Using Raoult's Law explain how the total vapour pressure over the solution is related to mole fraction of components in the following solution.
`CHCl_(3)" "(l)and CH_(2)Cl_(2)(l)" "(b)NaCl(s)and H_(2)O(l)`

Explanation

According to Raoult's law for any solution the partial vapour pressure of each volatile component in the solution in the solution is directly proportional to its mole fraciton.
`" "p_(1)=p_(1)^(@)x_(1)`
(a) `CHCl_(3) and CH_(2)Cl_(2)` (l) both components are volatile liquids, the total pressure will be `p=p_(1)=x_(1)p_(1)^(@)+x_(2)p_(2)^(@)`
`=x_(1)p_(1)^(@)+(1-x_(1))p_(2)^(@)=(p_(1)^(@)-p_(2)^(@))x_(1)+p_(2)^(@)`
where, p= total vapour pressure
`p_(1)`= partial vapour pressure of component 1
`p_(2)` = partial vapous pressure of component 2
(b) NaCl (s) and `H_(2)O(l)` both are non-valatile components.
Hence, for a solution containing non-volatile solute, the Raoult's law is applicable only to vaporisable compnent and total vapour pressure can be written as
`" "p=p_(1)=x_(1)p_(1)^(@)`