Question

An aqueous solution containing 34.1 g of an unknown molecular (non-electrolyte) compound in 159.9 g of water was found to have a freezing point of -1.5 ∘ C Calculate the molar mass of the unknown compound.

Answer 1

Let M be the molar mass of the compound. The number of moles of the unknown compound taken = (34.1 g/M) mole.

The depression in freezing point of the solution of the compound in water = ΔT_f = (T_f)_pure – (T_f)_soln = (0°C) – (-1.5°C) = 1.5°C.

Find the molality of the solution = (moles of solute)/(kg of water) = (34.1 g/M)/[(159.9 g)*(1 kg/1000 g)] = (34.1/M)/(0.1599) = 213.2583/M m

Use the relation

ΔT_f = K*m where K = 1.86°C/m is the freezing point depression constant of water and m = molality of the solution.

Plug in values

1.5°C = (1.86°C/m)*(213.2583/M) m = 396.6604/M

===> M = 396.6604/1.5 = 264.4403 ≈ 264.44

The molar mass of the unknown compound is 264.44 g/mol (ans).