JEE MAIN - Chemistry (2025 - 23rd January Evening Shift - No. 4)
Explanation
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 $\frac{\mathrm{P}_2^{\mathrm{o}} - \mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}$.
Comments (0)
