Question

Dry air was passed successively through a solution of 5 g of a solute in 80 g of water and then through pure water. The loss in mass of solution was 2.5 g and that of pure solvent 0.04 g. What is the molecular mass of solute?

Options

• 71.43

• 7.143

• 714.3

• 80

Answer 1

71.43

Explanation:

Given: Mass of solute (w) = 5 g

Mass of solvent (water) (W) = 80 g

Molar mass of water `(M_"water")` = 18 g mol⁻¹

Loss in weight of solution (m₁) = 2.5 g

Loss in weight of pure solvent (m₂) = 0.04 g

By Ostwald-Walker method:

`(P^circ - P)/P^circ = "Loss from pure solvent"/("Loss from solution" + "Loss from pure solvent") = m_2/(m_1 + m_2)`

⇒ `(P^circ - P)/P^circ = 0.04/(2.50 + 0.04)`

= `0.04/2.54`

= `2/127`

= 0.01575

For a dilute non-volatile solute, Raoult’s law gives:

`(P^circ - P)/P^circ = n_"solute"/n_"solvent" = (w/M)/(W/M_"water") = (w * M_"water")/(W * M)`    ...[M = Molar mass of solute]

⇒ `0.01575 = (5 xx 18)/(80 xx M)`

⇒ `M = (5 xx 18)/(80 xx 0.01575)`

= `90/1.26`

= 71.428

≈ 71.43