The boiling point of an aqueous solution is 101.36 ∘C. What is the freezing point?

Answer 1

Normal boiling point of water is 100ºC. Therefore, the change in the boiling point is 1.36ºC

∆T = imK

∆T = change in boling point = 1.36º

i = van't Hoff factor

m = molality

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

Since i and m will be the same whether you are determining the elevation in boiling point or the depression in the freezing point, the only variable is K

Kboiling = 0.512 m/º

Kfreezing = 1.86 m/º

∆T = imK

1.36 = im(0.512)

im = 2.66

For freezing point:

∆T = imK

∆T = 2.66 x 1.86

∆T = 4.95º

Freezing point of the solution = -4.95ºC