chemistry question

Question

The boiling point of an aqueous solution is 101.33 ∘C.  What is the freezing point? 

Answer 1

Normal boiling point of water = 100ºC

So, the change in boiling point = 1.33ºC

Use ∆T = imK to solve for m (molality) and then use that in the freezing point equation to find the freezing point

m = ∆T / (i)(K)

m = molality

∆T = 1.33º

i = van't Hoff factor which we can assign any value since it will be constant in both freezing and boiling =1

K = boiling point constant = 0.512º/m

m = 1.33 / (1)(0.512)

m = 2.6

Now use this to solve for change in freezing point:

∆T = imK

∆T = (1)(2.6)(1.86) (freezing point constant for water = 1.86º/m)

∆T = 4.8º

Freezing point of solution = -4.8ºC