Question

What is the molality of an aqueous solution of KBr having freezing point −3.72°C (K_f for water is 1.86 K kg mol⁻¹)?

Answer 1

Given:

Freezing point of solution = −3.72°C

Freezing point of pure water = 0°C

K_f (freezing point depression constant for water) = 1.86 K kg mol⁻¹ 

i (Van't Hoff factor for KBr) = 2 (since KBr completely dissociates into K⁺ and Br⁻ in water)

ΔT_f (change in freezing point) = 0 − (−3.72) = 3.72°C

Calculating Molality:

`m = (DeltaT_f)/(i xx K_f)`

`m = (3.72)/(2 xx 1.86)`

`m = (3.72)/(3.72)`

m = 1.00 mol/kg