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

For two miscible liquids,
`P_("total")` = Mole fraction of `A xx p_(A)^(0)+` Mole fraction of `B xx p_(B)^(0)`
No. of moles of `A = (28)/(140)=0.2`
Liquid B is water. Its mass is (100 28), i.e., 72.
No. of moles of `B = (72)/(18)=4.0`
Total no. of moles `= 0.2 +4.0 = 4.2`
Given, `P_("total")=160` mm
`p_(B)^(0)=150` mm
So, `160 = (0.2)/(4.2) xx p_(A)^(0)+(4.0)/(4.2) xx 150`
`p_(A)^(0)=(17.15xx4.2)/(0.2)=360.15` mm.