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)

• A. $9.35 \times 10^{-5}$
• B. $1.17 \times 10^{-4}$
• C. $9.35 \times 10^{5}$
• D. $1.23 \times 10^{-7}$

Hint:

Higher Henry’s constant means lower solubility of gas in liquid.

Answer 1

The Correct Option is A

Step 1: Partial pressure of N$_2$.
\[ P_{N_2} = 0.8 \times 10 = 8 \text{ atm} \]
Step 2: Using Henry’s law.
\[ x = \frac{P}{K_H} \]
Converting atm to mm Hg: $8 \times 760 = 6080$ mm Hg
\[ x = \frac{6080}{6.5 \times 10^7} = 9.35 \times 10^{-5} \]