Answer to Question #218707 in General Chemistry for sabc

Question #218707

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


1
Expert's answer
2021-07-19T04:05:00-0400

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 .


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