Question

The boiling point of an aqueous solution is 102.08 °C. What is the freezing point? Constants can be found here.

Answer 1

Colligative properties calculations are used for this type of problem. Calculations are as follows:


ΔT(freezing point) = (Kf)(molality)
ΔT(freezing point) = 1.86 °C kg / mol (molality)
Tf - 102.08 = 1.86m

Tf = 1.86m + 102.08

The concetration of the solution is needed in order to obtain a specific value.