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

Options

A. 10 g
B. 11.4 g
C. 9. 8 g
D. 12.8 g

Answer

A

Explanation

According to Raoult's law : `(p_(0) - p_s)/(p_(0)) = (w_(A)// m_(A))/((w_(A))/(m_(A)) + (w_(B))/(m_(B)))`
where `w_(A) , m_(A) to` weight dissolved and molecular mass of solute respectively .
`w_(B) , m_(B) to ` weight dissolved and molecular mass of solvent respectively .
`p_(0)` to vapour pressure of pure solvent
`p_(s) to` vapour pressure of solution .
`(p_(0) - p_(s))/(p_(0)) = (20)/(100) = 0.2`
`0.2 = ((w_(A))/(40))/((w_(A))/(40) + (114)/(114)) implies (w_(A) // 40)/((w_(A) + 40)/(40))`
`0.2 = (w_(A))/(w_(A) + 40)`
`0.2 w_(A) + 8 = w_(A)`
`8 = w_(A) - 0.2 w_(A) = 0.8 w_(A)`
`w_(A) = 10g`