Solution of sucrose (Molar Mass = 342 g mol⁻¹) is prepared by dissolving 34.2 g of it in 1000 g of water. Free

Question

Solution of sucrose (Molar Mass = 342 g mol⁻¹) is prepared by dissolving 34.2 g of it in 1000 g of water. Freezing point of the solution is: (K_f for water is 1.86 K kg mol⁻¹)

Accepted Answer

The freezing point depression ($\Delta T_f$) is calculated using the formula $\Delta T_f = K_f 	imes m$, where $m$ is the molality of the solution.

1.  **Calculate Molality ($m$):**
    *   Moles of sucrose ($n_2$) = $\frac{	ext{Mass}}{	ext{Molar Mass}} = \frac{34.2}{342} = 0.1 	ext{ mol}$.
    *   Mass of solvent (water) = $1000 	ext{ g} = 1 	ext{ kg}$.
    *   $m = \frac{0.1 	ext{ mol}}{1 	ext{ kg}} = 0.1 	ext{ mol kg}^{-1}$.

2.  **Calculate Depression in Freezing Point ($\Delta T_f$):**
    *   $\Delta T_f = 1.86 	ext{ K kg mol}^{-1} 	imes 0.1 	ext{ mol kg}^{-1} = 0.186 	ext{ K}$.

3.  **Calculate Freezing Point of Solution ($T_f$):**
    *   The freezing point of pure water ($T_f^0$) is typically taken as $273.15 	ext{ K}$, but in this specific problem context (matching the options), it is approximated as $273 	ext{ K}$.
    *   $T_f = T_f^0 - \Delta T_f = 273 	ext{ K} - 0.186 	ext{ K} = 272.814 	ext{ K}$.

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