Question

Using Raoult's law explain how the total vapour pressure over the solution is related to mole fraction of components in the following solutions.

$A$: $CHC{l}_3\left(l\right)\text{and}C{H}_{2}C{l}_2\left(l\right)$

$B$: $NaCl\left(s\right)\text{and}{H}_{2}O$

Answer 1

Raoult’s law

Raoult’s law states that for the solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in the solution.

Thus, for one of the two components of a binary solution;

$P_1\propto x_1$

$P_1=x_1{p}_1^0$

Where, $P_1$ is the partial vapour pressure of component $1\text{and}{p}_1^0$ is the vapour pressure of the pure solvent at the same temperature, and $x_1$ is the mole fraction of component 1.

Similarly, for component 2, its partial vapour pressure is denoted as $P_2$, where;

$P_2=x_2{p}_2^0$

Relation between total vapour pressure and mole fraction of components :

$\left(i\right)$
Chloroform $\left(CHC{l}_3\right)$ and dichloromethane $\left(C{H}_{2}C{l}_2\right)$are both volatile liquids. The total pressure $\left(P\right)$ will be the sum of the partial vapour pressures of the two volatile components.
$\Rightarrow P=P_1+P_2$
(Considering, $P_1\&P_2$ as partial vapour pressures of $CHC{l}_3\text{and}C{H}_{2}C{l}_2\text{and}{x}_1\&x_2$ as mole fractions of $CHC{l}_3\text{and}C{H}_{2}C{l}_2$ respectively.

According to Raoult's law,

$P_1=x_1{p}_1^0$
$P_2=x_2{p}_2^0$
$\Rightarrow P=x_1{p}_1^0+x_2{p}_2^0$
$\Rightarrow P=(1-x_2)p_1^0+x_2{p}_2^0$
(Where,$x_2+x_1=1$)
$\Rightarrow P=p_1^0-x_2{p}_1^0+x_2{p}_2^0$
$\Rightarrow P=(p_2^0-p_1^0)x_2+p_1^0$
Similarly, we can write in terms of $x_1$ as:

$P=(p_2^0-p_1^0)x_1+p_2^0$

$Nacl\left(s\right)\text{and}{H}_{2}O\left(l\right)$

For a solution containing non-volatile solute
i.e., $Nacl\left(s\right)\text{and}{H}_{2}O\left(l\right)$, the Raoult's law is applicable only to vaporisable component i.e., $H_{2}O\left(l\right)$ and total vapour pressure is written as:
$P=P_1=x_1{p}_1^0$

Where, $p_1^0$​ represents the vapour pressure of pure $H_{2}O\left(l\right)$.