Colligative Properties: Why Salt Melts Ice and Antifreeze Works

Colligative properties are the ones that only care how many particles you dissolve, not what they are. Dissolve anything in water and you lower its freezing point (ΔTf = i·Kf·m) and raise its boiling point (ΔTb = i·Kb·m). That single idea explains salted roads, antifreeze, and why pasta water with salt boils a hair hotter.

The two equations

Both effects have the identical shape:

• Freezing-point depression: ΔTf = i · Kf · m — subtract ΔTf from the pure freezing point.
• Boiling-point elevation: ΔTb = i · Kb · m — add ΔTb to the pure boiling point.

Three quantities feed both, computed by the freezing-point and boiling-point calculators.

The van’t Hoff factor (i)

This is the piece students most often forget. i is the number of particles a solute breaks into. A colligative property counts particles, so an ionic compound that splits into ions has a bigger effect than its formula-unit count suggests:

• Sugar (glucose) — stays as one molecule → i = 1
• NaCl → Na⁺ + Cl⁻ → i = 2
• CaCl₂ → Ca²⁺ + 2 Cl⁻ → i = 3

So 1 molal CaCl₂ depresses freezing about three times as much as 1 molal sugar, even though it’s “one solute.” (Real ionic solutions fall a little short of the ideal i because of ion pairing, but the whole number is the working value.)

Kf and Kb are solvent constants

Kf (cryoscopic) and Kb (ebullioscopic) depend only on the solvent. For water:

Constant	Value	Direction
Kf	1.86 °C·kg/mol	freezing point down
Kb	0.512 °C·kg/mol	boiling point up

Because Kf is ~3.6× larger than Kb, dissolving something drops the freezing point far more than it raises the boiling point — which is why de-icing is dramatic but “salt makes water boil hotter” is a barely measurable ~1 °C at kitchen concentrations.

Molality, not molarity

Colligative equations use molality (m = moles of solute per kilogram of solvent), not molarity (per litre of solution). Molality is used because it doesn’t change with temperature — and these calculations span from freezing to boiling. Convert to molality first (the molarity calculator helps with the mole count).

Worked example: salted water

1 mole of NaCl in 1 kg of water — how much does the freezing point drop?

ΔTf = i · Kf · m = 2 × 1.86 × 1 = 3.72 °C

So the water now freezes at −3.72 °C instead of 0 °C. The same solution boils at 100 + (2 × 0.512 × 1) = 101.02 °C. That asymmetry — a big freezing drop, a small boiling rise — is colligative behaviour in one sentence.

Quick summary

Colligative properties depend on the number of dissolved particles. Freezing point falls by ΔTf = i·Kf·m and boiling point rises by ΔTb = i·Kb·m, where i (the van’t Hoff factor) counts particles per formula unit and Kf/Kb are solvent constants (1.86 and 0.512 for water). This is why salt melts ice and antifreeze protects an engine at both ends. Work it out with the freezing-point depression and boiling-point elevation calculators.

Sources: standard general-chemistry treatment of colligative properties; cryoscopic and ebullioscopic constants for water. Educational information.