contributes no vapour.

Board exam focus

Expect the derivation/statement of relative lowering, identifying deviation type from an example, and a numerical for vapour pressure of a two-liquid mixture.

Cover the answer, recall, then check. 11 cards on Raoult's law and deviations.

Q1. State Raoult's law for a solution of two volatile liquids.
A1. The partial vapour pressure of each component equals its pure vapour pressure times its mole fraction: pᵢ = pᵢ°xᵢ.

Q2. What is the total vapour pressure of an ideal binary solution?
A2. p = p₁°x₁ + p₂°x₂ — the sum of the two partial pressures.

Q3. What does relative lowering of vapour pressure equal?
A3. The mole fraction of the non-volatile solute, x₂: (p₁° − p)/p₁° = x₂.

Q4. Define an ideal solution.
A4. One that obeys Raoult's law over the entire range, with ΔmixH = 0 and ΔmixV = 0.

Q5. Give two examples of nearly ideal solutions.
A5. Benzene + toluene, and n-hexane + n-heptane.

Q6. What causes positive deviation from Raoult's law?
A6. Weaker solute–solvent (A–B) interactions than in the pure liquids, giving higher vapour pressure and ΔmixH > 0.

Q7. Give an example of positive deviation.
A7. Ethanol + acetone (or ethanol + water) — A–B forces are weaker than A–A/B–B.

Q8. What causes negative deviation, with an example?
A8. Stronger A–B interactions (e.g. H-bonding); chloroform + acetone, giving lower vapour pressure and ΔmixH < 0.

Q9. What is an azeotrope?
A9. A binary mixture that boils at constant temperature with identical liquid and vapour composition, so it cannot be separated by distillation.

Q10. Which deviation forms a minimum-boiling azeotrope?
A10. Positive deviation (e.g. ethanol–water, ~95% ethanol).

Q11. Which deviation forms a maximum-boiling azeotrope?
A11. Negative deviation (e.g. nitric acid–water, ~68% HNO₃).

Why adding salt slows evaporation

Leave a glass of pure water and a glass of salty water side by side; the salty one evaporates more slowly. Every escaping molecule at the surface contributes to vapour pressure, and dissolving a solute lowers it. Raoult's Law turns this observation into a precise, testable equation.

Core idea: For a solution of volatile liquids, the partial vapour pressure of each component is proportional to its mole fraction. Adding a non-volatile solute lowers the solvent's vapour pressure in exact proportion to how much solvent is "diluted."

Deep explanation

Beginner: Raoult's Law for two volatile liquids

For a binary mixture of volatile liquids A and B:

p_A = p°_A · x_A and p_B = p°_B · x_B

where p° is the vapour pressure of the pure component. By Dalton's law the total pressure is:

p_total = p_A + p_B = p°_A·x_A + p°_B·x_B = p°_B + (p°_A − p°_B)·x_A

This is a straight line in x_A — a graph examiners love.

Intermediate: non-volatile solute and relative lowering

If the solute is non-volatile, only the solvent contributes vapour. Then p_solution = p°_solvent · x_solvent. Since x_solvent = 1 − x_solute, the relative lowering of vapour pressure equals the mole fraction of the solute:

(p° − p_solution) / p° = x_solute

This is the form used to find molar masses of unknown solutes.

Advanced: ideal vs non-ideal solutions

An ideal solution obeys Raoult's Law at all concentrations, with ΔH_mixing = 0 and ΔV_mixing = 0 (A–B interactions ≈ A–A and B–B). Example: benzene + toluene.

• Positive deviation (ethanol + acetone): A–B forces weaker than pure forces, so molecules escape more easily, p > Raoult prediction, ΔH_mix > 0.
• Negative deviation (chloroform + acetone, which H-bond): A–B forces stronger, escape harder, p < Raoult prediction, ΔH_mix < 0.

Solutions with large deviations form azeotropes — constant-boiling mixtures that cannot be separated by fractional distillation. Positive deviation gives a minimum-boiling azeotrope (95% ethanol–water); negative deviation gives a maximum-boiling azeotrope (nitric acid–water).

Worked example

The vapour pressure of pure water at 298 K is 23.8 mmHg. When 18 g of a non-volatile solute is dissolved in 178.2 g of water, vapour pressure falls to 23.5 mmHg. Find the molar mass of the solute.

Moles of water = 178.2/18 = 9.9. Let solute moles = n.
Relative lowering: (23.8 − 23.5)/23.8 = n/(n + 9.9).
0.3/23.8 = 0.01261 ≈ n/9.9 (dilute approximation) ⇒ n = 0.1248.
Molar mass = 18 g / 0.1248 = ≈ 144 g/mol.

Real-world / exam application

Relative lowering of vapour pressure is a colligative property used to determine molar masses. Azeotrope theory explains why "100% pure" ethanol cannot be obtained by simple distillation of the 95% azeotrope, and why fractional distillation of crude mixtures has limits. Graph-based MCQs on p_total vs x are common.

Exam tricks & shortcuts

• To identify deviation type: weaker A–B forces ⇒ positive deviation (higher p, endothermic mixing).
• Relative lowering equals x_solute, so it directly gives moles of solute in dilute cases.
• Mnemonic — "Positive deviation = Push apart" (weaker interactions push molecules into vapour).

Key formulas

• Raoult's law (volatile components): p_A = x_A·p°_A; p_total = x_A p°_A + x_B p°_B.
• Non-volatile solute: p_solution = x_solvent·p°_solvent.
• Relative lowering of vapour pressure: (p° − p)/p° = x_solute.
• Ideal solution: obeys Raoult's law; ΔH_mix = 0, ΔV_mix = 0.

Positive deviations (weaker A–B forces) raise vapour pressure; negative deviations lower it.

Worked Example

Problem: An ideal solution is made by mixing equal moles of two volatile liquids A and B. The pure-component vapour pressures are p°_A = 100 torr and p°_B = 200 torr. Find the total vapour pressure of the solution.

Solution:
Step 1 — State Raoult's law for a volatile component: its partial vapour pressure equals its mole fraction times its pure vapour pressure:
p_A = x_A p°_A and p_B = x_B p°_B.

Step 2 — Find the mole fractions. Since equal moles are mixed:
x_A = x_B = 0.5.

Step 3 — Compute the partial pressures:
p_A = 0.5 × 100 = 50 torr; p_B = 0.5 × 200 = 100 torr.

Step 4 — Add them for the total vapour pressure (Dalton's law):
p_total = p_A + p_B = 50 + 100 = 150 torr.

Answer: The total vapour pressure of the solution is 150 torr.