The boiling point of a benzene solution with a non-electrolyte solute is 90°C. Calculate the freezing point of the same solution.

Question

please help. I'm confused with this problem

Answer 1

Colligative properties:

Boiling point elevation and freezing point depression.

∆T = imK

∆T = change in boiling or freezing point = 9.9º (this is because I looked up the normal boiling point of benzene and found it to be 80.1ºC, so the change in b.p. would be 90º - 80.1º = 9.9º)

i = 1 for a non electrolyte

m = molality = ? (we will solve for this)

K = boiling constant for benzene (2.53º/m - I looked it up)

Solving for m we have..

m = ∆T / (i)(K) = 9.9 / (1)(2.53)

m = 3.91 molal

Now that we have the molality of the solution, we can find the freezing point as follows:

Normal freezing point of benzene = 5.5ºC (looked it up)

Freezing point constant (K) = 4.9º/m

∆T = imK

∆T = (1)(3.91)(4.9)

∆T = 19.2º

Freezing point of solution = 5.5 - 19.2 = -13.7ºC