Vapour Pressure of Solutions and Raoults Law Guide

Vapour pressure is the pressure exerted by a vapour in equilibrium with its liquid — it measures how eagerly molecules escape the liquid surface. When a non-volatile solute (like salt or sugar) dissolves, fewer solvent molecules sit at the surface, so fewer escape: the vapour pressure of the solution is lower than that of the pure solvent. This lowering is one of the colligative properties.

Raoult's law quantifies this exactly, and it is the basis for distillation, antifreeze, and even cooking (salted water boils hotter). It also tells us whether a mixture behaves ideally or shows positive/negative deviation — the key to understanding real liquid mixtures.

Key Concepts

• Vapour pressure (p): SI unit pascal (Pa); depends only on the liquid's nature and temperature, not on the container size.
• Raoult's law (volatile mixture): partial pressure of each component = its mole fraction × its pure vapour pressure: p₁ = x₁p₁°.
• Raoult's law (non-volatile solute): p(solution) = x(solvent) × p°(solvent); the lowering Δp = x(solute) × p°.
• Ideal solution: obeys Raoult's law at all compositions — solute-solvent attractions equal solute-solute and solvent-solvent attractions (e.g. benzene + toluene).
• Positive deviation: vapour pressure higher than Raoult predicts — solute-solvent attractions are weaker (e.g. ethanol + water breaks H-bonds).
• Negative deviation: vapour pressure lower than predicted — solute-solvent attractions are stronger (e.g. acetone + chloroform form H-bonds).

Formulas

• Raoult's law: p₁ = x₁ p₁°, p₂ = x₂ p₂°, total p = p₁ + p₂ = x₁p₁° + x₂p₂°. Units: Pa (or mmHg).
• Relative lowering: Δp/p° = x(solute) = n(solute)/[n(solute) + n(solvent)]. Unitless — a colligative property depending only on mole fraction.
• Vapour composition: y₁ = p₁/p(total) — the vapour is richer in the more volatile component (higher p°), which is why distillation works.
• Assumes ideal behaviour and equilibrium at constant temperature; deviations are measured against these equations.

Worked Example

At 25°C, pure water has p° = 23.8 mmHg. Dissolve 18.0 g glucose (180 g/mol, non-volatile) in 90.0 g water (18 g/mol). Find the solution's vapour pressure.

1. Moles glucose = 18.0/180 = 0.100 mol. Moles water = 90.0/18 = 5.00 mol.
2. Mole fraction of water x = 5.00/(5.00 + 0.100) = 0.9804.
3. p = x × p° = 0.9804 × 23.8 = 23.33 mmHg.
4. Lowering Δp = 23.8 − 23.33 = 0.47 mmHg, which also equals x(solute) × p° = 0.0196 × 23.8 = 0.47 mmHg. Checks out.

Common Misconceptions

• "Adding solute raises vapour pressure because there is more stuff." The opposite — non-volatile solute blocks surface molecules, so vapour pressure falls. (Volatile solutes are a different case handled by the full Raoult equation.)
• "Raoult's law works for all mixtures." It is exact only for ideal solutions. Real mixtures deviate, and strong deviations produce azeotropes that cannot be separated by simple distillation.
• "The vapour has the same composition as the liquid." No — the vapour is always richer in the more volatile component (y₁ ≠ x₁). This single fact makes fractional distillation possible.

Real-World Applications

• Distillation: petroleum refining, whisky making and solvent purification all exploit the vapour being richer in the volatile component.
• Antifreeze: ethylene glycol lowers water's vapour pressure (and freezing point) — colligative properties protecting car engines.
• Humidity control: desiccants and saturated salt solutions fix the vapour pressure (hence humidity) inside sealed chambers for labs and museums.
• Perfumery: fixatives are low-volatility solutes that lower the vapour pressure of fragrance oils, making perfume last longer.

In short, dissolved non-volatile solute lowers vapour pressure in proportion to its mole fraction (Δp/p° = x_solute), and Raoult's law p₁ = x₁p₁° describes ideal mixtures. Deviations reveal the real molecular attractions — and lead straight to azeotropes.