Answer to Question #186289 in General Chemistry for Mark Dave Cole

Question

1. Calculate the freezing and boiling points of a solution prepared by dissolving 15.5 g of Al(NO₃)₃ in 200.0 g of water.

Answer 1

Now, the equation that describes freezing-point depression looks like this

ΔTf=i⋅Kf⋅b

, whereΔTf- the freezing-point depression;

i - the van't Hoff factor

Kf- the cryoscopic constant of the solvent;

b - the molality of the solution.

In your case, the cryoscopic constant of water is said to be equal to

Kf=1.86°C

m=1.86°C kg mol^-1

i = 3

Molar Mass of Al(NO₃)₃ = 212.996

15.5/212.996

= 0.07277 M

Molality = 0.07277/0.2

= 3.186 × 0.364

= 1.16°C

Freezing point = 0-1.16 = -1.16°C

Boiling point = 100 - 1.16 = 98.85°C