An aqueous solution containing `28%` by mass of liquid A `(mol.mass = 140)` has a vapour pressure of `160 mm` at `30^@C`. Find the vapour pressure of the pure liquid A. (The vapour pressure of the water at `30^@C` is `150 mm`.)

Explanation

The given values are:
`W_("liquid") = 28 g`, `Mw_(1) = 140`
`W_(H_(2)O) = 72 g`, `Mw_(H_(2)O) = 18`
`p_(H_(2)O) = 0.10` bar, `p_("total")` = 0.2 bar
Mole fraction of liquid,`chi_(1) =((28)/(140))/((28)/(140)+(72)/(18)) = 0.048`
Mole fraction of water,`chi_(H_(2)O) = 1-0.048 = 0.952`
According to Raoult's law
`p_("total") = p_(1)@chi(1) + p_(H_(2)O)@chi(H_(2)O)`
`0.2 = p_(l)^(@) xx 0.048 + 0.1 xx 0.952`
or `p_(l)^(@) = (0.2 - 0.1 xx 0.952)/(0.048)=2.18` bar
Therefore, the vapour pressure of pure liquid is `2.18` bar