Calculate the mass of a non-volatile solute ( molecular mass `40`) which should be dissolved in `114 g` octane to reduce its vapour pressure to `80%`.

Explanation

`P_(s)=80% " of " P^(@)`
`=(80)/(100)P^(@)=0.8P^(@)`
Let Wg of solute is present in mixture.
Moles of solute present `=(W)/(40)` moles
Molar mass of octane, `C_(8)H_(18)`
`=8xx12+18=114 " g mol"^(-1)`
`therefore "Moles of octane"=(114)/(114)=1 " mol"`
Now, `(P^(@)-P_(s))/(P^(@))=x_(2)=(W//40)/((W)/(40)+1)`
`(P^(@)-0.80P^(@))/(P^(@))=(W//40)/(W//40+1)`
`1-0.80=(Wxx40)/(40(W+40))=(W)/(W+40)`
`0.20=(W)/(W+40)`
`0.2W+8=W` `8=W(1-0.2)`
8=0.8W
`therefore W=(8)/(0.8)=10 g`.