The vapour pressure of pure liquid solvent 0.50 atm When a non-volatile solute B is added to the solvent ,its vapour pressure drops to `0.30`atm Thus,mole fraction of the component B is

Options

A. `0.33`
B. `6.0`
C. `3.0`
D. `0.66`

Answer

C

Explanation

A is solute and B is solvent .
` p_(B) = p_(B)^(@) - (10)/(100) p_(B)^(@) = 0.9 p_(B)^(@)`
`M_(B) = (30)/(100) M_(A) = 0.30 M`
`(p_(B)^(@) - p_(B))/(p_(B)^(@)) = x_(A) = (w_(A) xx M_(B))/(M_(A) xx W_(B))`
`(p_(B)^(@) - 0.9p_(B)^(@))/(p_(B)^(@)) = (w_(A))/(w_(B)) xx (0.30 M_(A))/(M_(A))`
` 0.1 = (w_(A))/(w_(B)) xx (0.30 M _(A))/(M_(A))`
or ` " "(W_(B))/(W_(A)) = (0.3)/(0.1) = 3.0`