What is the concentration of `O_(2)` in a freshwater stream in equilibrium with air at `30^@C` and `2.0` bar ? Given, `K_(H)` (Henry's law constant) of `O_(2)=2.0xx10^(-3)mol//kg` bar at `30^@C`.

Options

A. `8.736xx10^(-3)g//kg`
B. `1.344xx10^(-3) g//kg`
C. `24.04g //kg`
D. `114.5 g//kg`

Explanation

To find the concentration of \( O_2 \) in a freshwater stream in equilibrium with air at \( 30^\circ C \) and \( 2.0 \) bar, we can use Henry's law, which states that the concentration of a gas in a liquid is proportional to the partial pressure of that gas above the liquid.

### Step-by-step Solution:

1. **Identify Henry's Law Constant**:
Given:
- Henry's law constant, \( K_H = 2.0 \times 10^{-3} \, \text{mol/kg bar} \)
- Temperature = \( 30^\circ C \)

2. **Determine the Partial Pressure of \( O_2 \)**:
The partial pressure of \( O_2 \) in air is approximately \( 21\% \) of the total air pressure. Since the total air pressure is \( 2.0 \, \text{bar} \):
\[ P_{O_2} = 0.21 \times 2.0 \, \text{bar} = 0.42 \, \text{bar} \]

3. **Apply Henry's Law**:
According to Henry's law, the concentration of \( O_2 \) in the freshwater stream can be calculated as:
\[ C_{O_2} = K_H \times P_{O_2} \]
Substituting the values:
\[ C_{O_2} = (2.0 \times 10^{-3} \, \text{mol/kg bar}) \times (0.42 \, \text{bar}) \]
\[ C_{O_2} = 8.4 \times 10^{-4} \, \text{mol/kg} \]

4. **Convert to grams per kg**:
To convert moles to grams, we use the molar mass of \( O_2 \), which is \( 32 \, \text{g/mol} \):
\[ C_{O_2} = 8.4 \times 10^{-4} \, \text{mol/kg} \times 32 \, \text{g/mol} = 2.688 \times 10^{-2} \, \text{g/kg} \]

5. **Final Result**:
The concentration of \( O_2 \) in the freshwater stream is:
\[ C_{O_2} \approx 2.688 \times 10^{-2} \, \text{g/kg} \quad \text{or} \quad 0.02688 \, \text{g/kg} \]