Question

Dry air contains 79\% \(\text{N}_2\) and 21\% \(\text{O}_2\). At temperature T(K), the Henry's law constants for \(\text{N}_2\) and \(\text{O}_2\) are \(8.57\times 10^4 \,\text{atm}\) and \(4.56\times 10^4 \,\text{atm}\), respectively. If this air is in contact with water at 1 atm, what is the ratio of the mole fractions \(\frac{X_{\text{N}_2}}{X_{\text{O}_2}}\) of \(\text{N}_2\) and \(\text{O}_2\) dissolved in water?

• A. 4 : 1
• B. 1 : 4
• C. 2 : 1
• D. 1 : 2

Hint:

For gases above water, the amount dissolved is proportional to partial pressure and inversely proportional to the Henry's constant. - Nitrogen's higher partial pressure and slightly higher Henry's constant still lead to about twice as much \(\text{N}_2\) dissolved as \(\text{O}_2\).

Answer 1

The Correct Option is C

Step 1: Partial Pressures of \(\text{N}_2\) and \(\text{O}_2\) - Total pressure = 1 atm. - \(\text{p}_{\text{N}_2} = 0.79 \,\text{atm}\), \(\text{p}_{\text{O}_2} = 0.21 \,\text{atm}\). 

Step 2: Henry's Law and Mole Fractions \[ X_{\text{gas}} = \frac{p_{\text{gas}}}{K_{\text{H}}(\text{gas})} \] Hence, \[ X_{\text{N}_2} = \frac{0.79}{8.57 \times 10^4}, \quad X_{\text{O}_2} = \frac{0.21}{4.56 \times 10^4}. \] 

Step 3: Ratio of \(\text{N}_2\) to \(\text{O}_2\) \[ \frac{X_{\text{N}_2}}{X_{\text{O}_2}} = \frac{\tfrac{0.79}{8.57 \times 10^4}}{\tfrac{0.21}{4.56 \times 10^4}} = \frac{0.79 \times 4.56}{8.57 \times 0.21} \approx 2 : 1. \] Thus, the ratio of the mole fractions is \(2:1\).