Question

A solution of glucose (molar mass = 180 g mol^-1) in water has a boiling point of 100.20°C. Calculate the freezing point of the same solution. Molal constants for water K_f and K_b are 1.86 K kg mol^-1 and 0.512 K kg mol^-1 respectively.

Answer 1

Given: T of glucose solution = 100.20°С

Formula: ΔT_f = K_f.m

ΔT = 100.20 – 100 = 0.20°C 

ΔT = K × m (where, m = molality) 

= `0.20/0.512` 

= 0.390 mol/kg 

ΔT = K_f × m

= 1.86 × 0.390

= 0.725°С

Freezing point of solution = 0 − 0.725

= 0.725°С