Question

The vapour pressure of acetone at 20°C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20°C, its vapour pressure was 183 torr. The molar mass (g mol⁻¹) of the substance is ______.

पर्याय

• 32

• 64

• 128

• 488

Answer 1

The vapour pressure of acetone at 20°C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20°C, its vapour pressure was 183 torr. The molar mass (g mol⁻¹) of the substance is 64.

Explanation:

Given: Vapour pressure of pure acetone, P° = 185 torr

Vapour pressure of solution, `P_"solution"` = 183 torr

Mass of solute = 1.2 g

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

Molar mass of acetone ≈ not needed here

Molar mass of the solute = ?

Using Raoult’s law

`(P^circ - P)/P^circ = n_"solute"/(n_"solvent" + n_"solute")` = `n_"solute"/n_"solvent"`    ...(Since solute is dilute)

`(185 - 183)/185 = (1.2/M)/(100/58)` 

`2/185 = 1.2/M xx 58/100`

`2/185 = (1.2 xx 58)/(100 xx M)`

`2/185 = 69.6/(100 M)`

`(2 xx 100 M)/185 = 69.6`

`(200 M)/185 = 69.6`

M = `(69.6 xx 185)/200`

M = 64.38 ≈ 64