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)^(@)`
`+`Mole fraction of `B xx P_(B)^(@)`
Number of moles of` A = 28/140=2`
Liquid B is water. Its mass is `(100-28)`,i.e., `72`.
Number of moles of `B=72/18 = 4.0`
Total number of moles `=0.2 + 4.0 =4.2`
Given ,`p_("total") = 160 mm`
`p_(B)^(@) = 150 mm`
So, `160=0.2/4.2 xx p_(A)^(@) + 4.0 /4.2 xx 150`
` p_(A)^(@)=(17.15 xx 4.2)/0.2 = 360.15 mm`