Depression of Freezing Point

Exactly the same reasoning that explains boiling-point elevation also explains the depression of freezing point, but applied to the equilibrium between solid solvent and liquid solvent instead of between liquid and vapour. A pure solvent freezes at the temperature at which the vapour pressure of the solid phase exactly equals the vapour pressure of the liquid phase (the two forms of the substance are then in equilibrium). Since dissolving a non-volatile solute lowers the liquid's vapour pressure at every temperature, the vapour-pressure curve of the solution meets the vapour-pressure curve of the pure solid solvent — which is unaffected by a solute that stays dissolved in the liquid and is excluded from the growing solid crystal — only at a temperature lower than the freezing point of the pure liquid solvent. The freezing point of the solution is therefore depressed relative to that of the pure solvent.

Just as with boiling-point elevation, this depression of freezing point, $\Delta T_f = T_f^{\circ}(\text{pure solvent}) - T_f(\text{solution})$, is found experimentally to be directly proportional to the molal concentration of the solution:

$K_f$, the molal depression constant or cryoscopic constant of the solvent, again depends only on the solvent, never the solute, and has units of $\text{K kg mol}^{-1}$. Water has $K_f = 1.86\ \text{K kg mol}^{-1}$, and benzene has the considerably larger value $K_f = 5.12\ \text{K kg mol}^{-1}$ — benzene's larger $K_f$ (and larger $K_b$) is exactly why benzene solutions are frequently favoured in laboratory molar-mass determinations, since a given amount of solute produces a more easily measured temperature change than the same amount would in water.

Rearranged for the solute's molar mass exactly as for boiling-point elevation, this cryoscopic method gives

$M_2 = \frac{1000\, K_f\, w_2}{\Delta T_f\, w_1}$ …