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

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

• $m = \frac{0.1 \text{ mol}}{1 \text{ kg}} = 0.1 \text{ mol kg}^{-1}$.

2. Calculate Depression in Freezing Point ($\Delta T_f$):

• $\Delta T_f = 1.86 \text{ K kg mol}^{-1} \times 0.1 \text{ mol kg}^{-1} = 0.186 \text{ K}$.

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