Under identical conditions of temperature and pressure the ratio of the rates of effusion of `O_(2)` and `CO_(2)` gases is given by

Options

A. `("rate of effusion of oxygen")/("rate of effusion of carbon dioxide") = 0.87`
B. `("rate of effusion of oxygen")/("rate of effusion of carbon dioxide") = 1.17`
C. `("rate of effusion of oxygen")/("rate of effusion of carbon dioxide") = 8.7`
D. `("rate of effusion of oxygen")/("rate of effusion of carbon dioxide") = 0.117`

Explanation

To find the ratio of the rates of effusion of \( O_2 \) and \( CO_2 \) gases under identical conditions of temperature and pressure, we can use Graham's law of effusion. According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

### Step-by-Step Solution:

1. **Identify the Molar Masses**:
- The molar mass of \( O_2 \) (oxygen) is \( 32 \, \text{g/mol} \).
- The molar mass of \( CO_2 \) (carbon dioxide) is calculated as follows:
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol (and there are 2 oxygen atoms in \( CO_2 \))
- Therefore, \( \text{Molar mass of } CO_2 = 12 + (2 \times 16) = 12 + 32 = 44 \, \text{g/mol} \).

2. **Apply Graham's Law**:
- According to Graham's law, the ratio of the rates of effusion of two gases is given by:
\[ \frac{\text{Rate of } O_2}{\text{Rate of } CO_2} = \sqrt{\frac{M_{CO_2}}{M_{O_2}}} \]
- Substituting the molar masses:
\[ \frac{\text{Rate of } O_2}{\text{Rate of } CO_2} = \sqrt{\frac{44}{32}} \]

3. **Calculate the Square Root**:
- Simplifying the fraction:
\[ \frac{44}{32} = 1.375 \]
- Now, take the square root:
\[ \sqrt{1.375} \approx 1.17 \]

4. **Final Ratio**:
- Thus, the ratio of the rates of effusion of \( O_2 \) to \( CO_2 \) is approximately \( 1.17 \).

### Conclusion:
The ratio of the rates of effusion of \( O_2 \) and \( CO_2 \) gases is approximately \( 1.17 \).