Question

An aqueous solution containing 15.9 g of an unknown molecular (nonelectrolyte) compound in 106.5 g of water was found to have a freezing point of -2.0 ∘C.

Calculate the molar mass of the unknown compound.

Answer 1

solvent= 106.5 g/1000 = 0.1065 Kg
Freezing point depression constant for water k= 1.86°C/m

Freezing point of water T_initial= 0°C
Freezing point of solution T_final= -2°C

depression in freezing point ∆T_f = 2°C.

m = molality = no. of moles of solute/mass of solute in Kg

m = n/mass

No. of moles = Mass/Molar mass

∆T_f = k . m
2°C = 1.86°C/m x m

m = 2/1.86

Molality = 1.075 m

Molaity = moles of solute/solvent in Kg

1.075 m = n/0.1065 kg

No. of moles = 1.075 x 0.1065 = 0.1144 moles

Given mass of solute = 15.9 g

N = mass/Molar mass

Molar mass = 15.9 /0.1144 =138.98 g/mol

Molar mass of unknown compound = 138.98 g/mol