Question

The boiling point of an aqueous solution is 101.02 °C. What is the freezing point?

Answer 1

\[\Delta T=i*Kb *m\] Here since nothing is given i can be safely assumed to be equal to1 .
So 1.02=Kb*m.
Kb for water is 0.512 so
m=1.02/0.512=1.8545
Now substitue in freezing point equation. Kf=1.86
T=1.86*1.8545=3.5.
Hence freezing poimt =-3.5 C