The mass of a non-volatile solute of molar mass `40g" "mol^(-1)` that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is

Options

A. 8g
B. 11.4g
C. 9.8 g
D. 12.8g

Answer

A

Explanation

`(p_(A)^(@) -p_(A))/(p_(A)^(@)) = (20)/(100) = x_(B)`
Let mass of non volatile solute be added = x g
`((x)/(40))/((x)/(40) + (114)/(114))= (20)/(100) = ((x)/(40))/((114)/(114))`
or `(x)/(40) xx 1 = (20)/(100) therefore x= (40 xx 20)/(100) = 8 g`