Show that the relative lowering of vapour pressure of a solution is equal to the mole fraction of the solute. How is this used to find the molar mass of a solute?

1. 1 · Start from Raoult's law. For the solvent, p₁ = x₁ p₁°.
2. 2 · Find the lowering. Δp₁ = p₁° − p₁ = p₁°(1 − x₁) = x₂ p₁°.
3. 3 · Divide by p₁°. (p₁° − p₁) ÷ p₁° = x₂: relative lowering equals the mole fraction of solute, so it is colligative.
4. 4 · Use the dilute-solution form. x₂ ≈ n₂ ÷ n₁ = (w₂ × M₁) ÷ (M₂ × w₁), so M₂ = (w₂ × M₁ × p₁°) ÷ [w₁ × (p₁° − p₁)].
5. 5 · Close in one line. So measuring the vapour pressures of the solvent and the solution gives the molar mass of a non-volatile solute.

Scoring tip: Write w₂, M₂ for the solute and w₁, M₁ for the solvent, and keep the same subscripts all through the answer.