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(D) Step $1$: Calculate the depression in freezing point $(\Delta T_f)$.$\Delta T_f = T_f^\circ - T_f = 0 	ext{ }^\circ	ext{C} - (-0.93 	ext{ }^\ci

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$0.4$

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(D) Step $1$: Calculate the depression in freezing point $(\Delta T_f)$.$\Delta T_f = T_f^\circ - T_f = 0 	ext{ }^\circ	ext{C} - (-0.93 	ext{ }^\circ	ext{C}) = 0.93 	ext{ K}$.Step $2$: Use the formula for depression in freezing point with the van't Hoff factor:$\Delta T_f = i 	imes K_f 	imes m$.Step $3$: Rearrange the formula to solve for molality $(m)$:$m = \frac{\Delta T_f}{i 	imes K_f}$.Step $4$: Substitute the given values:$m = \frac{0.93}{1.25 	imes 1.86} = \frac{0.93}{2.325} = 0.4 	ext{ m}$.Thus, the molality of the solution is $0.4 	ext{ m}$.

Calculate the molality of an aqueous solution of an electrolyte that freezes at $-0.93 \text{ }^\circ\text{C}$. Given that $K_f$ for water is $1.86 \text{ K kg mol}^{-1}$ and the van't Hoff factor $(i)$ is $1.25$. (Freezing point of pure water $= 0 \text{ }^\circ\text{C}$) (in $\text{ m}$)

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