Answer to Question #218707 in General Chemistry for sabc

Question

Calculate the freezing point and boiling point in each solution,

assuming complete dissociation of the solute.

a. 10.5 g FeCl3 in 1.50 * 102 g water

Answer 1

We Consider aqueous of FeCl3 .

Depression in freezing point = i . Kf . m

Elevation in Boilling point = i . Kb . m

i for FeCl₃ = 3 .

Kf for water = 1.86 K. Kg / mol .

Kb for water = 0.52 k.kg / mol .

Molar mass of FeCl₃ = 162.2 g / mol .

Moles of FeCl₃ = 10.5 / 162.2 = 0.065 mol .

Molality = 0.065 / 0.102 = 0.616 mol / kg .

Freezing point =0 - depression in frezing point = - 3 × 1.86 × 0.616 = -3.44 degree celcius.

Boulling point of Solution

= (373 + 3 × 0.52 × 0.616) K = 373.96 K .