$\Delta T_t = (273.15-269.15) \mathrm{K} = 4 \mathrm{K}$ Molar mass of sucrose $(\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11})$ $\; = (12 \times 12) + (22 \times 1) + (11 \times 16)$ $\; = 342 \mathrm{g} \mathrm{mol}^{-1}$ $10\%$ solution of sucrose in water means $10 \mathrm{g}$ of sucrose is present in $(100-10) \mathrm{g}$ of water. $\; \text{Number of moles of sucrose} \; = \; \frac{10}{342}$ $\; = 0.0292 \mathrm{mol}$ Therefore, molality of the solution $= \frac{0.0292 \times 1000}{90} = 0.3244 \mathrm{mol} \mathrm{kg}^{-1}$ We know that $\Delta T_t = K_f \times m$ $\Rightarrow K_f = \frac{\Delta T_t}{m} = \frac{4}{0.3244}$ $= 12.33 \mathrm{K} \mathrm{Kg} \mathrm{mol}^{-1}$ Molar mass of glucose $(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6)$ $= (6 \times 12) + (12 \times 1) + (6 \times 16)$ $= 180 \mathrm{g} \mathrm{mol}^{-1}$ $10\%$ solution of glucose in water means $10 \mathrm{g}$ of glucose is present in $(100-10) \mathrm{g}$ of water. $\; \text{Number of moles of glucose} \;$ $= \frac{10}{180} = 0.0555 \mathrm{mol}$ Therefore,molality of the solution $\; = \frac{0.0555 \times 1000}{90}$ $\; = 0.6166 \mathrm{mol} \mathrm{kg}^{-1}$ $\; \text{We know that} \; \Delta T_t = K_f \times m$ $\Rightarrow \; \Delta T_t = 12.33 \times 0.6166 = 7.60 \mathrm{K}$ So, the freezing point of $10\%$ glucose solution in water is $(273.15-7.60) \mathrm{K} = 265.55 \mathrm{K}$