What is the freezing point, in °C, of a solution made with 1.31 mol of CHCl₃ in 530.0 g of CCl₄ (Kf = 29.8 °C/m, normal freezing point, Tf = -22.9 °C)?

Answer 1

For these types of problems, we want to use the following expression:

∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor = 1 for CHCl3 (a non-electrolyte)

m = molality = mols CHCl3 / kg CCl4 = 1.31 mol / 0.530 kg = 2.47 m

K = freezing constant = 29.8º/m

Solving for ∆T we have ...

∆T = (1)(2.47)(29.8) = 73.6ºC

NOTE: this is the CHANGE in the freezing point, and NOT the new freezing point.

Since added substances DECREASE the freezing point (and raise the boiling point), we will need to subtract this from the normal freezing point to find the new freezing point. Or look at this as -73.6º and add it to the normal FP.

Freezing point of solution = -22.9 + -73.6 = -96.5ºC