Find the relative rates of diffusion of `CO_2` and `Cl_2` gases.

Answer

To find the relative rates of diffusion of carbon dioxide (CO₂) and chlorine (Cl₂) gases, we can use Graham's law of effusion, which states that 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 formula**: According to Graham's law, the relative rates of diffusion of two gases can be expressed as:
\[ \frac{\text{Rate of diffusion of } CO_2}{\text{Rate of diffusion of } Cl_2} = \sqrt{\frac{M_{Cl_2}}{M_{CO_2}}} \]
where \(M\) represents the molar mass of the gases.

2. **Calculate the molar mass of CO₂**:
- The molecular formula of carbon dioxide is CO₂.
- The atomic mass of carbon (C) is approximately 12 g/mol.
- The atomic mass of oxygen (O) is approximately 16 g/mol.
- Since there are two oxygen atoms in CO₂, the calculation is:
\[ M_{CO_2} = 12 + (2 \times 16) = 12 + 32 = 44 \text{ g/mol} \]

3. **Calculate the molar mass of Cl₂**:
- The molecular formula of chlorine gas is Cl₂.
- The atomic mass of chlorine (Cl) is approximately 35.5 g/mol.
- Since there are two chlorine atoms in Cl₂, the calculation is:
\[ M_{Cl_2} = 2 \times 35.5 = 71 \text{ g/mol} \]

4. **Substitute the molar masses into the formula**:
\[ \frac{\text{Rate of diffusion of } CO_2}{\text{Rate of diffusion of } Cl_2} = \sqrt{\frac{71}{44}} \]

5. **Calculate the square root**:
- First, calculate the fraction:
\[ \frac{71}{44} \approx 1.6136 \]
- Now, take the square root:
\[ \sqrt{1.6136} \approx 1.27 \]

6. **Interpret the result**:
- This means that the rate of diffusion of CO₂ is approximately 1.27 times faster than that of Cl₂.
- Therefore, the relative rates of diffusion can be expressed as:
\[ \text{Rate of } CO_2 : \text{Rate of } Cl_2 \approx 1.27 : 1 \]

### Final Answer:
The relative rates of diffusion of CO₂ and Cl₂ gases are approximately 1.27:1.

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