An aqueous solution has a freezing point of -13.8 ∘C, what is the boiling point of the solution? The Kf value for water is 1.86 ∘C/m, and the Kb value for water is 0.520 ∘C/m.

Question

An aqueous solution has a freezing point of -13.8 C, what is the boiling point of the solution? The Kf value for water is 1.86 C/m, and the Kb value for water is 0.520 C/m.

Answer 1

Use ∆T = imK

∆T = change in freezing or boiling point = 13.8º for the freezing point

i = van't Hoff factor = 1 for a non electrolyte and since it doesn't state what the solute is, we assume this value

m = molality = ?

K = freezing or boiling constant for water = 0.52 for freezing point

We can now solve for m:

m = ∆T / (i)(K) = 13.8 / (1)(0.52) = 26.5 molal

Using this, we can now find the change in boiling point:

∆T = (1)(26.5 m)(1.86) = 49.4ºC

Boiling point of the solution would be 100 + 49.4 = 149.4ºC