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. `XeF_(6)`
B. `XeF_(2)`
C. `XeF_(4)`
D. None of these

Explanation

To find the molecular formula of the unknown compound of xenon with fluorine, 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. Additionally, the rate of diffusion is directly proportional to the pressure of the gas.

### Step-by-Step Solution:

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

2. **Calculate the Rate of Diffusion:**
The rate of diffusion can be expressed as the number of moles per unit time. Since we have 1 mole of each gas, we can express the rates as:
- Rate of diffusion for N₂ (R₁) = 1 / t₁ = 1 / 38
- Rate of diffusion for XEFY (R₂) = 1 / t₂ = 1 / 57

3. **Apply Graham's Law of Diffusion:**
According to Graham's Law:
\[ \frac{R₁}{R₂} = \frac{P₁}{P₂} \cdot \sqrt{\frac{M₂}{M₁}} \]
Rearranging gives:
\[ R₁ \cdot t₂ = R₂ \cdot t₁ \cdot \frac{P₁}{P₂} \]

4. **Substituting the Values:**
Substitute the values into the equation:
\[ \frac{1/38}{1/57} = \frac{0.8}{1.6} \cdot \sqrt{\frac{M₂}{28}} \]
Simplifying gives:
\[ \frac{57}{38} = \frac{1}{2} \cdot \sqrt{\frac{M₂}{28}} \]

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

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

7. **Square Both Sides:**
Squaring both sides gives:
\[ \frac{M₂}{28} = \left(\frac{114}{38}\right)^2 \]

8. **Calculate M₂:**
\[ M₂ = 28 \cdot \left(\frac{114}{38}\right)^2 \]
Calculate the right side:
\[ M₂ = 28 \cdot \left(3\right)^2 = 28 \cdot 9 = 252 \text{ g/mol} \]

9. **Determine the Molecular Formula:**
Now we know the molar mass of the unknown compound is 252 g/mol. The atomic mass of xenon (Xe) is 131 g/mol, and that of fluorine (F) is 19 g/mol. Let Y be the number of fluorine atoms:
\[ 131 + 19Y = 252 \]
Rearranging gives:
\[ 19Y = 252 - 131 = 121 \]
Therefore:
\[ Y = \frac{121}{19} \approx 6.37 \approx 6 \]

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

### Final Answer:
The molecular formula of the compound is **XeF₆**.