The vapour pressure of a solvent decreased by 10 mm of Hg when a non-volatile solute was added to the solvent. The mole fraction of solute in solution is 0.2, what would be the mole fraction of solvent if the decrease in vapour pressure is 20 mm of Hg?

Options

A. `0.8`
B. `0.6`
C. `0.4`
D. `0.2`

Answer

B

Explanation

According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of solute, i.e.
`(p^(@)-p)/(p^(@))=(n)/(n+N)` or `(Delta p)/(p^(@))=(n)/(n+N)`
`(10)/(p^(@))=0.2`
`therefore p^(@)=50 mm`
For other solution of same solvent
`(20)/(p^(@))=(n)/(n+N)` or `(20)/(50)=(n)/(n+N)`
`0.4=(n)/(n+N)` (mole fraction of solute)
`because` Mole fraction of solvent + mole fraction of solute = 1
So, mole fraction of solvent `= 1-0.4=0.6`