JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Solutions

At 298 K, the mole percentage of

N_{2} (g)

in air is

80%

. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of

N_{2} (g)

in water at 298 K? (

K_{H}

for

N_{2}

6.5 \times 1 0^{7} mm Hg

)

(A) $1.23 \times 1 0^{- 7}$
(B) $1.17 \times 1 0^{- 4}$
(C) $9.35 \times 1 0^{5}$
(D) $9.35 \times 1 0^{- 5}$ ✓ Correct

Answer: (D)

Solution

Henry's law gives

p_{N_{2}} = K_{H} x_{N_{2}}

. The partial pressure of nitrogen is its mole fraction in air times the total pressure,

p_{N_{2}} = 0.80 \times 10 = 8 atm = 8 \times 760 = 6080 mm Hg

. Therefore

x_{N_{2}} = \frac{p _{N_{2}}}{K _{H}} = \frac{6080}{6.5 \times 1 0 ^{7}} = 9.35 \times 1 0^{- 5}

. The mole fraction of

N_{2}

in water is

9.35 \times 1 0^{- 5}

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option

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