Question

An aqueous dilute solution of a non-volatile solute boils at 373.052K. Find the freezing point of the solution. For water Kb = 0.52 K kgmol-1 and Kf = 1.86 K kgmol-1. Normal boiling point of water = 373K and normal freezing point = 273K.

Answer 1

it has given that, an aqueous dilute solution of a non - volatile solute boils at 373.052 K.

Kb = 0.52 K kg/mol, Kf = 1.86 K kg/mol, Normal boiling point of water = 373 K and normal freezing point = 273 K.

To find : The freezing point of the solution.

solution : change in boiling points, ∆Tb = (373.052K- 373K) = 0.052 K

using formula, ∆Tb = m × Kb

⇒0.052 K = m × 0.52 K Kg/mol

⇒m = 0.1 mol/kg

now for freezing point using formula, ∆Tf = m × Kf

⇒ (273K - T) = 0.1 mol/kg × 1.86 K kg/mol

⇒273 K - T = 0.186 K

⇒T = 273 K - 0.186 K = 272.814 K

Therefore the freezing point of the solution is 272.814 K.

Answer 2

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