This question concerns freezing point depression.

Question

Predict the freezing point of a solution of 456 g ethanol and 75.0 g KCl.

Answer 1

Molar mass KCl = 74.56 g / mol

moles KCl used = 75.0 g x 1 mol / 74.56 g = 1.01 mols

Kg of ethanol = 456 g x 1 kg / 1000 g = 0.456 kg

To determine the freezing point, we will need to know the freezing point of PURE ethanol. Looking it up on the internet, I find it to be -114.6ºC (you may have a different value).

Use ∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor for KCl = 2 (K⁺ and Cl^-; 2 particles)

m = molality = mols KCl per kg ethanol = 1.01 mol / 0.456 kg = 2.21 m

K = freezing point depression constant for ethanol = -1.99º/m

Solving for ∆T...

∆T = (2)(2.21)(1.99) = -8.82º

Freezing point of solution = -114.6º + (-8.82º) = -123ºC