The freezing point of a 1.00 m aqueous solution of HF is found to be $$-1.91°C$$. The freezing point constant of water, $$K_f$$ is 1.86 K kg mol$$^{-1}$$. The percentage dissociation of HF at this concentration is

The depression in freezing point is related to the van’t Hoff factor $$i$$ by the formula
$$\Delta T_f = i\,K_f\,m$$
where $$\Delta T_f$$ is the fall in freezing point, $$K_f$$ is the cryoscopic constant of the solvent and $$m$$ is the molality of the solution.

Given data:
$$\Delta T_f = 1.91\;{\text{K}}$$ (because $$-1.91^{\circ}\text{C}$$ is a drop of 1.91 K)
$$K_f = 1.86\;{\text{K kg mol}}^{-1}$$
$$m = 1.00\;{\text{mol kg}}^{-1}$$

Calculate the van’t Hoff factor:
$$i = \frac{\Delta T_f}{K_f\,m} = \frac{1.91}{1.86 \times 1.00} = 1.02688 \;\approx\; 1.027$$

HF dissociates as $$HF \rightleftharpoons H^{+} + F^{-}$$
For one mole of HF, let the degree of dissociation be $$\alpha$$. Initial moles: $$1$$ (HF only) After dissociation: $$(1-\alpha)\;{\text{mol HF}},\; \alpha\;{\text{mol }}H^{+},\; \alpha\;{\text{mol }}F^{-}$$ Total moles of solute particles $$=\,(1-\alpha)+\alpha+\alpha = 1+\alpha$$

The general relation for a binary (1 → 2) dissociation is $$i = 1 + \alpha$$. Equating with the experimental value:
$$1 + \alpha = 1.027$$ $$\alpha = 0.027$$

Percentage dissociation $$= \alpha \times 100\% = 0.027 \times 100\% = 2.7\%$$

Option D which is: 2.7%