Colligative properties: relative lowering of vapour pressure
Colligative properties depend on the number of solute particles, not their nature. The relative lowering of vapour pressure equals the mole fraction of the solute.
What 'colligative' means
Colligative means 'bound together'. These properties depend on how many solute particles are present, not on which particles they are. A 0.1 mol kg⁻¹ solution of glucose and a 0.1 mol kg⁻¹ solution of urea in water lower the vapour pressure by the same amount. The simple formulas apply to dilute solutions of a non-volatile, non-electrolyte solute. Because each property depends on the number of solute particles, measuring it lets us find the molar mass of the solute.
Relative lowering from Raoult's law
- 1The solvent obeys Raoult's law: p₁ = x₁ p₁°.
- 2Lowering of vapour pressure: Δp₁ = p₁° − p₁ = p₁° − x₁ p₁° = p₁°(1 − x₁).
- 3In a binary solution 1 − x₁ = x₂, so Δp₁ = x₂ p₁°.
- 4★Divide by p₁°: (p₁° − p₁) ÷ p₁° = x₂ = n₂ ÷ (n₁ + n₂).
- 5In a dilute solution n₂ is much smaller than n₁, so (p₁° − p₁) ÷ p₁° ≈ n₂ ÷ n₁.
- 6Put n = w ÷ M: (p₁° − p₁) ÷ p₁° = (w₂ × M₁) ÷ (M₂ × w₁). So, the molar mass of the solute is M₂ = (w₂ × M₁ × p₁°) ÷ [w₁ × (p₁° − p₁)]. Here w₂, M₂ belong to the solute and w₁, M₁ to the solvent.
★ key step
Worked example: vapour pressure of a glucose solution
18 g of glucose (M = 180 g mol⁻¹) is dissolved in 178.2 g of water at 298 K. The vapour pressure of pure water at 298 K is 23.8 mm Hg. Find the vapour pressure of the solution.
- n₂ = 18 ÷ 180 = 0.1 mol; n₁ = 178.2 ÷ 18 = 9.9 mol.
- x₂ = 0.1 ÷ (9.9 + 0.1) = 0.1 ÷ 10 = 0.01.
- Lowering: p₁° − p₁ = x₂ × p₁° = 0.01 × 23.8 = 0.238 mm Hg.
- p₁ = 23.8 − 0.238 = 23.562 mm Hg, about 23.56 mm Hg.
About 23.56 mm Hg
Worked example: molar mass from vapour pressure
At a certain temperature, pure water has a vapour pressure of 30.0 mm Hg. A solution of 9.0 g of a non-volatile non-electrolyte in 90 g of water has a vapour pressure of 29.7 mm Hg. Find the molar mass of the solute, treating the solution as dilute.
- Relative lowering = (30.0 − 29.7) ÷ 30.0 = 0.3 ÷ 30.0 = 0.010.
- Dilute solution: (p₁° − p₁) ÷ p₁° = (w₂ × M₁) ÷ (M₂ × w₁).
- 0.010 = (9.0 × 18) ÷ (M₂ × 90).
- M₂ = (9.0 × 18) ÷ (0.010 × 90) = 162 ÷ 0.90 = 180 g mol⁻¹.
M₂ = 180 g mol⁻¹ (glucose has this molar mass)
Points to remember
- The relative lowering depends only on the mole fraction of solute, so it is colligative. The lowering itself, p₁° − p₁ = x₂ p₁°, also depends on the solvent's own p₁°.
- The formula is for a non-volatile solute. A volatile solute adds its own vapour pressure.
- For an electrolyte, every ion counts as a particle. This needs the van't Hoff factor, i, covered at the end of this chapter.
Honey and the thick sugar syrup of gulab jamun dry out far more slowly than a cup of water left open in the kitchen. So much dissolved sugar lowers the vapour pressure of the water in them.
2 marks: “What are colligative properties? Name the four colligative properties.” Properties of dilute solutions that depend only on the number of solute particles, not on their nature: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure.
3 marks: “Pure benzene (M = 78 g mol⁻¹) has a vapour pressure of 500 mm Hg at a certain temperature. When 1.6 g of a non-volatile solute is dissolved in 39 g of benzene, the vapour pressure falls to 490 mm Hg. Find the molar mass of the solute.” Relative lowering = 10 ÷ 500 = 0.02 = (w₂ × M₁) ÷ (M₂ × w₁) = (1.6 × 78) ÷ (M₂ × 39). M₂ = (1.6 × 78) ÷ (0.02 × 39) = 124.8 ÷ 0.78 = 160 g mol⁻¹.
Relative lowering = x₂, the mole fraction of solute. It counts the guests (particles), not who they are.