An aqueous solution has a boiling point of 102.45°C. What is its freezing point? ( Kf = 1.86°C m-1 and Kb = 0.51°C m-1 )

The freezing point for the solution is -8.9°C

Explanation:

We must combine the two colligative properties of freezing point depression and boiling point elevation to solve this excersise.

ΔT = Kb . m

ΔT = T° boiling point of solution - T° boiling of pure solvent

Kb = Ebullioscopic constant

m = molality

102.45°C - 100°C = 0.51°C/m . m

2.45°C / 0.51 m /°C = m → 4.80 mol/kg

We found out the molality in the boiling point elevation to replace at the freezing point depression formula

ΔT = Kf . m

ΔT = T° freezing pure solvent - T° freezing solution

Kf = Cryoscopic constant

0°C - T° freezing solution = 1.86°C/m . 4.80 m

T° freezing solution = - 8.9°C