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. `11.4g`
B. `9.8g`
C. `12.8g`
D. 10g

Answer

D

Explanation

`"Given, "p^(@)=100,p=100-20=80`,
`M_(2)=40" gmol"^(-1),M_(1)("octane")=C_(8)H_(18)=114" gmol"^(-1)`
`w_(2)=?,w_(1)("octane")=114g`
`(p^(@)-p)/(p)=(w_(2))/(M_(2))xx(M_(1))/(w_(1)`
`(100-80)/(80)=(w_(2))/(40)xx(114)/(114)`
`rArrw_(2)=10g`