Given: Mass of solute (sucrose, $w_2$) = $68.5 \text{ g}$ Molar mass of sucrose ($M_2$) = $342 \text{ g mol}^{-1}$ Mass of solvent (water, $w_1$) = $1000 \text{ g} = 1 \text{ kg}$ $K_f$ for water = $1.86 \text{ K kg mol}^{-1}$

First, we calculate the molality ($m$) of the solution: $m = \frac{w_2}{M_2 \times w_1 \text{ (in kg)}} = \frac{68.5}{342 \times 1} \approx 0.2 \text{ mol kg}^{-1}$

Now, calculate the depression in freezing point ($\Delta T_f$): $\Delta T_f = K_f \times m = 1.86 \times 0.2 = 0.372^\circ\text{C}$

The freezing point of pure water is $0^\circ\text{C}$. Therefore, the freezing point of the solution ($T_f$) is: $T_f = T_f^\circ - \Delta T_f = 0^\circ\text{C} - 0.372^\circ\text{C} = -0.372^\circ\text{C}$.