Given: Sucrose solution $=4\%(\mathrm{w}/\mathrm{w})$ $M=342\,\mathrm{g}\,\mathrm{mol}^{-1}$ Freezing point of solution $=271.15\,\mathrm{K}$ Freezing point of pure water $=273.15\,\mathrm{K}$ Glucose solution $=5\%$ $M=180\,\mathrm{g}\,\mathrm{mol}^{-1}$ To calculate: Freezing point of $5\%$ glucose solution. Formula: $\Delta T_f\;= i \times K_f \times m$ $m\;=\;\frac{\;\text{moles of solute}\;}{\;\mathrm{Kg}\;\text{ of solvent}\;}$ $\;\text{moles}\;\;=\;\frac{\;\text{mass}\;}{\;\text{molar mass}\;}$ Sucrose solution is $4\%(\mathrm{w}/\mathrm{w})$ which means there is 4.0 grams of sucrose dissolved in $100\,\mathrm{g}$ of solution. Mass of solution $=$ mass of solute + mass of solvent $100.0\,\mathrm{g}=4.0\,\mathrm{g}+$ mass of solvent. Hence, mass of solvent $=100.0-4.0=96.0\,\mathrm{g}$ $96\,\mathrm{g} \times \frac{1\,\mathrm{kg}}{1000\,\mathrm{g}} = 0.096\,\mathrm{kg}$ Moles of solute $\;\text{moles}\;\;=\;\frac{\;\text{mass}\;}{\;\text{molar mass}\;} = \frac{4.0\,\mathrm{g}}{342\,\mathrm{g}\,\mathrm{mol}^{-1}}$ $=0.011695 \;\text{moles}\;$ Molality of solution: $m = \frac{\;\text{moles of solute}\;}{\;\mathrm{kg}\;\text{ of solvent}\;}$ $= \frac{0.011695\,\mathrm{moles}}{0.096\,\mathrm{kg}\;\text{water}} = 0.1218\,\mathrm{m}$ Sucrose is a non-electrolyte, hence $i=1$ $\Delta T_f\;=\;\text{Freezing point of solvent}$ $- \;\text{Freezing point of solution}$ $\Delta T_f\;=273.15\,\mathrm{K}-271.15\,\mathrm{K}=2.00\,\mathrm{K}$ $\Delta T_f\;= i \times K_f \times m$ $\;2.00\,\mathrm{K}=i \times K_f \times 0.1218$ $\; K_f=16.42\,\mathrm{Km}^{-1}$ For glucose solution, Glucose solution is $5\%(\mathrm{w}/\mathrm{w})$ which means there is 5.0 grams of glucose dissolved in $100\,\mathrm{g}$ of solution. Mass of solution $=$ mass of solute + mass of solvent $100.0\,\mathrm{g}=5.0\,\mathrm{g}+$ mass of solvent. Hence, mass of solvent $=100.0-5.0=95.0\,\mathrm{g}$ $95.0\,\mathrm{g} \times \frac{1\,\mathrm{kg}}{1000\,\mathrm{g}} = 0.095\,\mathrm{kg}$ Moles of solute $\;\text{moles}\;=\;\frac{\;\text{mass}\;}{\;\text{molar mass}\;}$ $= \frac{5.0\,\mathrm{g}}{180\,\mathrm{g}\,\mathrm{mol}^{-1}} = 0.0277 \;\text{moles}\;$ Molality of solution: $m = \frac{\;\text{moles of solute}\;}{\;\mathrm{kg}\;\text{ of solvent}\;}$ $= \frac{0.0277 \;\text{moles}\;}{0.095\,\mathrm{kg}\;\text{water}} = 0.2923\,\mathrm{m}$ Glucose is a non - electrolyte, hence $i=1$ $\Delta T_f = i \times K_f \times m$ $\Delta T_f\;=1 \times 16.42 \times 0.2923$$\Delta T_f\;=4.801\,\mathrm{Km}^{-1}$$\Delta T_f\;=\;\text{Freezing point of solvent}$$- \;\text{Freezing point of solution}$$4.801\;=273.15\,\mathrm{K}-\;\text{Freezing point of solution}$$\;\text{Freezing point of solution}\;\;=273.15\,\mathrm{K}-4.801\,\mathrm{K}$$=268.35\,\mathrm{K}$Thus, the freezing point of the glucose solution is $268.35\,\mathrm{K}$.