Question

The vapour pressure of a solution of 5 g of a non-electrolyte in 100 g of water at a particular temperature is 2985 Nm⁻². The vapour pressure of pure water at that temperature is 3000 Nm⁻². The molecular weight of solute is ______.

Options

• 180

• 90

• 270

• 200

Answer 1

The vapour pressure of a solution of 5 g of a non-electrolyte in 100 g of water at a particular temperature is 2985 Nm⁻². The vapour pressure of pure water at that temperature is 3000 Nm⁻². The molecular weight of solute is 180.

Explanation:

Given: Vapour pressure of solution = 2985 Nm⁻²

Vapour pressure of pure water = 3000 Nm⁻²

Mass of solute = 5 g

Mass of solvent (water) = 100 g = 0.1 kg

The solute is non-electrolyte

By using Raoult’s law

`(Delta P)/P^circ = (p^circ - p)/p^circ = n_"solute"/n_"solvent"`

= `(3000 - 2985)/3000`

= `15/3000`

= 0.005

Molar mass of water = 18 g/mol

`n_"solvent" = 100/18`

= 5.56

Let molecular mass of solvent be M.

∴ `5/M = 0.005 xx 5.56`

`5/M = 0.0278`

`M = 5/0.0278`

M = 180