Calculate the molality of the solution of nonvolatile solute if it freezes at $-0.36^{\circ} \mathrm{C}$. [ $\mathrm{K}_{\mathrm{f}}$ for solvent $=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ ]

Molality determination via freezing point depression

The freezing point depression $\Delta T_f = T_f^\circ - T_f = 0^\circ \mathrm{C} - (-0.36^\circ \mathrm{C}) = 0.36 \ \mathrm{K}$, where $T_f^\circ = 0^\circ \mathrm{C}$ is the normal freezing point of water.

Using the colligative property relation $\Delta T_f = K_f \cdot m$, where $K_f = 1.86 \ \mathrm{K} \ \mathrm{kg} \ \mathrm{mol}^{-1}$ for water, the molality is $m = \frac{\Delta T_f}{K_f} = \frac{0.36}{1.86} \approx 0.1935 \ \mathrm{mol} \ \mathrm{kg}^{-1}$.

Rounded to three significant figures, $m = 0.193 \ \mathrm{mol} \ \mathrm{kg}^{-1}$, which corresponds to option B.