Determine the freezing point depression of 2 kg of water when 2 mol salt is added to it. The kf of water is 1.86 degrees C/m. Show all your work.

Answer 1

Colligative properties:

Freezing point depression.

∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor = 2 for NaCl (assuming that is the "salt" they are using) b/c NaCl = Na+ and Cl-

m = molality = moles salt / kg water = 2 mol / 2 kg = 1 molal

K = freezing point constant = 1.86º/m

Solving for ∆T we have...

∆T = (2)(1)(1.86) = 3.72º

Freezing point depression is -3.72º

New freezing point would be -3.72ºC