A solution containing $$0.85$$ g of $$\text{ZnCl}_2$$ in $$125.0$$ g of water freezes at $$-0.23°$$C. The apparent degree of dissociation of the salt is ($$K_f$$ for water $$= 1.86$$ K kg mol$$^{-1}$$, atomic mass: Zn $$= 65.3$$ and Cl $$= 35.5$$)

Molar mass of $$\text{ZnCl}_2$$: $$M = 65.3 + 2(35.5) = 136.3\ \text{g mol}^{-1}$$.

Moles of solute present: $$n = \frac{0.85\ \text{g}}{136.3\ \text{g mol}^{-1}} = 0.006235\ \text{mol}$$.

Mass of water (solvent) $$= 125.0\ \text{g} = 0.1250\ \text{kg}$$.

Molality of the solution: $$m = \frac{n}{\text{kg of solvent}} = \frac{0.006235}{0.1250} = 0.04988\ \text{mol kg}^{-1}$$.

Observed depression in freezing point: $$\Delta T_f = 0.23\ \text{K}$$.

For freezing‐point depression, $$\Delta T_f = i\,K_f\,m$$.
Hence $$i = \frac{\Delta T_f}{K_f\,m} = \frac{0.23}{1.86 \times 0.04988} = 2.48$$ (to three significant figures).

Dissociation of $$\text{ZnCl}_2$$: $$\text{ZnCl}_2 \rightarrow \text{Zn}^{2+} + 2\text{Cl}^-$$.
Total particles after complete dissociation, $$v = 3$$.

Relation between van’t Hoff factor $$i$$ and degree of dissociation $$\alpha$$:
$$i = 1 + \alpha(v - 1) \; \Longrightarrow \; i = 1 + 2\alpha$$.

Therefore $$\alpha = \frac{i - 1}{2} = \frac{2.48 - 1}{2} = 0.739 \approx 0.735$$.

Percentage dissociation $$= \alpha \times 100\% \approx 73.5\%$$.

Option B which is: $$73.5\%$$