QJEEM25Ch6774: JEE Main 2025 Chemistry - Class 12 JEEM25Ch6774 Solution: Consider a binary solution of two volatile liquid components 1…

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 $\frac{\mathrm{P}_2^{\mathrm{o}} - \mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}$ .