A 0.100 mol kg^-1 aqueous solution of a weak monobasic acid HA has K_a=1.2x10^-2. Assume molality equals molarity numerically, use the full quadratic expression K_a=(Calpha^2)/(1-alpha) (do not neglect alpha against 1), and take K_f=1.86 K kg mol^-1. The freezing point of the solution is closest to

JEE Advanced levelSingle correct optionDifficulty 3 of 3so201

A $0.100 mol k g^{- 1}$ aqueous solution of a weak monobasic acid HA has $K_{a} = 1.2 \times 1 0^{- 2}$ . Assume molality equals molarity numerically, use the full quadratic expression $K_{a} = \frac{C α ^{2}}{1 - α}$ (do not neglect $α$ against 1), and take $K_{f} = 1.86 K kg mo l^{- 1}$ . The freezing point of the solution is closest to

0.100 α^{2} + 0.012 α - 0.012 = 0

, giving

α = 0.2916

. Then

i = 1 + α = 1.2916

and

Δ T_{f} = 1.86 \times 0.100 \times 1.2916 = 0.240 K

, so

T_{f} = - 0.24 0^{\circ} C

. The shortcut

α = K_{a} / C = 0.3464

is not valid here because

α

is not small and would give

- 0.25 0^{\circ} C