JEE Main 2025 — Chemistry Physical Chemistry

2025mcqmedium

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:

Official previous-year question

Held on 23 Jan 2025 · Verified 6 Jul 2026.

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

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Solution

Given mole fraction of liquid 1 in liquid and vapour phase is $x_1$ and $y_1$ respectively. JEE Main 2025 Chemistry, Solutions — solution figure

$\begin{aligned} & \mathrm{P}_1=\mathrm{P}_{\mathrm{T}} y_1=\mathrm{P}_1^0 x_1 \\ & \frac{\mathrm{P}_{\mathrm{T}}}{x_1}=\frac{\mathrm{P}_1^0}{y_1} \\ & \frac{\mathrm{P}_2^0+x_1\left(\mathrm{P}_1^0+\mathrm{P}_2^0\right)}{x_1}=\frac{\mathrm{P}_1^0}{y_1} \\ & \frac{\mathrm{P}_2^0}{x_1}+\left(\mathrm{P}_1^0+\mathrm{P}_2^0\right)=\frac{\mathrm{P}_1^0}{y_1} \\ & \frac{\mathrm{P}_2^0}{x_1}=\frac{\mathrm{P}_1^0}{y_1}-\left(\mathrm{P}_1^0+\mathrm{P}_2^0\right) \\ & \frac{1}{x_1}=\frac{1}{y_2}\left(\frac{\mathrm{P}_1^0}{\mathrm{P}_2^0}\right)-\frac{\left(\mathrm{P}_1^0+\mathrm{P}_2^0\right)}{\mathrm{P}_2^0}\end{aligned}$ $\begin{aligned} & \frac{1}{x_1}=\frac{1}{y_2}\left(\frac{P_1^0}{P_2^0}\right)+\frac{\left(P_2^0-P_1^0\right)}{P_2^0} \\ & y=m x+c \\ & \text { Slope }=\frac{P_1^0}{P_2^0} ; \text { Intercept }=\frac{P_2^0-P_1^0}{P_2^0}\end{aligned}$

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Options

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

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

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

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

Solution

Given mole fraction of liquid 1 in liquid and vapour phase is $x_1$ and $y_1$ respectively. JEE Main 2025 Chemistry, Solutions — solution figure

JEE Main 2025 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2025 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

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JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

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