A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
The freezing point of 5% glucose solution is 269 K (or 269.1 K).
Explanation
This problem involves comparing freezing point depression for two solutions with same mass percentage but different solutes. Since both solutions have same mass of solute (5g) and same mass of solvent (95g water), the ratio of freezing point depressions is inversely proportional to the ratio of their molar masses. The depression for glucose will be higher because it has lower molar mass, resulting in more moles and higher molality compared to sucrose.
Solution Steps
Step 1: Calculate depression in freezing point for cane sugar solution. ΔTf = 273.15 – 271 = 2.15 K
Step 2: Calculate molality of cane sugar solution. For 5% solution: 5 g solute in 95 g water (0.095 kg) Molar mass of cane sugar (C₁₂H₂₂O₁₁) = 342 g mol⁻¹ Molality = (5/342) / 0.095 = 0.154 mol kg⁻¹
Step 3: Calculate Kf for water. Kf = ΔTf / molality = 2.15 / 0.154 = 13.96 K kg mol⁻¹
Step 4: Calculate molality of glucose solution. Molar mass of glucose (C₆H₁₂O₆) = 180 g mol⁻¹ Molality = (5/180) / 0.095 = 0.292 mol kg⁻¹
Step 5: Calculate depression in freezing point for glucose solution. ΔTf = Kf × molality = = 4.08 K
Step 6: Calculate freezing point of glucose solution. Freezing point = 273.15 – 4.08 = 269.07 K 269 K