Question

an aqueous solution of lithium chloride (LiCk: f MM=42.4 g/mole) is 27% licl by Mass. (27g of LiCk +73 g of H2O) it has a density of 1.127 grams per mL.
a) determine the molality
b) determine the freezing point of the solution. kf=1.86°C/m
c) determine the molarity of the solution.

Answer 1

(a) Molality = moles of solute/mass of solvent in kgs

moles of solute = mass/molar mass

mass of LiCl = 27g

Molar Mass of LiCl = 42.4g/mol

moles = 27g/42.4g/mol = 0.637mols

Mass of solvent in kgs = 73g = 0.073kg

Molality = 0.637/0.073 = 8.73m

(b) depression in freezing point is given by,

delta T = Kfm

Where delta T = change in freezing point of the solute

Kf = molal freezing point depression constant

m = molality of solute

So,

DeltaT= 1.86C/m *8.73m

Delta T = 16.24C

freezing point of pure LiCl = 605C

So freezing point of LiCl solution = 605-16.24 = 588.76C

c) Molarity = moles of solute/ Vol.of solvent in litres

moles of LiCl = 0.637

Vol= Mass/Density

Vol= 73g/1.127g/ml

Vol = 64.774mL =0.065L

Molarity = 0.637/0.065 =9.8M