JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Solutions

The weak acid HA ionises as $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$, so one mole of acid gives $(1+\alpha )$ moles of particles and the van't Hoff factor is $i=1+\alpha =1+0.3=1.3$. Because freezing-point depression is a colligative property, the observed depression is $i$ times the value expected for the undissociated acid, $(\Delta T_f{)}_{obs}=i{K}_{f}m$ against $(\Delta T_f{)}_{calc}=K_{f}m$. The fractional rise is therefore $\frac{(\Delta T_f{)}_{obs}-(\Delta T_f{)}_{calc}}{(\Delta T_f{)}_{calc}}\times 100=(i-1)\times 100=(1.3-1)\times 100$. The observed freezing-point depression is 30 per cent higher than the theoretical value.