The vapour pressure at a given temperature of an ideal solution containing `0.2 mol` of non-volatile solute and `0.8 mol` of a solvent is `60 mm` of `Hg`. The vapour pressure of the pure solvent at the same temperature will be

Options

A. `120 mm Hg`
B. `60 mm Hg`
C. `150 mmHg`
D. `75 mm Hg`

Answer

3

Explanation

According to second Raoult's law, we have
`(P_(1)^(0)-P^(0))/P_(1)^(0)= X_(2)=n_(2)/(n_(1)+n_(2))`
Thus, `(P_(1)^(0)-60 mm Hg)/P_(1)^(0)=0.2/(0.2+0.8)=0.2/1.0`
`=0.2`
or `P_(1)^(0)-60 mm Hg=0.2 P_(1)^(0)`
`P_(1)^(0)-0.2 P_(1)^(0)=60 mm Hg`
`P_(1)^(0)(1-0.2)=60 mm Hg`
`0.8 P_(1)^(0) 60 mm Hg`
`P_(1)^(0)=60/0.8 mm Hg`
`75 mm Hg`