24 mL `H_(2)` , diffuses in 100 s. How much volume of `SO_(2)` , will diffuse during the same time?

Options

A. 11.312 mL
B. 5.656 mL
C. 8 mL
D. none of these

Explanation

To solve the problem, we will 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 Values:**
- Volume of `H2` (V1) = 24 mL
- Time for diffusion (t) = 100 s (this will be the same for both gases)
- Molar mass of `H2` (M1) = 2 g/mol
- Molar mass of `SO2` (M2) = 32 g/mol (for sulfur) + 16 g/mol (for oxygen) * 2 = 64 g/mol

2. **Use Graham's Law:**
According to Graham's law:
\[ \frac{V1}{V2} = \sqrt{\frac{M2}{M1}} \]
where V1 is the volume of `H2`, V2 is the volume of `SO2`, M1 is the molar mass of `H2`, and M2 is the molar mass of `SO2`.

3. **Substitute the Known Values:**
\[ \frac{24 \text{ mL}}{V2} = \sqrt{\frac{64}{2}} \]

4. **Calculate the Right Side:**
\[ \sqrt{\frac{64}{2}} = \sqrt{32} = 4\sqrt{2} \]

5. **Set Up the Equation:**
\[ \frac{24}{V2} = 4\sqrt{2} \]

6. **Cross-Multiply to Solve for V2:**
\[ 24 = V2 \cdot 4\sqrt{2} \]
\[ V2 = \frac{24}{4\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \]

7. **Rationalize the Denominator:**
\[ V2 = 3\sqrt{2} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{3 \cdot 2}{2} = 3 \cdot 1 = 3 \text{ mL} \]

8. **Final Result:**
The volume of `SO2` that will diffuse in the same time is approximately 3 mL.

### Summary:
- Volume of `SO2` that diffuses in 100 seconds is **3 mL**.