Question

The boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1°C higher than that of the pure ether. The molecular weight of the substance will be ____________.

(K_b = 2.16 K kg mol⁻¹)

Options

• 148 g mol⁻¹

• 158 g mol⁻¹

• 168 g mol⁻¹

• 178 g mol⁻¹

Answer 1

The boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1°C higher than that of the pure ether. The molecular weight of the substance will be 158 g mol⁻¹.

Explanation:

The molecular weight of substance = M₂

Molal elevation constant = K_b = 2.16 K kg mol⁻¹

Mass of solute = W₂ = 0.11 g

Mass of solvent = W₁ = 15 g

Elevation in boiling point = ∆T_b = 0.1°C = 0.1 K

M₂ = `("K"_"b" xx "W"_2)/(∆"T"_"b" xx "W"_1)`

= `(2.16  "K kg mol"^-1 xx 0.11  "g")/(0.1  "K" xx 15  "g")`

= 0.158 kg mol⁻¹

= 158 g mol⁻¹