31 g of ethylene glycol (C2H6O2) is mixed with 500 g of solvent (Kf of the solvent is 2 K kg/mol). What is the freezing point of the solution in K? (freezing point of solvent = 273 K)
Answer & explanation
Correct answer: option 2
Depression in freezing point, ΔTf = \(\frac{K_f × W_b × 1000}{W_a × M_b}\)
Given, Cryoscopic constant, \(K_f = 2 K\text{ kg / mol}\)
\(W_b = 31\)
\(W_a = 5000\)
\(M_b = 62\)
∴ ΔTf = \(\frac{2 × 31 × 1000}{500 × 62}\) = 2 K
Freezing point of solution = 273 - 2 = 271 K