A sample of `O_(2)` gas is collected over water at `23^(@)C` at a barometric pressure of 751 mm Hg (vapour pressure of water at `23^(@)C` is 21 mm Hg). The partical pressure of `O_(2)` gas in the sample collected is

Options

A. 21 mm Hg
B. 751 mm Hg
C. 0.96 atm
D. 1.02 atm

Explanation

To find the partial pressure of the oxygen gas collected over water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Barometric pressure (P_total) = 751 mm Hg
- Vapor pressure of water (P_water) at 23°C = 21 mm Hg

2. **Use the Formula for Partial Pressure:**
The partial pressure of the dry gas (O₂) can be calculated using the formula:
\[ P_{O_2} = P_{total} - P_{water} \]

3. **Substitute the Values into the Formula:**
\[ P_{O_2} = 751 \, \text{mm Hg} - 21 \, \text{mm Hg} \]

4. **Calculate the Partial Pressure of O₂:**
\[ P_{O_2} = 730 \, \text{mm Hg} \]

5. **Convert the Pressure to atm (if required):**
To convert mm Hg to atm, use the conversion factor (1 atm = 760 mm Hg):
\[ P_{O_2} \, \text{(in atm)} = \frac{730 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} \approx 0.964 \, \text{atm} \]

### Final Answer:
The partial pressure of O₂ gas in the sample collected is **730 mm Hg** or approximately **0.964 atm**.

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