Question

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 273.15K.

Answer 1

Molar mass of cane sugar

C₁₂H₂₂O₁₁ = 342 g mol^-1 

Molality of sugar = \(\frac{5\times1000}{342\times 100}\)= 0.146 

∆T₂ for sugar solution = 273.15 – 271 = 2.15°

∆T_f = K_f x m 

K_f = \(\frac{2.15}{0.146}\) 

Molality of glucose solution = \(\frac{5}{180}\) x \(\frac{1000}{100}\)= 0.278 m 

∆T_f (Glucose) = \(\frac{2.15}{0.146}\) x 0.278 = 4.09°K 

Freezing point of glucose solution = 273.15 – 4.09 = 269.06 K