4 gm of sulphur dioxide gas diffuses from a container in 8 min. Mass of helium gas diffusing from the same container over the same time interval is :

Options

A. 0.5 g
B. 1 g
C. 2 g
D. None of these

Explanation

To solve the problem of how much helium gas diffuses from the same container as 4 g of sulfur dioxide in 8 minutes, 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 Given Data:**
- Mass of sulfur dioxide (SO₂) = 4 g
- Time taken for diffusion = 8 minutes
- Molar mass of sulfur dioxide (SO₂) = 64 g/mol
- Molar mass of helium (He) = 4 g/mol

2. **Use Graham's Law of Effusion:**
Graham's law states:
\[ \frac{\text{Rate of diffusion of } SO₂}{\text{Rate of diffusion of } He} = \sqrt{\frac{M_{He}}{M_{SO₂}}} \]
where \(M\) is the molar mass of the gas.

3. **Calculate the Rate of Diffusion:**
The rate of diffusion can be expressed in terms of mass and time:
\[ \frac{m_{SO₂}}{t} \text{ and } \frac{m_{He}}{t} \]
Since the time \(t\) is the same for both gases, we can simplify the equation:
\[ \frac{m_{SO₂}}{m_{He}} = \sqrt{\frac{M_{He}}{M_{SO₂}}} \]

4. **Substitute the Known Values:**
Substitute the values of the molar masses:
\[ \frac{4 \text{ g}}{m_{He}} = \sqrt{\frac{4 \text{ g/mol}}{64 \text{ g/mol}}} \]

5. **Calculate the Right Side:**
\[ \sqrt{\frac{4}{64}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \]

6. **Set Up the Equation:**
Now we have:
\[ \frac{4}{m_{He}} = \frac{1}{4} \]

7. **Cross Multiply to Solve for \(m_{He}\):**
\[ 4 \cdot 4 = m_{He} \]
\[ m_{He} = 16 \text{ g} \]

8. **Find the Mass of Helium:**
Since we are looking for the mass of helium that diffuses in the same time interval, we need to find how much helium corresponds to the same rate of diffusion:
\[ m_{He} = \frac{4 \text{ g}}{4} = 1 \text{ g} \]

### Final Answer:
The mass of helium gas diffusing from the same container over the same time interval is **1 gram**.

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