Colligative properties are the strange consequence that solutions don’t care what you dissolve, only how many particles end up floating around. A mole of glucose lowers water’s freezing point by the same amount as a mole of urea, because neither dissociates and neither does anything chemically interesting to the water structure. NaCl, on the other hand, doubles the effect — because it splits into Na⁺ and Cl⁻, and the water counts particles, not formula units. That counting is why you spread salt on icy steps and why marathon runners cramp when they sweat out their electrolytes.

The four properties and their equations

Boiling point elevation: ΔT_b = i × K_b × m

Freezing point depression: ΔT_f = i × K_f × m

Vapor pressure lowering (Raoult’s law): P_solution = X_solvent × P°_solvent

Osmotic pressure: π = i × M × R × T

The constants i, K_b, K_f, m, X carry all the chemistry:

i — van’t Hoff factor, the number of particles per formula unit of dissolved solute
K_b, K_f — solvent-specific constants, in °C·kg/mol
m — molality, mol solute per kg solvent (not molarity — see below)
X — mole fraction

Common solvents:

Solvent	K_b	K_f	BP (°C)	FP (°C)
Water	0.512	1.86	100.0	0.0
Benzene	2.53	5.12	80.1	5.5
Camphor	5.95	37.7	204	179.8

Camphor’s enormous K_f (37.7) is why it became the classic solvent for molar mass determination by freezing point depression — even tiny amounts of solute give measurable temperature changes.

Working a freezing point depression problem

Dissolve 10.0 g of NaCl in 500 g of water. What’s the new freezing point?

NaCl dissociates into Na⁺ + Cl⁻, so i = 2.

Moles NaCl = 10.0 / 58.44 = 0.171 mol

Molality = 0.171 / 0.500 kg = 0.342 m

ΔT_f = i × K_f × m = 2 × 1.86 × 0.342 = 1.27 °C

New freezing point = 0.0 − 1.27 = −1.27 °C

Always subtract for freezing point. The solution freezes at a lower temperature than the pure solvent. Add for boiling point — solutions boil higher.

Worked examples by property

Boiling point elevation with glucose

50.0 g of glucose (180.16 g/mol) in 200.0 g of water:

Glucose is non-electrolyte: i = 1
Moles: 50.0 / 180.16 = 0.2775
Molality: 0.2775 / 0.200 = 1.388 m
ΔT_b = 1 × 0.512 × 1.388 = 0.711 °C
New BP = 100.7 °C

Antifreeze: how much CaCl₂ for −5.0 °C?

CaCl₂ → Ca²⁺ + 2Cl⁻, so i = 3. You want ΔT_f = 5.0 °C in 1.00 kg of water.

m = ΔT_f / (i × K_f) = 5.0 / (3 × 1.86) = 0.896 m

Moles needed: 0.896 × 1.00 kg = 0.896 mol

Mass: 0.896 × 110.98 g/mol = 99.4 g

CaCl₂ is a better deicer than NaCl for two reasons baked into this calculation: higher i (3 vs 2) and lower freezing point of the brine. That’s why airports use CaCl₂ on runways instead of road salt.

Vapor pressure with glycerol

30.0 g of glycerol (92.09 g/mol) in 100.0 g of water at 25 °C, where pure water has P° = 23.8 mmHg.

Moles glycerol: 30.0 / 92.09 = 0.3258
Moles water: 100.0 / 18.015 = 5.551
X_water: 5.551 / (5.551 + 0.3258) = 0.9446
P = 0.9446 × 23.8 = 22.5 mmHg

A 1.3 mmHg drop. Small but measurable, and it’s the reason saltwater evaporates more slowly than fresh.

Molar mass from freezing point depression

A 5.00 g sample of an unknown non-electrolyte dissolved in 100.0 g of benzene depresses the freezing point by 1.56 °C:

i = 1
m = 1.56 / (1 × 5.12) = 0.3047 m
Moles: 0.3047 × 0.100 = 0.03047 mol
Molar mass: 5.00 / 0.03047 = 164 g/mol

Useful for unknowns that are hard to characterize spectroscopically — works on anything that dissolves and doesn’t dissociate.

Easy ways to mess this up

Forgetting i for ionic compounds. NaCl looks like a single thing on paper but acts like two particles in solution. Skip i and your prediction for any salt will be off by a factor of 2 or more. Real solutions also show “ion pairing” at higher concentrations — i for NaCl is closer to 1.9 in practice, not exactly 2 — but for textbook problems use the formula-unit count.

Molality vs molarity. Colligative formulas use molality (per kg solvent), not molarity (per L solution). Molality is temperature-independent, which matters because you’re calculating temperature changes. Using molarity by mistake will give you the wrong answer at any non-trivial concentration.

Sign direction. Freezing point goes down, boiling point goes up. Always. If your solution boils colder than pure solvent, you made an arithmetic error.

Mass of solvent vs mass of solution. Molality denominator is just the solvent mass. Don’t include the solute mass in the denominator.

Practice

Run these against the Solution Concentration Calculator:

Boiling point of 25.0 g NaCl in 500 g water.
Freezing point of 100.0 g ethylene glycol (62.07 g/mol) in 500 g water.
Vapor pressure at 25 °C of 60.0 g sucrose (342.3 g/mol) in 200.0 g water (P°_water = 23.8 mmHg).
A 3.00 g unknown non-electrolyte in 50.0 g water depresses freezing point by 0.620 °C — molar mass?
Grams of KCl needed to raise the boiling point of 2.00 kg of water by 1.50 °C.

How to Calculate Colligative Properties