If oxygen diffuses at a rate of `6cm^(3),"sec"^(-1)` through a fine hole, find the rate of diffusion of carbon dioxide gas under the same conditions.  

Answer

To solve the problem of finding the rate of diffusion of carbon dioxide gas when the rate of diffusion of oxygen is given, we can use Graham's law of effusion. According to Graham's law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Rate of diffusion of oxygen (O₂) = 6 cm³/sec
- Molar mass of oxygen (O₂) = 32 g/mol (since O = 16 g/mol, O₂ = 2 × 16)
- Molar mass of carbon dioxide (CO₂) = 44 g/mol (C = 12 g/mol, O₂ = 32 g/mol, CO₂ = 12 + 32)

2. **Apply Graham's Law**:
According to Graham's law:
\[ \frac{\text{Rate of diffusion of O}_2}{\text{Rate of diffusion of CO}_2} = \sqrt{\frac{\text{Molar mass of CO}_2}{\text{Molar mass of O}_2}} \]
Let the rate of diffusion of CO₂ be \( x \). Thus, we can write:
\[ \frac{6}{x} = \sqrt{\frac{44}{32}} \]

3. **Square Both Sides**:
Squaring both sides to eliminate the square root gives:
\[ \left(\frac{6}{x}\right)^2 = \frac{44}{32} \]
Which simplifies to:
\[ \frac{36}{x^2} = \frac{44}{32} \]

4. **Cross-Multiply**:
Cross-multiplying gives:
\[ 36 \times 32 = 44 \times x^2 \]
Simplifying this:
\[ 1152 = 44x^2 \]

5. **Solve for \( x^2 \)**:
\[ x^2 = \frac{1152}{44} \]
\[ x^2 = 26.18 \]

6. **Calculate \( x \)**:
Taking the square root of both sides:
\[ x = \sqrt{26.18} \approx 5.12 \text{ cm}^3/\text{sec} \]

7. **Conclusion**:
The rate of diffusion of carbon dioxide (CO₂) is approximately **5.12 cm³/sec**.