At 760 torr pressure and `20^(@)C` tempreature , 1 L of water dissolves 0.04 gm of pure oxygen or 0.02 gm of pure nitrogen. Assuming that dry air is compound of 20% oxygen and 80% nitrogen (by volume), the masses (in g/L) of oxygen and nitrogen dissolved by 1 L of water at `20^(@)C` exposed to air at a total pressure of 760 torr are respectively :

Options

A. 0.008, 0.016
B. 0.016, 0.008
C. 0.16, 0.08
D. 0.04, 0.02

Answer

To solve the problem, we need to determine the masses of oxygen and nitrogen dissolved in 1 L of water when exposed to air at a total pressure of 760 torr. We will use Henry's law and the information given in the question.

### Step-by-Step Solution:

1. **Understanding the Given Data**:
- At 760 torr and 20°C, 1 L of water dissolves:
- 0.04 g of pure oxygen (O₂)
- 0.02 g of pure nitrogen (N₂)
- Dry air is composed of 20% O₂ and 80% N₂ by volume.

2. **Using Henry's Law**:
- Henry's law states that the solubility of a gas in a liquid is proportional to the partial pressure of that gas above the liquid.
- The formula is:
\[ C = k_H \cdot P \]
where \( C \) is the concentration (solubility), \( k_H \) is Henry's constant, and \( P \) is the partial pressure of the gas.

3. **Calculating Henry's Constants**:
- For oxygen:
\[ k_{H_{O_2}} = \frac{0.04 \, \text{g/L}}{760 \, \text{torr} - P_{H_2O}} \]
- For nitrogen:
\[ k_{H_{N_2}} = \frac{0.02 \, \text{g/L}}{760 \, \text{torr} - P_{H_2O}} \]

4. **Calculating Partial Pressures in Air**:
- The partial pressure of oxygen in dry air:
\[ P_{O_2} = 0.2 \times 760 \, \text{torr} - P_{H_2O} \]
- The partial pressure of nitrogen in dry air:
\[ P_{N_2} = 0.8 \times 760 \, \text{torr} - P_{H_2O} \]

5. **Substituting into Henry's Law**:
- For oxygen:
\[ C_{O_2} = k_{H_{O_2}} \cdot P_{O_2} = \frac{0.04}{760 - P_{H_2O}} \cdot (0.2 \times 760 - P_{H_2O}) \]
- For nitrogen:
\[ C_{N_2} = k_{H_{N_2}} \cdot P_{N_2} = \frac{0.02}{760 - P_{H_2O}} \cdot (0.8 \times 760 - P_{H_2O}) \]

6. **Simplifying the Equations**:
- The vapor pressure of water at 20°C is approximately 17.5 torr (this value can be looked up).
- Substitute \( P_{H_2O} \) into the equations:
- For oxygen:
\[ C_{O_2} = \frac{0.04}{760 - 17.5} \cdot (0.2 \times 760 - 17.5) \]
- For nitrogen:
\[ C_{N_2} = \frac{0.02}{760 - 17.5} \cdot (0.8 \times 760 - 17.5) \]

7. **Calculating the Values**:
- Calculate \( C_{O_2} \):
\[ C_{O_2} = \frac{0.04}{742.5} \cdot (152 - 17.5) = \frac{0.04}{742.5} \cdot 134.5 \approx 0.008 \, \text{g/L} \]
- Calculate \( C_{N_2} \):
\[ C_{N_2} = \frac{0.02}{742.5} \cdot (608 - 17.5) = \frac{0.02}{742.5} \cdot 590.5 \approx 0.016 \, \text{g/L} \]

8. **Final Results**:
- The masses of oxygen and nitrogen dissolved in 1 L of water at 20°C exposed to air at a total pressure of 760 torr are:
- Oxygen: 0.008 g/L
- Nitrogen: 0.016 g/L