Using Raoult's law,explain how the total vapour pressure over the solution is related to the mole fraction of components in the following solutions:
$(a)$ $CHCl_{3(l)}$ and $CH_{2}Cl_{2(l)}$
$(b)$ $NaCl_{(s)}$ and $H_{2}O_{(l)}$

(N/A) For a binary solution where both components are volatile liquids ($CHCl_{3}$ and $CH_{2}Cl_{2}$),the total vapour pressure $(p_{total})$ is given by the sum of the partial vapour pressures of the components:
$p_{total} = p_{1} + p_{2} = x_{1}p_{1}^{0} + x_{2}p_{2}^{0}$
Since $x_{1} + x_{2} = 1$,we can write $p_{total} = x_{1}p_{1}^{0} + (1 - x_{1})p_{2}^{0} = x_{1}(p_{1}^{0} - p_{2}^{0}) + p_{2}^{0}$.
Thus,the total vapour pressure varies linearly with the mole fraction of one component.
$(b)$ For a solution containing a non-volatile solute $(NaCl_{(s)})$ and a volatile solvent $(H_{2}O_{(l)})$,the solute does not contribute to the vapour pressure.
According to Raoult's law,the total vapour pressure of the solution is equal to the partial vapour pressure of the solvent:
$p_{total} = p_{solvent} = x_{solvent}p_{solvent}^{0}$
Here,the total vapour pressure is directly proportional to the mole fraction of the solvent $(H_{2}O_{(l)})$.