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.

Explanation

Given that the boiling point `(T_b)` of glucose solution = `100.20^@C`
Apply the relation :
` Delta T_b = K_b xx m`
`m = (Delta T_b)/(K_b) = (0.20)/(0.512)`
m = 0.390 mol/kg
Applying it to the freezing point
` Delta T_f = K_f xx m`
` Delta T_f = 1.86 K kg "mol"^(-1) xx 0.390 "mol" kg^(-1) = 0.725 K`
Freezing point of the solution = 273.15 K - 0.725 K = 272.425 K