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Chemistry · Solutions and Colligative Properties

JEE Main 2026 — 24 January, Evening Shift — Question 64

At 298 K , the mole percentage of N2( g)\mathrm{N}_{2}(\mathrm{~g}) in air is 80%80 \%. Water is in equilibrium with air at a pressure of 10 atm . What is the mole fraction of N2( g)\mathrm{N}_{2}(\mathrm{~g}) in water at 298 K ? ( KH\mathrm{K}_{\mathrm{H}} for N2\mathrm{N}_{2} is 6.5×107 mmHg6.5 \times 10^{7} \mathrm{~mm} \mathrm{Hg} )

  1. Option A:

    1.23×10−71.23 \times 10^{-7}

  2. Option B:

    1.17×10−41.17 \times 10^{-4}

  3. Option C:

    9.35×1059.35 \times 10^{5}

  4. Option D:

    9.35×10−59.35 \times 10^{-5}

    Correct

Answer: D

Step-by-step solution

PN2=KH⋅XN2\mathrm{P}_{\mathrm{N}_{2}}=\mathrm{K}_{\mathrm{H}} \cdot \mathrm{X}_{\mathrm{N}_{2}} PN2=0.8×10=8 atm\mathrm{P}_{\mathrm{N}_{2}}=0.8 \times 10=8 \mathrm{~atm} 8×760=6.5×107×XN28 \times 760=6.5 \times 10^{7} \times \mathrm{X}_{\mathrm{N}_{2}} XN2=8×7606.5×107\mathrm{X}_{\mathrm{N}_{2}}=\frac{8 \times 760}{6.5 \times 10^{7}} XN2=9.35×10−5\mathrm{X}_{\mathrm{N}_{2}}=9.35 \times 10^{-5}.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Evening Shift
Subject
Chemistry
Chapter
Solutions and Colligative Properties
Topic
Gas in Liquid Solutions (Henry's Law)

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