Freezing Point of an Antifreeze Solution

Medium

Problem

An antifreeze mixture contains ethylene glycol (a nonelectrolyte) dissolved in water at a molality of 2.00 m. Calculate the freezing point of this solution. (Kf for water = 1.86 C/m.)

Given

Ethylene glycol molality m = 2.00 mol/kg
van't Hoff factor i = 1 (ethylene glycol is a nonelectrolyte)
Kf(water) = 1.86 C/m
Freezing point of pure water = 0.00 C.

Find

The freezing point of the solution in degrees Celsius.

Principle

Freezing point depression is a colligative property: it depends on the number of dissolved particles, not their identity. The depression equals i x Kf x m, where i counts particles per formula unit (1 for a nonelectrolyte like ethylene glycol), Kf is the solvent's cryoscopic constant, and m is the molality.

Formula

\Delta T_f = i K_f m

delta Tf = i x Kf x m; Tf(solution) = Tf(pure) - delta Tf

Solution

Identify the van't Hoff factor: ethylene glycol does not dissociate, so i = 1.
Compute the depression: delta Tf = i x Kf x m = 1 x 1.86 C/m x 2.00 m = 3.72 C.
Subtract from the pure solvent's freezing point: Tf = 0.00 C - 3.72 C = -3.72 C.

The solution freezes at -3.72 C

Unit check

Kf has units C/m (degrees per molal), so Kf x m gives C x (mol/kg)/(mol/kg) = C, a temperature difference. Subtracting from 0.00 C yields a temperature in C, as required.

What this means physically

Dissolving 2.00 mol of ethylene glycol per kilogram of water lowers the freezing point by 3.72 C, so the coolant stays liquid down to -3.72 C. Real antifreeze is more concentrated, pushing the freezing point far lower — the same colligative principle, just a bigger molality.

Common mistake

Two classic errors: using molarity instead of molality (the constant Kf is defined per molal), and adding the depression to 0 C instead of subtracting it — dissolved solute lowers the freezing point, never raises it.