One mole of nitrogen gas at 0.8 atm takes 38 s to diffuse through a pinhole, whereas one mole of an unknown compound of xenon with fluorine at 1.6 atm takes 57s to diffuse through the same hole. Calculate the molecular formula of the compound.

Options

A. (a) `XeF_(6)`
B. (b) `XeF_(2)`
C. (c) `XeF_(4)`
D. (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 and directly proportional to its pressure.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- For nitrogen gas (N₂):
- Pressure (P₁) = 0.8 atm
- Time taken (t₁) = 38 s
- For the unknown compound (XCFY):
- Pressure (P₂) = 1.6 atm
- Time taken (t₂) = 57 s

2. **Calculate the Rates of Diffusion:**
- The rate of diffusion (R) can be expressed as the number of moles divided by time:
- R₁ (for N₂) = 1 mole / 38 s
- R₂ (for XCFY) = 1 mole / 57 s

3. **Apply Graham's Law:**
- According to Graham's law:
\[ \frac{R₁}{R₂} = \frac{P₁}{P₂} \cdot \sqrt{\frac{M₂}{M₁}} \]
- Here, M₁ is the molar mass of nitrogen (N₂), which is approximately 28 g/mol, and M₂ is the molar mass of the unknown compound (XCFY).

4. **Substitute the Known Values:**
- Substitute R₁ and R₂:
\[ \frac{\frac{1}{38}}{\frac{1}{57}} = \frac{0.8}{1.6} \cdot \sqrt{\frac{M₂}{28}} \]
- Simplifying the left side gives:
\[ \frac{57}{38} = \frac{1}{2} \cdot \sqrt{\frac{M₂}{28}} \]

5. **Cross-Multiply and Rearrange:**
- Cross-multiplying gives:
\[ 57 \cdot 2 = 38 \cdot \sqrt{\frac{M₂}{28}} \]
- This simplifies to:
\[ 114 = 38 \cdot \sqrt{\frac{M₂}{28}} \]

6. **Isolate the Square Root:**
- Divide both sides by 38:
\[ \sqrt{\frac{M₂}{28}} = \frac{114}{38} \]
- Calculate the right side:
\[ \sqrt{\frac{M₂}{28}} \approx 3 \]

7. **Square Both Sides:**
- Squaring both sides gives:
\[ \frac{M₂}{28} = 9 \]
- Thus, we find:
\[ M₂ = 9 \cdot 28 = 252 \text{ g/mol} \]

8. **Determine the Molecular Formula:**
- The molecular mass of the unknown compound (XCFY) is given by:
\[ M₂ = M_{Xe} + Y \cdot M_{F} \]
- Where:
- M_{Xe} (Xenon) = 131 g/mol
- M_{F} (Fluorine) = 19 g/mol
- Plugging in the values:
\[ 252 = 131 + Y \cdot 19 \]
- Rearranging gives:
\[ Y \cdot 19 = 252 - 131 = 121 \]
- Solving for Y:
\[ Y = \frac{121}{19} \approx 6.368 \approx 6 \]

9. **Final Molecular Formula:**
- The molecular formula of the compound is:
\[ \text{XCF}_6 \]

### Final Answer:
The molecular formula of the compound is **XCF₆**.