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

The vapour pressure is reduced to 80% on the addition of a non - volatile solute so that.
`(p_(A)^(@) p_(A))/(p_(A)^(@)) = (100-80)/(100) = (1)/(5) = 0.2`
Now , according to Raoult.s law .
`(p_(A)^(@) - p_(A))/(p_(A)^(@)) = x_(B)`
Let mass of non - volatile solute to be a dded = x g .
`therefore " " (p_(A)^(@) - p_(A))/(p_(A)^(@))= ((x)/(40))/((114)/(114) + (x)/(40))`
(Molar mass of octane = 114)
` 0.2 = (x//40)/(1+x//40)`

`therefore " " 0.2 = (1 + (x)/(40)) = (x)/(40)`
or `" " 0.2 = (x)/(40) = (0.2x)/(40)`
`therefore " " 0.2 = (0.8x)/(40)`
` or " "0.2 = (x)/(40) - (0.2x)/(40)`
`therefore " "0.2 = (0.8x)/(40g)`
or `" " x = 10 g`
`therefore ` Mass of solute to be added = 10 g .