JEE Main 2025 — Solutions Question with Solution

For a binary solution of two volatile liquid components labeled 1 and 2, let $x_1$ and $y_1$ represent the mole fractions of component 1 in the liquid and vapor phases, respectively. The linear relationship between the inverse of these mole fractions is plotted as $\frac{1}{x_1}$ versus $\frac{1}{y_1}$.

To derive the slope and intercept of this linear plot, consider the following calculations:

Using Raoult's Law for a Liquid Solution:

For a liquid solution with volatile components 1 and 2:

${\mathrm{P}}_1={\mathrm{P}}_{\mathrm{T}}\cdot y_1={\mathrm{P}}_1^{\mathrm{o}}\cdot x_1$

Therefore:

$\frac{{\mathrm{P}}_{\mathrm{T}}}{x_1}=\frac{{\mathrm{P}}_1^{\mathrm{o}}}{y_1}$

Rearranging the Equation:

By substituting and rearranging, we have:

$\frac{{\mathrm{P}}_2^{\mathrm{o}}+x_1({\mathrm{P}}_1^{\mathrm{o}}-{\mathrm{P}}_2^{\mathrm{o}})}{x_1}=\frac{{\mathrm{P}}_1^{\mathrm{o}}}{y_1}$

Simplifying further:

$\frac{{\mathrm{P}}_2^{\mathrm{o}}}{x_1}+({\mathrm{P}}_1^{\mathrm{o}}-{\mathrm{P}}_2^{\mathrm{o}})=\frac{{\mathrm{P}}_1^{\mathrm{o}}}{y_1}$

Expressing $\frac{1}{x_1}$:

Solving for $\frac{1}{x_1}$, we obtain:

$\frac{1}{x_1}=\left(\frac{{\mathrm{P}}_1^{\mathrm{o}}}{{\mathrm{P}}_2^{\mathrm{o}}}\right)\left(\frac{1}{y_1}\right)+\left(\frac{{\mathrm{P}}_2^{\mathrm{o}}-{\mathrm{P}}_1^{\mathrm{o}}}{{\mathrm{P}}_2^{\mathrm{o}}}\right)$

Determining the Slope and Intercept:

The slope of the line is:

$\text{Slope}=\frac{{\mathrm{P}}_1^{\mathrm{o}}}{{\mathrm{P}}_2^{\mathrm{o}}}$

The intercept of the line is:

$\text{Intercept}=\frac{{\mathrm{P}}_2^{\mathrm{o}}-{\mathrm{P}}_1^{\mathrm{o}}}{{\mathrm{P}}_2^{\mathrm{o}}}$

In summary, for the plot of $\frac{1}{x_1}$ against $\frac{1}{y_1}$, the slope is $\frac{{\mathrm{P}}_1^{\mathrm{o}}}{{\mathrm{P}}_2^{\mathrm{o}}}$ and the intercept is $\frac{{\mathrm{P}}_2^{\mathrm{o}}-{\mathrm{P}}_1^{\mathrm{o}}}{{\mathrm{P}}_2^{\mathrm{o}}}$.