Abnormal molar masses and the van't Hoff factor
Electrolytes dissociate and some solutes associate, so colligative effects come out larger or smaller than expected. The van't Hoff factor, i, corrects this: ΔTf = i Kf m.
Why some molar masses come out wrong
Colligative properties count particles. In water, KCl splits into K⁺ and Cl⁻, so 1 mol of KCl gives nearly 2 mol of particles. The observed ΔTf is nearly double, and the molar mass worked out from it is about half the true value. In benzene, ethanoic (acetic) acid molecules pair up into dimers through hydrogen bonds, so there are fewer particles. The molar mass then comes out about double: near 120 g mol⁻¹ instead of 60. A molar mass that is higher or lower than the true value is an abnormal molar mass.
Values of i and what they mean
| Solute (solvent) | What happens | i (ideal value) | Molar mass found from colligative data |
|---|---|---|---|
| Glucose, urea, sucrose (water) | Nothing | 1 | Normal |
| NaCl, KCl, MgSO₄ (water) | Dissociates into 2 ions | 2 | About half the true value |
| K₂SO₄, CaCl₂ (water) | Dissociates into 3 ions | 3 | About one-third |
| Ethanoic acid, benzoic acid (benzene) | Associates into dimers | 0.5 | About double |
Finding the degree of dissociation or association
- 1Dissociation: one formula unit gives n ions. Start with 1 mol; a fraction α dissociates.
- 2Moles of particles at equilibrium = (1 − α) + nα = 1 + (n − 1)α.
- 3★So, i = 1 + (n − 1)α, and α = (i − 1) ÷ (n − 1).
- 4Association: n molecules join to form one particle, (A)ₙ. Start with 1 mol; a fraction α associates.
- 5Moles of particles = (1 − α) + α ÷ n, so i = 1 − α(1 − 1/n).
- 6For dimers (n = 2): i = 1 − α/2, so α = 2(1 − i).
★ key step
Worked example: dissociation of NaCl
A 0.10 mol kg⁻¹ aqueous solution of NaCl shows a freezing point depression of 0.3534 K. Find the van't Hoff factor and the degree of dissociation of NaCl (Kf = 1.86 K kg mol⁻¹).
- Calculated ΔTf, with no dissociation = Kf × m = 1.86 × 0.10 = 0.186 K.
- i = observed ÷ calculated = 0.3534 ÷ 0.186 = 1.9.
- NaCl → Na⁺ + Cl⁻, so n = 2.
- α = (i − 1) ÷ (n − 1) = (1.9 − 1) ÷ (2 − 1) = 0.9.
i = 1.9; α = 0.9, so NaCl is 90% dissociated
Worked example: association of benzoic acid
2.44 g of benzoic acid (C₆H₅COOH, M = 122 g mol⁻¹) in 100 g of benzene lowers its freezing point by 0.64 K. Kf of benzene = 5.12 K kg mol⁻¹. Find (a) the observed molar mass, (b) i and (c) the percentage of the acid that forms dimers.
- (a) M(observed) = (Kf × w₂ × 1000) ÷ (ΔTf × w₁) = (5.12 × 2.44 × 1000) ÷ (0.64 × 100).
- M(observed) = 12492.8 ÷ 64 = 195.2 g mol⁻¹.
- (b) i = normal molar mass ÷ observed molar mass = 122 ÷ 195.2 = 0.625.
- (c) 2C₆H₅COOH ⇌ (C₆H₅COOH)₂, so n = 2 and i = 1 − α/2.
- α = 2(1 − i) = 2 × (1 − 0.625) = 2 × 0.375 = 0.75.
(a) 195.2 g mol⁻¹; (b) i = 0.625; (c) 75% of the acid is present as dimers
Exam points
- Use i only for solutes that dissociate or associate. For glucose, urea and sucrose, i = 1.
- Measured values of i for salts are below the ideal value, because oppositely charged ions attract each other. They come closer to the ideal value on dilution.
- To compare colligative effects, compare i × m. For 0.1 m solutions: K₂SO₄ (about 0.3) > NaCl (about 0.2) > glucose (0.1).
- The largest i × m gives the highest boiling point, the lowest freezing point and the highest osmotic pressure.
ORS (oral rehydration solution) is designed by counting particles: each NaCl gives two, so its salts are weighed to give the right osmotic effect in the gut. Salt melts ice on hill roads better than the same mass of sugar for the same reason.
2 marks: “What is the van't Hoff factor? Give its value for K₂SO₄ in a dilute aqueous solution, assuming complete dissociation.” i = observed colligative property ÷ calculated colligative property (or normal molar mass ÷ observed molar mass). K₂SO₄ → 2K⁺ + SO₄²⁻, so i = 3.
3 marks: “Arrange 0.1 mol kg⁻¹ aqueous solutions of glucose, NaCl and K₂SO₄ in increasing order of freezing point, with reasons.” Compare i × m: K₂SO₄ 0.3, NaCl 0.2, glucose 0.1. A larger i × m gives a larger ΔTf, so freezing point: K₂SO₄ < NaCl < glucose.
3 marks: “Ethanoic acid shows a molar mass of about 120 g mol⁻¹ in benzene. Explain.” In benzene, two ethanoic acid molecules join by hydrogen bonds between their COOH groups to form a dimer. The number of particles is halved (i ≈ 0.5), so ΔTf is halved and the calculated molar mass doubles.
i = how many particles one formula unit really becomes. Split up into ions: i > 1, molar mass looks smaller. Pair up into dimers: i < 1, molar mass looks bigger.