An aqueous solution containing 28% by mass of a liquid A (molecular mass = 140) has a vapour pressure of 160 mm at `37^(@)C`. Find the vapour pressure of pure liquid A (the vapour pressure of water at `37^(@)C` is 150 mm)

Explanation

For two miscible liquids,
`P_("total")=P_A+P_B=P_A^@X_A+P_B^@X_B`
`n_A=(28)/(140)=0.2`
`n_B=(72)/(18)=4.0`
Liquid B is water. It mass is (100-28) i.e. 72
Total number of moles =0.2 +4.0 =4.2
Given `P_("total")=160 mm , P_B^0 =150 mm`
So `160 =(0.2)/(4.2)xxP_A^0+(4.0)/(4.2)xx5.0`
`P_A^0=(17.15xx4.2)/(0.2)=360.15mm`