What are colligative properties?
Some properties of a dilute solution of a non-volatile solute depend only on the number of solute particles, not on their kind. These are colligative properties (from Latin 'bound together'). One mole of glucose and one mole of urea in 1 kg water have the same effect.
1. Relative lowering of vapour pressure
A non-volatile solute takes some of the surface, so fewer solvent molecules escape. From Raoult's law, p = x₁·p₁°, so the fall Δp = p₁° − p = x₂·p₁°. Therefore
(p₁° − p) ÷ p₁° = x₂
The relative lowering equals the mole fraction of the solute. For a dilute solution x₂ ≈ n₂/n₁ = (w₂M₁)/(M₂w₁), which lets us find M₂.
Elevation of boiling point
A liquid boils when its vapour pressure equals the outside pressure (1 atm). A solution has lower vapour pressure, so it must be heated more. The rise is
ΔTb = Tb − Tb° = Kb · m
m is molality; Kb is the boiling point elevation constant (molal elevation constant, ebullioscopic constant), unit K kg mol⁻¹. For water Kb = 0.52 K kg mol⁻¹.
Since m = (w₂ × 1000) ÷ (M₂ × w₁), we get M₂ = (Kb × w₂ × 1000) ÷ (ΔTb × w₁), with w₁ in grams.
Depression of freezing point
At the freezing point, the liquid and solid have the same vapour pressure. The solution's vapour pressure is lower, so it matches the solid's only at a colder temperature. The fall is
ΔTf = Tf° − Tf = Kf · m
Kf is the freezing point depression constant (cryoscopic constant). For water Kf = 1.86 K kg mol⁻¹, for benzene 5.12 K kg mol⁻¹.
M₂ = (Kf × w₂ × 1000) ÷ (ΔTf × w₁)
Uses: salt on icy roads; antifreeze (ethylene glycol) in car radiators.
Osmosis and osmotic pressure
A semipermeable membrane (SPM) lets solvent molecules through but not solute particles (examples: animal bladder, cell membrane, copper ferrocyanide film). Osmosis is the flow of solvent through an SPM from the pure solvent (or dilute solution) into the more concentrated solution.
Osmotic pressure (π) is the extra pressure that must be put on the solution to just stop osmosis. For dilute solutions:
π = C R T = (n₂ ÷ V) R T, so M₂ = (w₂ R T) ÷ (π V)
R = 0.083 L bar mol⁻¹ K⁻¹. Osmotic pressure is the best method for proteins and polymers: even tiny concentrations give a measurable π at room temperature, and molarity is used so no heating is needed.
Isotonic, hypertonic and hypotonic
- Isotonic: same osmotic pressure (0.9% NaCl ≈ blood cells). No net flow.
- Hypertonic: higher osmotic pressure outside; the cell loses water and shrinks.
- Hypotonic: lower outside; the cell takes in water and swells.
Reverse osmosis
If a pressure larger than π is put on the solution side, pure solvent is pushed out through the membrane. This is reverse osmosis, used to make drinking water from sea water (cellulose acetate membranes).
Try it (at home)
Put a few raisins in plain water and a few in very salty water overnight. Predict first. The ones in plain water swell (water goes in: hypotonic). Cut two potato strips; the one in salty water turns soft and bendy (water leaves: hypertonic). In the 3D, step 4 shows the same flow through the green membrane.
Finding molar mass from colligative properties
Every colligative property counts moles of solute. If we know the mass of solute we added and measure the effect, we can find its molar mass M₂:
- Vapour pressure: (p° − p)/p° = (w₂M₁)/(M₂w₁)
- Boiling point: M₂ = Kb·w₂·1000 / (ΔTb·w₁)
- Freezing point: M₂ = Kf·w₂·1000 / (ΔTf·w₁)
- Osmotic pressure: M₂ = w₂RT / (πV)
Board tip: one numerical from this unit (3 marks) is almost certain. Write the formula, list data with units, convert g → kg or mL → L, then substitute.
Abnormal molar mass and the van 't Hoff factor
If the solute splits (dissociates) into ions, there are more particles, so the effect is bigger and the molar mass we calculate comes out smaller than the true one. If the solute pairs up (associates), there are fewer particles and the calculated molar mass comes out bigger. Such values are called abnormal molar masses.
The van 't Hoff factor:
i = normal molar mass ÷ abnormal molar mass = observed colligative property ÷ calculated colligative property
- i > 1: dissociation (NaCl → Na⁺ + Cl⁻, i ≈ 2; K₂SO₄, i ≈ 3)
- i < 1: association (2 CH₃COOH ⇌ (CH₃COOH)₂ in benzene, i ≈ 0.5)
- i = 1: no change (glucose, urea)
Modified formulas: Δp/p° = i·x₂, ΔTb = i·Kb·m, ΔTf = i·Kf·m, π = i·CRT.
Degree of dissociation and association
For a solute giving n particles, with degree of dissociation α: i = 1 + (n − 1)α. For association into groups of n, with degree α: i = 1 − α + α/n.
Key formulas and definitions
- Relative lowering: (p₁° − p) ÷ p₁° = x₂ ≈ (w₂M₁) ÷ (M₂w₁)
- ΔTb = i · Kb · m; Kb(water) = 0.52 K kg mol⁻¹
- ΔTf = i · Kf · m; Kf(water) = 1.86 K kg mol⁻¹
- π = i · C R T; R = 0.083 L bar mol⁻¹ K⁻¹
- M₂ = (K × w₂ × 1000) ÷ (ΔT × w₁)
- i = normal molar mass ÷ abnormal molar mass
- Dissociation: i = 1 + (n − 1)α; Association: i = 1 − α + α/n
Worked examples
1. 0.5 mol of glucose is dissolved in 1 kg of water. Find the boiling point (Kb = 0.52 K kg mol⁻¹).
m = 0.5 mol kg⁻¹. ΔTb = 0.52 × 0.5 = 0.26 K. Boiling point = 373.15 + 0.26 = 373.41 K (100.26 °C).
2. Find the freezing point of a solution with 6 g urea (M = 60) in 250 g of water (Kf = 1.86).
Moles = 6 ÷ 60 = 0.1. m = 0.1 ÷ 0.25 = 0.4 mol kg⁻¹. ΔTf = 1.86 × 0.4 = 0.744 K. Freezing point = 273.15 − 0.744 = 272.41 K (−0.744 °C).
3. The vapour pressure of pure water is 3.17 kPa. A solution contains 0.1 mol of a non-volatile solute in 0.9 mol of water. Find the vapour pressure of the solution.
x₂ = 0.1 ÷ 1.0 = 0.1. Δp = 0.1 × 3.17 = 0.317 kPa. p = 3.17 − 0.317 = 2.85 kPa.
4. Dissolving 1.8 g of a non-volatile solute in 100 g of water raises the boiling point by 0.052 K. Find the molar mass (Kb = 0.52).
M₂ = (Kb × w₂ × 1000) ÷ (ΔTb × w₁) = (0.52 × 1.8 × 1000) ÷ (0.052 × 100) = 936 ÷ 5.2 = 180 g mol⁻¹.
5. Find the osmotic pressure at 27 °C of a solution containing 2 g of a protein of molar mass 40 000 g/mol in 200 mL of solution (R = 0.083 L bar mol⁻¹ K⁻¹).
n = 2 ÷ 40 000 = 5 × 10⁻⁵ mol. V = 0.2 L. C = 2.5 × 10⁻⁴ mol L⁻¹. T = 300 K. π = 2.5 × 10⁻⁴ × 0.083 × 300 = 6.2 × 10⁻³ bar (about 623 Pa).
6. 0.1 mol of NaCl is dissolved in 1 kg water. Assuming full dissociation, find ΔTf (Kf = 1.86).
NaCl gives 2 ions, so i = 2. ΔTf = i·Kf·m = 2 × 1.86 × 0.1 = 0.372 K. Freezing point = −0.372 °C.
7. A solution of acetic acid in benzene shows a molar mass of 120 g/mol from freezing point data. True M = 60. Find i and the degree of association (acid forms dimers).
i = 60 ÷ 120 = 0.5. For dimers, i = 1 − α + α/2 = 1 − α/2. 0.5 = 1 − α/2 → α = 1, i.e. 100% association.
8. 0.01 m K₂SO₄ solution freezes at −0.0521 °C. Find i and the degree of dissociation (Kf = 1.86).
Calculated ΔTf (no dissociation) = 1.86 × 0.01 = 0.0186 K. i = 0.0521 ÷ 0.0186 = 2.8. K₂SO₄ → 2K⁺ + SO₄²⁻, n = 3. i = 1 + 2α → 2.8 = 1 + 2α → α = 0.9 (90%).
Common mistakes
- Using molarity in ΔTb or ΔTf. Both use molality (per kg of solvent). Only π uses molarity.
- Forgetting the van 't Hoff factor for salts: NaCl doubles the effect (i = 2), CaCl₂ nearly triples it (i = 3).
- Putting temperature in °C in π = CRT. Always use kelvin: T = °C + 273.
- Mixing up the direction of osmosis: solvent moves from the dilute side to the concentrated side, not the other way.