A large cylinder of helium filled at 1000 pascal had a thin orifice through which helium escaped into an evacuated space at the rate of 6.4 m mol/h. How long would it take for 10 m mol `SO_(2)` to leak through a similar orific if the `SO_(2)` were confined at the same pressure ?

Options

A. 6.25 h
B. 0.39 h
C. 4.42 h
D. 1.00 h

Explanation

To solve the problem, we need to determine how long it would take for 10 m mol of sulfur dioxide (SO₂) to leak through an orifice, given the rate at which helium (He) escapes through a similar orifice.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Rate of helium escape (R₁) = 6.4 m mol/h
- Molar mass of helium (M₁) = 4 g/mol
- Molar mass of sulfur dioxide (M₂) = 64 g/mol
- Amount of SO₂ to leak = 10 m mol

2. **Apply Graham's Law of Effusion:**
According to Graham's law, the rates of effusion of two gases are inversely proportional to the square root of their molar masses:
\[ \frac{R₁}{R₂} = \sqrt{\frac{M₂}{M₁}} \]
Here, R₂ is the rate of SO₂ escape that we need to find.

3. **Substitute Known Values:**
\[ \frac{6.4}{R₂} = \sqrt{\frac{64}{4}} \]
Simplifying the right side:
\[ \sqrt{\frac{64}{4}} = \sqrt{16} = 4 \]
So, we have:
\[ \frac{6.4}{R₂} = 4 \]

4. **Solve for R₂:**
Rearranging the equation gives:
\[ R₂ = \frac{6.4}{4} = 1.6 \text{ m mol/h} \]

5. **Calculate Time Required for SO₂ to Leak:**
We need to find out how long it takes for 10 m mol of SO₂ to leak at the rate of 1.6 m mol/h:
\[ \text{Time} = \frac{\text{Amount of SO₂}}{R₂} = \frac{10 \text{ m mol}}{1.6 \text{ m mol/h}} = 6.25 \text{ hours} \]

### Final Answer:
It would take **6.25 hours** for 10 m mol of SO₂ to leak through the orifice.

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