A 5% solution (by mass) of cane sugar in water has a freezing point of 271 K. The freezing point of a 5% solution (by mass) of glucose in water is: [freezing point of pure water: 273.15 K]

Step-by-step solution

Depression in freezing point is given by T_ f = K_ f m For 5 % (by mass) cane sugar solution: Mass of cane sugar = 5 g, Mass of water = 95 g Molar mass of cane sugar = 342 g/mol T_ f1 = 273.15 - 271 = 2.15 K 2.15 = K_ f 5 1000 342 95 For 5 % (by mass) glucose solution: Mass of glucose = 5 g, Mass of water = 95 g Molar mass of glucose = 180 g/mol T_ f2 = K_ f 5 1000 180 95 Taking the ratio of T_ f2 and T_ f1 : T_ f2 2.15 = 342 180 = 1.9 T_ f2 = 2.15 1.9 = 4.085 K Freezing point of glucose solution = 273.15 - 4.085 =