Colligative Properties and the van't Hoff Factor
A colligative property depends only on how many solute particles are dissolved, never on what they are. That single sentence is the whole topic, and it is also why the van't Hoff factor exists: the moment a solute dissociates into ions or associates into dimers, the number of particles stops matching the number of formula units you weighed out, and every prediction must be corrected. This article develops the four properties, defines the correction properly, and works through the numerical types that appear in exams.
The four colligative properties
2. Boiling point elevation: ΔTb = i · Kb · m
3. Freezing point depression: ΔTf = i · Kf · m
4. Osmotic pressure: π = i · M · R · T
Note which concentration scale each uses. Boiling and freezing point changes use molality m (mol solute per kg solvent), which does not vary with temperature; osmotic pressure uses molarity M, since it is measured at a fixed temperature. Mixing the two up is the commonest source of a wrong answer here.
All four share one cause: dissolving a non-volatile solute lowers the chemical potential of the liquid solvent by RT ln xsolvent, a negative quantity. The liquid's potential then equals the vapour's at a higher temperature (boiling point rises) and the solid's at a lower one (freezing point falls). Osmotic pressure is the mechanical pressure needed to restore the pure-solvent potential across a semipermeable membrane.
| Solvent | Kf (K kg mol⁻¹) | Kb (K kg mol⁻¹) |
|---|---|---|
| Water | 1.86 | 0.512 |
| Benzene | 5.12 | 2.53 |
| Acetic acid | 3.90 | 3.07 |
| Chloroform | — | 3.63 |
| Carbon tetrachloride | — | 5.03 |
| Camphor | ≈ 40 | — |
Camphor's enormous cryoscopic constant is why Rast's method uses it for molar mass determination: a tiny amount of solute produces a large, easily measured temperature drop.
Defining the van't Hoff factor
equivalently i = (normal molar mass) ÷ (observed molar mass)
equivalently i = (actual number of particles in solution) ÷ (number of formula units dissolved)
The second form explains the phrase abnormal molar mass. A colligative measurement returns a particle count, which you convert to a molar mass. If the solute split into two ions you counted twice as many particles as expected, so the calculated molar mass comes out at half the true value; if it dimerised, at twice.
Association of n molecules into one, degree α: i = 1 − α(1 − 1/n) → i < 1
| Solute | Behaviour in water | Ideal i |
|---|---|---|
| Glucose, urea, sucrose | Non-electrolyte | 1 |
| NaCl, KBr | 2 ions | 2 |
| K₂SO₄, BaCl₂, CaCl₂ | 3 ions | 3 |
| K₃[Fe(CN)₆], FeCl₃ | 4 ions | 4 |
| Al₂(SO₄)₃ | 5 ions | 5 |
| [Co(NH₃)₆]Cl₃ | 4 ions (ligands do not ionise) | 4 |
| Benzoic or acetic acid in benzene | Dimerises by hydrogen bonding | towards 0.5 |
Worked examples
Example 1 — freezing point of a strong electrolyte. Find the freezing point of 0.100 m aqueous NaCl, assuming complete dissociation (Kf = 1.86 K kg mol⁻¹).
i = 2 for complete dissociation.
ΔTf = i Kf m = 2 × 1.86 × 0.100 = 0.372 K
Freezing point = 0.000 − 0.372 = −0.372 °C.
The measured value is slightly smaller, corresponding to i ≈ 1.87. Nothing is wrong with the theory: at finite concentration interionic attraction and ion pairing make the ions less than fully independent, and the ideal i is reached only on extrapolation to infinite dilution.
Example 2 — association, and an abnormal molar mass. 1.00 g of acetic acid (M = 60.0 g/mol) in 100 g of benzene lowers the freezing point by 0.512 K. Kf(benzene) = 5.12 K kg mol⁻¹. Find i, the degree of association, and the apparent molar mass.
Molality: m = (1.00 ÷ 60.0) ÷ 0.100 kg = 0.01667 ÷ 0.100 = 0.1667 mol/kg
Expected depression with i = 1: ΔTf = 5.12 × 0.1667 = 0.8533 K
i = observed ÷ expected = 0.512 ÷ 0.8533 = 0.600
i < 1 confirms association. For dimerisation, n = 2, so i = 1 − α(1 − ½) = 1 − α/2:
0.600 = 1 − α/2 ⟹ α/2 = 0.400 ⟹ α = 0.800, i.e. 80% of the acetic acid is present as dimers.
Apparent molar mass = normal ÷ i = 60.0 ÷ 0.600 = 100 g/mol — between the monomer's 60 and the dimer's 120, as an 80% dimerised mixture should be. Acetic acid dimerises in benzene through a pair of hydrogen bonds in a cyclic eight-membered ring; in water it does not, because water competes for those bonds.
Example 3 — a weak acid, and extracting Ka. A 0.100 m aqueous solution of monochloroacetic acid (ClCH₂COOH) freezes at −0.2046 °C. Find the degree of dissociation and the acid dissociation constant.
Expected depression with i = 1: 1.86 × 0.100 = 0.186 K
i = 0.2046 ÷ 0.186 = 1.100
The acid gives 2 particles, so n = 2 and i = 1 + (2 − 1)α = 1 + α:
α = 1.100 − 1 = 0.100, i.e. 10.0% dissociated.
Treating molality as molarity for a dilute aqueous solution,
Ka = Cα² ÷ (1 − α) = 0.100 × (0.100)² ÷ 0.900 = 0.100 × 0.0100 ÷ 0.900 = 0.00100 ÷ 0.900 = 1.11 × 10⁻³
Chloroacetic acid's accepted Ka is close to 1.4 × 10⁻³, so the cryoscopic route lands in the right place — and historically this was one of the main ways of establishing that weak electrolytes are only partly ionised.
Example 4 — osmotic pressure of a coordination compound. Calculate the osmotic pressure of 0.0100 M K₃[Fe(CN)₆] at 300 K, assuming complete dissociation (R = 0.0821 L atm K⁻¹ mol⁻¹).
The compound gives 3 K⁺ plus one [Fe(CN)₆]³⁻ ion. The cyanide ligands stay bound to iron, so i = 4, not 10.
π = iMRT = 4 × 0.0100 × 0.0821 × 300
0.0821 × 300 = 24.63 ; × 0.0100 = 0.2463 ; × 4 = 0.985 atm
The same solution would give ΔTf ≈ 4 × 1.86 × 0.0100 = 0.0744 K. Nearly one atmosphere against seven hundredths of a degree: osmotic pressure is by far the most sensitive of the four at low concentration.
Example 5 — molar mass of a macromolecule by osmometry. A solution containing 1.00 g of a neutral protein per litre develops an osmotic pressure of 2.57 × 10⁻³ atm at 300 K. Find its molar mass.
For a neutral, non-dissociating solute i = 1, and π = (w ÷ MV)RT, so M = wRT ÷ (πV):
M = (1.00 × 0.0821 × 300) ÷ (2.57 × 10⁻³ × 1.00) = 24.63 ÷ 0.00257 = 9.58 × 10³ g/mol
By freezing point depression the same solution would give ΔTf ≈ 1.86 × (1.00 ÷ 9583) ≈ 1.9 × 10⁻⁴ K — far below the resolution of an ordinary thermometer. That is the standard justification for osmometry in polymer and protein chemistry.
Example 6 — reading i backwards from an abnormal molar mass. A colligative measurement on aqueous BaCl₂ returns an apparent molar mass of 69.4 g/mol. The true value is 137.33 + 2 × 35.45 = 208.23 g/mol, so i = 208.23 ÷ 69.4 = 3.00 — exactly right for Ba²⁺ + 2 Cl⁻. Whenever an abnormal molar mass is a simple fraction of the true one, the denominator is the particle count.
Mistakes that cost marks
- Using molarity in ΔTf or ΔTb. Those two equations take molality; only osmotic pressure uses molarity.
- Counting ligands as free ions. [Co(NH₃)₆]Cl₃ gives i = 4, not 10 — the ammonia stays coordinated. Ionisation isomers such as [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br both give i = 2 but different precipitation tests.
- Assuming i is always greater than 1. Association makes i less than 1, and a solute that both dissociates and associates can sit anywhere.
- Forgetting the solvent when judging association. Benzoic acid dimerises in benzene and ionises in water — same solute, opposite deviation.
- Applying the formulae to concentrated solutions. All four expressions are dilute-solution limits; interionic attraction pulls the observed i below the ideal value as concentration rises.
- Using a volatile solute. Raoult's-law lowering of vapour pressure assumes the solute contributes no vapour pressure of its own.
- Mixing up n and α. n is how many particles one formula unit could give; α is the fraction that actually did.
Summary
| Quantity | Equation | Concentration scale | Best used for |
|---|---|---|---|
| Vapour pressure lowering | (p° − p)/p° = i·xsolute | Mole fraction | Volatile solvents, moderate concentrations |
| Boiling point elevation | ΔTb = i·Kb·m | Molality | Thermally stable solutes |
| Freezing point depression | ΔTf = i·Kf·m | Molality | Small molecules; Kf is larger than Kb |
| Osmotic pressure | π = i·M·R·T | Molarity | Very dilute solutions, proteins, polymers |
| van't Hoff factor | i = 1 + (n−1)α or i = 1 − α(1 − 1/n) | — | Dissociation and association respectively |
Check your i-factor working in one step. The van't Hoff Factor calculator handles the whole family — degree of dissociation and association, corrected freezing point depression and boiling point elevation, osmotic pressure and abnormal molar mass — so you can verify each of the examples above before trusting the method under exam conditions.
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