Air contains `79% N_(2)` and `21% O_(2)` by volume. If the barometric pressure is `750 mm Hg`. The partial pressure of oxygen is

Options

A. 157.5 mm of Hg
B. 175.5 mm of Hg
C. 3125.0 mm of Hg
D. None of these

Explanation

To find the partial pressure of oxygen in the air, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Composition of Air**:
- The air contains 79% nitrogen (N₂) and 21% oxygen (O₂) by volume.

2. **Identify Total Barometric Pressure**:
- The total barometric pressure is given as 750 mm Hg.

3. **Calculate the Mole Fraction of Oxygen**:
- The mole fraction (X) of a gas in a mixture is the ratio of the volume (or moles) of that gas to the total volume (or total moles) of the mixture.
- For oxygen (O₂), the mole fraction can be calculated as:
\[ X_{O_2} = \frac{\text{Volume of } O_2}{\text{Total Volume}} = \frac{21}{100} = 0.21 \]

4. **Calculate the Partial Pressure of Oxygen**:
- The partial pressure (P) of a gas in a mixture can be calculated using the formula:
\[ P_{O_2} = X_{O_2} \times P_{\text{total}} \]
- Substituting the values we have:
\[ P_{O_2} = 0.21 \times 750 \text{ mm Hg} \]
- Performing the multiplication:
\[ P_{O_2} = 157.5 \text{ mm Hg} \]

5. **Conclusion**:
- The partial pressure of oxygen in the air is 157.5 mm Hg.

### Final Answer:
The partial pressure of oxygen (O₂) is **157.5 mm Hg**.

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