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

To find the mole fraction of N₂(g) in water at 298 K, we can use Henry's Law. According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. Mathematically, this relationship is expressed as:

\(C = K_H \cdot P\)

where:

• \(C\) is the concentration (solubility) of the gas in the liquid (in mole fraction).
• \(K_H\) is Henry's law constant (in atm). Here, \(K_H\) is given in mm Hg, so we need to convert it to atm.
• \(P\) is the partial pressure of the gas (in atm).

We are given:

• Mole percentage of N₂ in air = 80%
• Total pressure = 10 atm
• \(K_H\) for N₂ = \(6.5 \times 10^7\) mm Hg

The partial pressure of N₂ can be found by multiplying its mole fraction in air by the total pressure:

\(P_{N_2} = 0.80 \times 10 = 8 \text{ atm}\)

Next, we convert \(K_H\) from mm Hg to atm:

1 atm = 760 mm Hg, so:

\(K_H = \frac{6.5 \times 10^7}{760} \text{ atm}^{-1} = 8.55 \times 10^4 \text{ atm}^{-1}\)

Using Henry's Law:

\(C = K_H \times P_{N_2} = \frac{1}{8.55 \times 10^4} \times 8\)

\(C = \frac{8}{8.55 \times 10^4} \approx 9.35 \times 10^{-5}\)

Therefore, the mole fraction of N₂ in water at 298 K is approximately \(9.35 \times 10^{-5}\), which matches the correct answer.