Question

A 5% solution (by mass) of cane-sugar in water has freezing point of 271 K. Calculate the freezing point of 5% solution (by mass) of glucose in water if the freezing point of pure water is 273.75 K. [Molecular masses : Glucose C₆H₁₂O₆ : 180 amu; Cane x sugar C₁₂H₁₂O₁₁ : 342 amu]

Answer 1

5% by mass cane sugar = 5 g cane sugar (Mol. wt = 342 amu in 100 g solution)

5% by mass glucose = 5 g glucose (Mol. wt =180 amu) in 100 g solution

Freezing point of glucose solution,

T_f = 269.065 K