For an ideal solution of two components  $\mathrm{A}$ and  $\mathrm{B}$, If  ${\mathrm{x}}_{\mathrm{A}}$and  ${\mathrm{y}}_{\mathrm{A}}$ are mole fractions of component ' $\mathrm{A}$ ' in solution and vapour phase respectively, then the slope of linear line in the graph drawn between$\frac{1}{{\mathrm{x}}_{\mathrm{A}}}$ and $\frac{1}{{\mathrm{y}}_{\mathrm{A}}}$ is

Let's first find the Relation between ${\mathrm{x}}_{\mathrm{A}}$and ${\mathrm{y}}_{\mathrm{A}}$.

We know that

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

${\mathrm{P}}_{\mathrm{A}}={\mathrm{P}}_{\mathrm{A}}^0{\mathrm{X}}_{\mathrm{A}}$  and  ${\mathrm{P}}_{\mathrm{B}}={\mathrm{P}}_{\mathrm{B}}^0{\mathrm{X}}_{\mathrm{B}}$

Where ${\mathrm{P}}_{\mathrm{A}}^0$ and ${\mathrm{P}}_{\mathrm{B}}^0$ is the vapour pressure of pure $\mathrm{A}$ and $\mathrm{B}$.

We also know that

According to Dalton's law of partial pressure,

${\mathrm{P}}_{\mathrm{A}}={\mathrm{y}}_{\mathrm{A}}{\mathrm{P}}_{\text{Total }}$

${\mathrm{P}}_{\text{Total }}={\mathrm{P}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}$

$\Rightarrow {\mathrm{Y}}_{\mathrm{A}}=\frac{{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}}$

Now,

Substituting, values of ${\mathrm{P}}_{\mathrm{A}}$ and ${\mathrm{P}}_{\mathrm{B}}$

${\mathrm{Y}}_{\mathrm{A}}=\frac{{\mathrm{P}}_{\mathrm{A}}^0{\mathrm{x}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{A}}^0{\mathrm{X}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}^0{\mathrm{X}}_{\mathrm{B}}}$

Since, ${\mathrm{x}}_{\mathrm{A}}+{\mathrm{x}}_{\mathrm{B}}=1$

$\Rightarrow {\mathrm{x}}_{\mathrm{B}}=1-{\mathrm{x}}_{\mathrm{A}}$

Putting ${\mathrm{x}}_{\mathrm{B}}$ in ${\mathrm{y}}_{\mathrm{A}}$ we get,

${\mathrm{y}}_{\mathrm{A}}=\frac{{\mathrm{P}}_{\mathrm{A}}^0{\mathrm{x}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{A}}^0{\mathrm{x}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}^0\left(1-{\mathrm{x}}_{\mathrm{A}}\right)}$

${\mathrm{y}}_{\mathrm{A}}=\frac{{\mathrm{P}}_{\mathrm{A}}^0{\mathrm{x}}_{\mathrm{A}}}{{{\displaystyle \mathrm{x}}}_{\mathrm{A}}\left({\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0\right)+{\mathrm{P}}_{\mathrm{B}}^0}$

${\mathrm{y}}_{\mathrm{A}}=\frac{{\mathrm{P}}_{\mathrm{A}}^0{\mathrm{x}}_{\mathrm{A}}}{{{\displaystyle \mathrm{x}}}_{\mathrm{A}}\left({\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0\right)}+\frac{{\mathrm{P}}_{\mathrm{A}}^0{\mathrm{x}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{B}}^0}$

$y_{\mathrm{A}}=\frac{{\mathrm{P}}_{\mathrm{A}}^0}{{\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0}+\frac{{\mathrm{P}}_{\mathrm{A}}^0}{{\mathrm{P}}_{\mathrm{B}}^0}{x}_{\mathrm{A}}\cdots \text{ (i) }$

Now, let's find Relation between $\frac{1}{{\mathrm{x}}_{\mathrm{A}}}$ and $\frac{1}{{\mathrm{y}}_{\mathrm{A}}}$

Reciprocating the equation (i) we get,

$\frac{1}{{\mathrm{y}}_{\mathrm{A}}}=\frac{{\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0}{{\mathrm{P}}_{\mathrm{A}}^0}+\frac{{\mathrm{P}}_{\mathrm{B}}^0}{{\mathrm{P}}_{\mathrm{A}}^0}\frac{1}{{{\displaystyle \mathrm{x}}}_{\mathrm{A}}}$

This equation is in the form of $\mathrm{y}=\mathrm{m} \mathrm{x}+\mathrm{c}$

From above equation we can see,

Plot of $\frac{1}{{\mathrm{y}}_{\mathrm{A}}}\mathrm{VS}\frac{1}{{\mathrm{x}}_{\mathrm{A}}}$ will be straight line with

Slope $\left(\mathrm{m}\right)=\frac{{\mathrm{P}}_{\mathrm{B}}^0}{{\mathrm{P}}_{\mathrm{A}}^0}$

and intercept(c)$=\frac{{\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0}{{\mathrm{P}}_{\mathrm{A}}^0}$

Hence, option (b) is the answer.