Gases Liquids and Solids

Question

A certain liquid X has a normal freezing point of 5.50°C and a freezing point depression constant = Kf 1.83·°C·kgmol−1. A solution is prepared by dissolving some iron(III) chloride (FeCl3) in 550.g

 of X. This solution freezes at 4.4°C. Calculate the mass of FeCl3 that was dissolved.

Answer 1

Freezing point depression: Colligative properties:

∆T = imK

∆T = change in freezing point = 5.50º - 4.4º = 1.1º

i = van't Hoff factor = 4 for FeCl₃ since it dissociates to 1Fe³⁺ and 3Cl^- (4 particles)

m = molality = moles FeCl₃/kg solvent = ?

K = freezing constant = 1.83ºC/m

Solving for m we have m = ∆T/(i)(K) = 1.1/((4)(1.83)

m = 0.150 m = 0.150 moles/kg

The solution had 550 g of solvent = 0.550 kg, therefore 0.150 mol/kg x 0.550 kg = 0.0827 moles FeCl₃

mass FeCl₃ = 0.0827 moles x 162 g/mol = 13.4 g FeCl₃