The boiling point of an aqueous solution is 101.02 °c. what is the freezing point?
h2o kf=1.86 normal freezing pt= 0.00 kb=0.512 normal bp= 100.00

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

ΔT(boiling point) = 101.02 °C - 100.0 °C= 1.02 °C
ΔT(boiling point)  = (Kb)m
m =  1.02 °C / 0.512 °C kg / mol
m = 1.99 mol / kg

ΔT(freezing point)  = (Kf)m
ΔT(freezing point)  = 1.86 °C kg / mol (1.99 mol / kg)
ΔT(freezing point)  = 3.70 °C
Tf - T = 3.70 °C
T = -3.70 °C

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