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

`( P^(0) - P) prop X_(B)`
`((P^(0)-P)_(2))/((P^(0)-P)_(1))= (X_(B_(2)))/(X_(B_(1)))`
`( 20)/( 10) = ( X_(B_(2)))/(0.2)`
`X_(B_(2))=0.4 `
`:.` Mole fraction of solvent `=1 -0.4 = 0.6 `