Using Raoult's law explain how the total vapour pressure over the solution is related to mole fraction of components in the following solutions.
(a) `CHCl_(3(l))`
(b) `CH_(2)Cl_(2(l))`

Explanation

(i) `CHCl_(3)` : For a binary solution having both components as volatile liquids (viz. `CHCl_(3)` and `CH_(2)Cl_(2)`), the total pressure will be
`p=x_(1)p_(1)^(0)+x_(2)p_(2)^(0)`
`= p_(1)^(0)+1-x_(2)p_(2)^(0)`
`= (p_(1)^(0)-p_(2)^(0))x_(1)+p_(2)^(0)`
p = total vapour pressure of the given mixture / binary solution of the given volatile liquids.
`p_(1)=` partial vapour presure of component 1 (i.e., `CHCl_(3)`)
`p_(2)=` partial vapour pressure of component 2 (i.e., `CH_(2)Cl_(2)`)
(ii) `NaCl_((s))` and `H_(2)O_((l))` : For a solutioncontaining non - volatile solute i.e., `NaCl_((s))`and `H_(2)O_((l))`, the Raoult.s law is applicable only to vaporisable component (1) i.e., `H_(2)O_((l))` and total vapour pressure is written as `p=p_(1)=x_(1)p_(1)^(0)` where `p_(1)^(0)p_(1)` oreresents the vapour pressure of pure `H_(2)O_((l))`