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JEE Main 202523 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Evening Shift
Subject: Chemistry
Chapter: Solutions

JEE Main 2025 · 23 January · Evening Shift · Chemistry · Solutions

Consider a binary solution of two volatile liquid components 1 and...

Consider a binary solution of two volatile liquid components 1 and

2 . x_1

and

y_1

are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of

\frac{1}{x_1}

vs

\frac{1}{y_1}

are given respectively as:

Options

A $\frac{\mathrm{P}_2^0}{\mathrm{P}_1^0}, \frac{\mathrm{P}_2^0-\mathrm{P}_1^0}{\mathrm{P}_2^0}$
B $\frac{\mathrm{P}_1^0}{\mathrm{P}_2^0}, \frac{\mathrm{P}_2^0-\mathrm{P}_1^0}{\mathrm{P}_2^0}$
C $\frac{\mathrm{P}_1^0}{\mathrm{P}_2^0}, \frac{\mathrm{P}_1^0-\mathrm{P}_2^0}{\mathrm{P}_2^0}$
D $\frac{\mathrm{P}_2^0}{\mathrm{P}_1^0}, \frac{\mathrm{P}_1^0-\mathrm{P}_2^0}{\mathrm{P}_2^0}$

Correct answer

B. $\frac{\mathrm{P}_1^0}{\mathrm{P}_2^0}, \frac{\mathrm{P}_2^0-\mathrm{P}_1^0}{\mathrm{P}_2^0}$

Step-by-step solution

Given mole fraction of liquid 1 in liquid and vapour phase is

x_1 and y_1

respectively.

\begin{aligned} & \mathrm{P}_1=\mathrm{P}_{\mathrm{T}} y_1=\mathrm{P}_1^0 x_1 \\ & \frac{\mathrm{P}_{\mathrm{T}}}{x_1}=\frac{\mathrm{P}_1^0}{y_1} \\ & \frac{\mathrm{P}_2^0+x_1\left(\mathrm{P}_1^0+\mathrm{P}_2^0\right)}{x_1}=\frac{\mathrm{P}_1^0}{y_1} \\ & \frac{\mathrm{P}_2^0}{x_1}+\left(\mathrm{P}_1^0+\mathrm{P}_2^0\right)=\frac{\mathrm{P}_1^0}{y_1} \\ & \frac{\mathrm{P}_2^0}{x_1}=\frac{\mathrm{P}_1^0}{y_1}-\left(\mathrm{P}_1^0+\mathrm{P}_2^0\right) \\ & \frac{1}{x_1}=\frac{1}{y_2}\left(\frac{\mathrm{P}_1^0}{\mathrm{P}_2^0}\right)-\frac{\left(\mathrm{P}_1^0+\mathrm{P}_2^0\right)}{\mathrm{P}_2^0}\end{aligned}

\begin{aligned} & \frac{1}{x_1}=\frac{1}{y_2}\left(\frac{P_1^0}{P_2^0}\right)+\frac{\left(P_2^0-P_1^0\right)}{P_2^0} \\ & y=m x+c \\ & \text{ Slope }=\frac{P_1^0}{P_2^0} ; \text{ Intercept }=\frac{P_2^0-P_1^0}{P_2^0}\end{aligned}

$

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