Question

The ratio of \( N_2 \) and \( O_2 \) gases in the atmosphere is \( 4:1 \). The ratio of the mole fractions of the dissolved gases \( N_2:O_2 \) in rain water will be approximately --------. At \( 293\,K \), \( K_H \) for Nitrogen and Oxygen in kbar units are \( 76.48 \) and \( 34.86 \) respectively.

• A. \( 3:1 \)
• B. \( 2:1 \)
• C. \( 4:1 \)
• D. \( 1:4 \)

Hint:

In Henry's law, solubility is inversely proportional to \( K_H \). A gas with lower \( K_H \) is more soluble in liquid.

Answer 1

The Correct Option is B

Step 1: Use Henry's law.
According to Henry's law:
\[ p = K_H x \]
So:
\[ x = \frac{p}{K_H} \]

Step 2: Write ratio of mole fractions.
\[ \frac{x_{N_2}}{x_{O_2}} = \frac{\frac{p_{N_2}}{K_{H,N_2}}}{\frac{p_{O_2}}{K_{H,O_2}}} \]

Step 3: Rearrange the expression.
\[ \frac{x_{N_2}}{x_{O_2}} = \frac{p_{N_2}}{p_{O_2}} \times \frac{K_{H,O_2}}{K_{H,N_2}} \]

Step 4: Substitute given values.
Given:
\[ \frac{p_{N_2}}{p_{O_2}} = \frac{4}{1} \]
\[ K_{H,N_2}=76.48,\quad K_{H,O_2}=34.86 \]
So:
\[ \frac{x_{N_2}}{x_{O_2}} = 4 \times \frac{34.86}{76.48} \]

Step 5: Calculate the ratio.
\[ \frac{34.86}{76.48} \approx 0.456 \]
\[ 4 \times 0.456 = 1.824 \]

Step 6: Approximate the value.
\[ 1.824:1 \approx 2:1 \]

Step 7: Final conclusion.
\[ \boxed{2:1} \]