Answer to Question #210659 in Chemistry for amrit

Question

A solution containing 6.35 g nonelectrolyte dissolved in 500g of water freezes at -0.465°C. Determine the molecular weight of the solute.

Answer 1

The depression in the freezing point ΔKf = ​0−(−0.465) = 0.465

ΔTf​=Kf x m

0.465=1.86 m

m = 0.25 mol/kg

Let M be the molecular weight of solute.

The number of moles is the ratio of mass to molecular weight.

Number of moles of solute = 6.35 / M

​Molality of solution is the number of moles of solute in 1000 g of water.

Molality = (6.35 × 1000) / (M × 500) = 0.25

⟹M=50.8 g/mol