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 is `0.2`, what would be the mole fraction of solvent if the decrease in vapour pressure is `20 mm` of `Hg`.

Options

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

Answer

C

Explanation

`("P"_("solvent")^(0)-"P"_("solution"))/("P"_("solvent")^(0))="x"_("solute")`
`10/"P"_("solvent")^(0)=0.2`
`"P"_("solvent")`=50 mm Hg
Now again
`("P"_("solvent")^(0)-"P"_("solution"))/("P"_("solvent"))="x"_("solute")`
`20/50="x"_("solute")`
`"x"_("solute")=1-(2)/(5)=0.6`