JEE Main 2026 — Jan 24, Shift 2 — Chemistry solution

JEE Main2026Jan 24, Shift 2Chemistry

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 $N_{2}$ is $6.5 \times 1 0^{7}$ mm Hg )

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

Solution

Henry's law: $P_{N_{2}} = K_{H} \cdot X_{N_{2}}$ .

Partial pressure of $N_{2}$ in air: $P_{N_{2}} = 0.80 \times 10 atm = 8 atm = 8 \times 760 mm Hg$ .

$X_{N_{2}} = \frac{P _{N_{2}}}{K _{H}} = \frac{8 \times 760}{6.5 \times 1 0 ^{7}} \approx 9.35 \times 1 0^{- 5}$ .

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