At 298 K, the mole percentage of in air is 80%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of in water at 298 K? [2026]

Question

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?

$(K_{H} for N_{2} = 6.5 \times 10^{7} mm Hg)$ [2026]

Smart Achievers · Accepted Answer

(4)

$P_{N_{2}} = K_{H} \cdot X_{N_{2}}$

$P_{N_{2}} = 0.8 \times 10 = 8 atm$

$8 \times 760 = 6.5 \times 10^{7} \times X_{N_{2}}$

$X_{N_{2}} = \frac{8 \times 760}{6.5 \times 10^{7}}$

$X_{N_{2}} = 9.35 \times 10^{- 5}$

Solution

Q.
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

$(K_{H} for N_{2} = 6.5 \times 10^{7} mm Hg)$ [2026]

1 $1.17 \times 10^{- 4}$

2 $9.35 \times 10^{5}$

3 $1.23 \times 10^{- 7}$

4 $9.35 \times 10^{- 5}$

Ans.
(4)

$P_{N_{2}} = K_{H} \cdot X_{N_{2}}$

$P_{N_{2}} = 0.8 \times 10 = 8 atm$

$8 \times 760 = 6.5 \times 10^{7} \times X_{N_{2}}$

$X_{N_{2}} = \frac{8 \times 760}{6.5 \times 10^{7}}$

$X_{N_{2}} = 9.35 \times 10^{- 5}$

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Solution

Q. At 298 K, the mole percentage of N2(g) in air is 80%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of N2(g) in water at 298 K (KH for N2=6.5×107 mm Hg) [2026]

1 1.17×10-4

2 9.35×105

3 1.23×10-7

4 9.35×10-5

Ans. (4) PN2=KH·XN2 PN2=0.8×10=8 atm 8×760=6.5×107×XN2 XN2=8×7606.5×107 XN2=9.35×10-5