Question

What is the freezing point of an aqueous solution made by dissolving 0.118 mole of MgCl2 in 100.0 g of water? Kf = 1.86 oC/m

Answer 1

Answer:

Freezing point of solution is -4.39°C

Explanation:

Freezing point depression's formula:

ΔT = Kf . m . i

ΔT = Freezing point of pure solvent - Freezing point of solution

0°C - X = 1.86°C/m . m . 2

i = Van't Hoff factor, ions dissolved in solution

MgCl₂ → Mg²⁺  + 2Cl⁻     i = 3  →  1 mol of Mg and 2 mol of chloride

Let's think the molality (mol/kg)

First of all, let's convert the mass of solvent from g to kg

100 g / 1000 = 0.1 kg

Molality (mol/kg) = 0.118 mol / 0.1 kg = 1.18 m

0°C - X = 1.86°C/m . 1.18 m . 2

X = - 4.39°C