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 is directly proportional to its mole fraction.
`p_(1)=p_(1)^(@)x_(1)`
a. `CHCl_(3)(1)` and `CH_(2)Cl_(2)` (1) both are volative components.
Hence, for a binary in which both components are volatile liquids,the total pressure will be `p=p_(1)+p_(2)=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_(1)=` partial vapour pressure of component 1
`p_(2)=` partial vapour pressure of component?
b. NaCl(s) and `H_(2)O(1)` bot are non-volatile components.
Hence, for a solution containing non-volatile solute, the Raoult’s law is applicable only to vaporizable component and total vapour pressure can be written as
`p=p_(1)=x_(1)p_(1)^(@)`