A metallic carbonyl `M (CO)_(x)` is in gaseous state. The rate of diffusion of `CH_(4)` is 3.31 time faster than this gaseous carbonyl under identical conditions. If atomic mass of metal is 63.29, the closest integer value of X is___________.

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 Information:**
- The rate of diffusion of methane (CH₄) is 3.31 times faster than that of the metallic carbonyl M(CO)ₓ.
- The atomic mass of the metal (M) is 63.29 g/mol.

2. **Set Up Graham's Law:**
According to Graham's law:
\[ \frac{R_{CH_4}}{R_{M(CO)_x}} = \sqrt{\frac{M_{M(CO)_x}}{M_{CH_4}}} \]
Where:
- \( R_{CH_4} \) = rate of diffusion of CH₄
- \( R_{M(CO)_x} \) = rate of diffusion of the metallic carbonyl
- \( M_{CH_4} \) = molar mass of CH₄ = 16 g/mol (C: 12 g/mol + H: 4 g/mol)

3. **Substituting the Known Values:**
Since the rate of diffusion of CH₄ is 3.31 times that of the metallic carbonyl, we can write:
\[ \frac{R_{CH_4}}{R_{M(CO)_x}} = 3.31 \]
Therefore, we can substitute this into Graham's law:
\[ 3.31 = \sqrt{\frac{M_{M(CO)_x}}{16}} \]

4. **Square Both Sides:**
To eliminate the square root, we square both sides:
\[ (3.31)^2 = \frac{M_{M(CO)_x}}{16} \]
\[ 10.9561 = \frac{M_{M(CO)_x}}{16} \]

5. **Calculate the Molar Mass of the Metallic Carbonyl:**
Rearranging gives:
\[ M_{M(CO)_x} = 10.9561 \times 16 \]
\[ M_{M(CO)_x} = 175.2976 \text{ g/mol} \]

6. **Determine the Molar Mass of the Carbonyl:**
The molar mass of the metallic carbonyl can be expressed as:
\[ M_{M(CO)_x} = M + x \times M_{CO} \]
Where:
- \( M_{CO} = 12 + 16 = 28 \text{ g/mol} \)
- Therefore, we have:
\[ 175.2976 = 63.29 + x \times 28 \]

7. **Solve for x:**
Rearranging gives:
\[ x \times 28 = 175.2976 - 63.29 \]
\[ x \times 28 = 112.0076 \]
\[ x = \frac{112.0076}{28} \approx 4 \]

8. **Conclusion:**
The closest integer value of \( x \) is **4**.