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Question
At 298 K , the mole percentage of $N_{2}(\text{ }g)$ in air is $80\%$. Water is in equilibrium with air at a pressure of 10 atm . What is the mole fraction of $N_{2}(\text{ }g)$ in water at 298 K ? ( $K_{H}$ for $N_{2}$ is $6.5 \times 10^{7}\text{ }mmHg$ )
Options
A.$1.23 \times 10^{- 7}$
B.$1.17 \times 10^{- 4}$
C.$9.35 \times 10^{5}$
D.$9.35 \times 10^{- 5}$
Solution
$P_{N_{2}} = K_{H} \cdot X_{N_{2}}$
$${P_{N_{2}} = 0.8 \times 10 = 8\text{ }atm }{8 \times 760 = 6.5 \times 10^{7} \times X_{N_{2}} }{X_{N_{2}} = \frac{8 \times 760}{6.5 \times 10^{7}} }$$$X_{N_{2}} = 9.35 \times 10^{- 5}$.
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