If 25 mL of `CO_(2)` diffuses out of a vessel in 75 seconds, what volume of `SO_(2)` would diffuse out in the same time under the same conditions ?

Explanation

To solve the problem of how much volume of SO₂ would diffuse out in the same time as 25 mL of CO₂, 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 Given Data**:
- Volume of CO₂ (V₁) = 25 mL
- Time (t) = 75 seconds
- Molar mass of CO₂ (M₁) = 44 g/mol
- Molar mass of SO₂ (M₂) = 64 g/mol

2. **Write Graham's Law**:
According to Graham's law:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
where \(R_1\) and \(R_2\) are the rates of diffusion of CO₂ and SO₂, respectively, and \(M_1\) and \(M_2\) are their molar masses.

3. **Express the Rates of Diffusion**:
The rate of diffusion can be expressed as:
\[ R = \frac{\text{Volume}}{\text{Time}} \]
Therefore:
\[ R_1 = \frac{25 \text{ mL}}{75 \text{ s}} \quad \text{and} \quad R_2 = \frac{V_2}{75 \text{ s}} \]
where \(V_2\) is the volume of SO₂ that we need to find.

4. **Set Up the Equation**:
Substituting the expressions for \(R_1\) and \(R_2\) into Graham's law gives:
\[ \frac{\frac{25}{75}}{\frac{V_2}{75}} = \sqrt{\frac{64}{44}} \]
Simplifying this, we get:
\[ \frac{25}{V_2} = \sqrt{\frac{64}{44}} \]

5. **Calculate the Square Root**:
Calculate the right side:
\[ \sqrt{\frac{64}{44}} = \sqrt{\frac{16}{11}} = \frac{4}{\sqrt{11}} \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ V_2 = 25 \cdot \frac{\sqrt{11}}{4} \]

7. **Calculate \(V_2\)**:
Now we can calculate \(V_2\):
\[ V_2 \approx 25 \cdot 0.9487 \approx 23.67 \text{ mL} \]

### Final Answer:
The volume of SO₂ that would diffuse out in the same time is approximately **23.67 mL**.