Answer:
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