31 g of ethylene glycol (C₂H₆O₂) is mixed with 500 g of solvent (K_f of the solvent is 2 K kg/mol). What is the freezing point of the solution in K? (freezing point of solvent = 273 K)

271

Depression in freezing point, ΔT_f = \(\frac{K_f × W_b × 1000}{W_a × M_b}\)

Given, Cryoscopic constant, \(K_f = 2 K\text{ kg / mol}\)

\(W_b  = 31\)

\(W_a = 5000\)

\(M_b = 62\)

∴ ΔT_f = \(\frac{2 × 31 × 1000}{500 × 62}\) = 2 K

Freezing point of solution = 273 - 2 = 271 K