State and explain :
Raoult's law for volatile solute.

Explanation

Let A and B be two components of a solution at a particular temperature. Their partial vapour pressure be `p_(A)andp_(B)`. Let the vapour pressure in pure state are `p_(A)^(@)andp_(B)^(@)`.
According to Raoult.s law, `p_(A)=x_(A)p_(A)^(@)andp_(B)=x_(B)p_(B)^(@)`
If p is total pressure of the system, then by Dalton.s law of partial pressures,
`p=p_(A)+p_(B)=x_(A)p_(A)^(@)+x_(B)p_(B)^(@)orp=(1-x_(B))^(@)p_(A)^(@)+x_(B)p_(B)^(@)=(p_(B)^(@)-p_(A)^(@))x_(B)+p_(A)^(@)`
Since `p_(A)^(@)andp_(B)^(@)` are constant at a particular temperature.
When `x_(A)=1`, i.e. liquid is pure A
`p=p_(A)^(@)and`
When `x_(B)=1`, i.e. liquid is pure B,
`p=p_(B)^(@)`