Calculate the mass of a non-volatile solute (molar mass 40 g `mol^(-1)`) which should be dissolved in 114g octane to reduce its vapour pressure to 80%.

Explanation

Molar mass of solvent (octane, `C_(8)H_(18)`)
`=12xx8+1xx18=114"g mol"^(-1)`
Molecular mass of non-volatile solute `=40 " g mol"^(-1)`
Let the mass of non-volatile solute dissolved = a g
Mass of octane = 114 g
Moles of solute `=(a)/(40)` mol
Moles of octane `=(114)/(114)=1.0` mol
Let the vapour pressure of pure solvent = 1.0 bar
Vapour pressure of solution `=(1xx80)/(1.00)=0.8` bar
Now `(p^(@)-p)/(p^(@))=x_(B)`
`:. (1-0.8)/(1.00)=(a//40)/(1//40+1.0)`
or `0.2=(a//40)/(a//40+1.0) or 0.2(a//40+1.0)=a//40`
or `0.2+(a)/(20)=(a)/(40) or (a)/(20)-(a)/(200)=0.2`
On solving, a = 10 g
Alternatively, considering the solution to be very dilute,
`0.2=(a//40)/(a//40+1.0)~~(a//40)/(1), a=8g`