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. 150 mm of Hg
B. 60 mm of Hg
C. 75 mm of Hg
D. 120 mm of Hg

Answer

C

Explanation

We know that, according to Raoult's law
`(p^(@)-p)/(p^(@))=chi_(B)`
`(p^(@)-60)/(p^(@))=(n_(B))/(n_(A)+n_(B))=(0.2)/(0.2+0.8)`
`=(0.2)/(1.0)=(2)/(10)=(1)/(5)`
`p^(@)-60=(p^(@))/(5)`
`rArr p^(@)-(p^(@))/(5)=60`
`(5p^(@)-p^(@))/(5)=60`
`4p^(@)=60xx5`
`p^(@)=(60xx5)/(4)=(300)/(4)=75` mm of Hg