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Chemistry · Solutions and Colligative Properties

JEE Main 2025 — 23 January, Evening Shift — Question 40

Consider a binary solution of two volatile liquid components 1 and 2x12 \mathrm{x}_{1} and y1\mathrm{y}_{1} are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of 1x1\frac{1}{x_{1}} vs 1y1\frac{1}{y_{1}} are given respectively as :

  1. Option A:

    P10P20,P20−P10P20\frac{\mathrm{P}_{1}^{0}}{\mathrm{P}_{2}^{0}}, \frac{\mathrm{P}_{2}^{0}-\mathrm{P}_{1}^{0}}{\mathrm{P}_{2}^{0}}

    Correct
  2. Option B:

    P20P10,P10−P20P20\frac{\mathrm{P}_{2}^{0}}{\mathrm{P}_{1}^{0}}, \frac{\mathrm{P}_{1}^{0}-\mathrm{P}_{2}^{0}}{\mathrm{P}_{2}^{0}}

  3. Option C:

    P10P20,P10−P20P20\frac{\mathrm{P}_{1}^{0}}{\mathrm{P}_{2}^{0}}, \frac{\mathrm{P}_{1}^{0}-\mathrm{P}_{2}^{0}}{\mathrm{P}_{2}^{0}}

  4. Option D:

    P20P10,P20−P10P20\frac{\mathrm{P}_{2}^{0}}{\mathrm{P}_{1}^{0}}, \frac{\mathrm{P}_{2}^{0}-\mathrm{P}_{1}^{0}}{\mathrm{P}_{2}^{0}}

Answer: A

Step-by-step solution

∵\because For liquid solution of two liquids ' 1 ' and ' 2 '

P1=PTy1=P1ox1\mathrm{P}_{1}=\mathrm{P}_{\mathrm{T}} \mathrm{y}_{1}=\mathrm{P}_{1}^{\mathrm{o}} \mathrm{x}_{1}

∴PTx1=P10y1\therefore \frac{\mathrm{P}_{\mathrm{T}}}{\mathrm{x}_{1}}=\frac{\mathrm{P}_{1}^{0}}{\mathrm{y}_{1}}

∴P2o+x1(P1o−P2o)x1=P1oy1\therefore \frac{\mathrm{P}_{2}^{\mathrm{o}}+\mathrm{x}_{1}\left(\mathrm{P}_{1}^{\mathrm{o}}-\mathrm{P}_{2}^{\mathrm{o}}\right)}{\mathrm{x}_{1}}=\frac{\mathrm{P}_{1}^{\mathrm{o}}}{\mathrm{y}_{1}}

∴P20x1+(P1o−P2o)=P10y1\therefore \frac{\mathrm{P}_{2}^{0}}{\mathrm{x}_{1}}+\left(\mathrm{P}_{1}^{\mathrm{o}}-\mathrm{P}_{2}^{\mathrm{o}}\right)=\frac{\mathrm{P}_{1}^{0}}{\mathrm{y}_{1}}

∴1x1=(P1oP2o)(1y1)+(P2o−P1oP2o)\therefore \frac{1}{\mathrm{x}_{1}}=\left(\frac{\mathrm{P}_{1}^{\mathrm{o}}}{\mathrm{P}_{2}^{\mathrm{o}}}\right)\left(\frac{1}{\mathrm{y}_{1}}\right)+\left(\frac{\mathrm{P}_{2}^{\mathrm{o}}-\mathrm{P}_{1}^{\mathrm{o}}}{\mathrm{P}_{2}^{\mathrm{o}}}\right)

∴\therefore Slope =(P1oP2o)=\left(\frac{\mathrm{P}_{1}^{\mathrm{o}}}{\mathrm{P}_{2}^{\mathrm{o}}}\right)

∴\therefore Intercept =(P2o−P1oP2o)=\left(\frac{\mathrm{P}_{2}^{\mathrm{o}}-\mathrm{P}_{1}^{\mathrm{o}}}{\mathrm{P}_{2}^{\mathrm{o}}}\right)

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
23 January, Evening Shift
Subject
Chemistry
Chapter
Solutions and Colligative Properties
Topic
Liquid in Liquid Solutions (Raoult's Law)

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