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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2026 · 24 January · Evening Shift · Chemistry · Solutions

At 298 K, the mole percentage of N_ 2 (~g) in air is 80 \% . Water...

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})

Options

  1. A1.23×10−71.23 \times 10^{-7}
  2. B1.17×10−41.17 \times 10^{-4}
  3. C9.35×1059.35 \times 10^{5}
  4. D9.35×10−59.35 \times 10^{-5}

Correct answer

D. 9.35×10−59.35 \times 10^{-5}

Step-by-step solution

Using Henry's Law: Mole fraction = PKH\frac{P}{K_H}
N₂ mole percentage in air = 80%, total pressure = 10 atm
Partial pressure of N₂ = 0.80 × 10 = 8 atm = 8 × 760 = 6080 mmHg
KHK_H for N₂ = 6.5 × 10⁷ mmHg
Mole fraction of N₂ = 60806.5×107=6.08×1036.5×107=0.935×10−4=9.35×10−5\frac{6080}{6.5 \times 10^7} = \frac{6.08 \times 10^3}{6.5 \times 10^7} = 0.935 \times 10^{-4} = 9.35 \times 10^{-5}
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