Answer to Question #350438 in General Chemistry for Roy

Question

Calculate the freezing point and boiling point of each aqueous solution, assuming complete dissociation of the solute.

a. 0.100 m K₂S

b. 21.5 g of CuCl₂ in 4.50 x10² g water

c. 5.5% NaNO₃ by mass (in water)

Answer 1

The freezing points of each of the solutions are as follows;

0.100 m K2S -  - 0.558oC

21.5 g of CuCl2 in 4.50 * 102 g water -  -2oC

5.5% NaNO3 by mass (in water)​ -   - 2.6oC

What is freezing point?

The freezing point is the point at which liquid changes to solid. Let us now look at the freezing point of each solution.

a)

Since;

ΔT = K m i

K = 1.86 oC m-1

m = 0.100 m

i = 3

ΔT = 1.86 oC m-1 * 0.100 m * 3 = 0.558oC

Freezing point = 0oC - 0.558oC = - 0.558oC

b) Number of moles of CuCl2 = 21.5 g/134.45 g/mol = 0.16 moles

molality = 0.16 moles/0.45 Kg = 0.36 m

ΔT = K m i

ΔT = 1.86 oC m-1 * 0.36 m * 3 = 2oC

Freezing point = 0oC - 2 = -2oC

c) Number of moles of NaNO3 = 5.5g/85 g/mol = 0.065 moles

molality of the solution = 0.065 moles/0.0945 Kg = 0.69 m

ΔT = 1.86 oC m-1 * 0.69 m * 2 = 2.6oC

Freezing point = 0oC - 2.6oC = - 2.6oC