Question

The boiling point of an aqueous solution is 101.77 °C. What is the freezing point? Constants can be found here.

Answer 1

You are not providing the constants, but I'll assume that the constants would be for the water because you are talking about aqueous solution.

The general formula is:

= K * m m is molality

Now for the boiling point = Ts - Tb where Ts is the boiling point of solution and Tb the b.p. of water. so:

For the freezing point = Tf - Ts

Kb for water is 0.52 and Kf is 1.86

101.77 - 100 = 0.52 m

m = 3.4038 m

With this, we can calculate the freezing point:

0 - Tf = 1.86 * 3.4038

-Tf = 6.33

Tf = -6.33 °C

Hope this helps