Question

What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of in 115 g of water? [Use these Molar Masses: Ca = 40, N = 14, O = 16. Also, the molal freezing point depression constant for water is 1.86 °C/m.]

Answer 1

Given data

Mass of Ca(NO3)2 = 11.3 g

Mass of water = 115 g x 1kg/1000g = 0.115 kg

From freezing-point depression equation

Tf - Ts = i x Kf x m

where

Tf = freezing-point of water

Ts = Freezing point of solution
i = the van't Hoff factor
Kf = molal freezing point depression constant = 1.86 °C/m.
m = molality of solution

Calcium nitrate is a soluble ionic compound

The dissociation reaction

Ca(NO3)2(s) = Ca2+(aq) + 2NO3- (aq)

2 moles of NO3- ions and 1 mol of Ca2+ ion

i = 3

Molecular weight of Ca(NO3)2 = 40 + 2*(14 + 16*3) = 164 g/mol

Moles of Ca(NO3)2 = mass/molecular weight

= (11.3 g) / (164 g/mol)

= 0.0689 mol

Molality of solution = moles / kg of water

= (0.0689 mol ) / (0.115 kg)

= 0.5992 mol/kg

Tf - Ts = i x Kf x m

0 - Ts = 3 x 1.86 x 0.5992

Ts = - 3.34 °C