This question concerns freezing point depression.

Question

Predict the freezing point of a solution of 155 g LiCl in 575 g water.

Answer 1

molar mass LiCl = 42.39 g / mol

∆T = imK

∆T = change in freezing point = ?

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

m = molality = mols LiCl / kg = see below

K = freezing point depression constant for water = 1.86º/m

Calculating m (molality)

moles = 155 g LiCl x 1 mol / 42.39 g = 3.657 mols

kg of water = 575 g x 1 kg / 1000 g = 0.575 kg

m = 3.675 mol / 0.575 kg = 6.359 m

Solving for ∆T

∆T = (2)( 6.359)(1.86)

∆T = 23.7º

Freezing point of solution = 0º - 23.7º = -23.7ºC