FeCl₃ has a van't Hoff factor of 3.400. What is the freezing point (in °C) of an aqueous solution made with 0.7100 m FeCl₃? (Kf for water is 1.860 °C/m)

Question

FeCl₃ has a van't Hoff factor of 3.400. What is the freezing point (in °C) of an aqueous solution made with 0.7100 m FeCl₃? (Kf for water is 1.860 °C/m)

Answer 1

∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor = 3.400

m = molality = moles/kg = 0.7100

K = constant = 1.860º/m

Solving for ∆T, we have...

∆T = (3.400)(0.7100)(1.860)

∆T = 4.490ºC

Freezing point of solution = -4.490ºC (b/c normal freezing point is 0ºC