At identical temperature and pressure, the rate of diffusion of hydrogen gas is `3sqrt(3)` times that of a hydrocarbon having molecular formula `C_(n)H_(2n-n)`. What is the value of n?

Options

A. 1
B. 4
C. 3
D. 8

Explanation

To solve the problem, we need to determine the value of \( n \) in the hydrocarbon with the molecular formula \( C_nH_{2n-2} \) given that the rate of diffusion of hydrogen gas is \( 3\sqrt{3} \) times that of the hydrocarbon at identical temperature and pressure.

### Step-by-Step Solution:

1. **Understand the Rate of Diffusion**: According to Graham's law of effusion, the rate of diffusion of gases is inversely proportional to the square root of their molar masses. This can be expressed as:
\[ \frac{R_{H_2}}{R_{C_nH_{2n-2}}} = \sqrt{\frac{M_{C_nH_{2n-2}}}{M_{H_2}}} \]
where \( R \) is the rate of diffusion and \( M \) is the molar mass.

2. **Set Up the Equation**: From the problem, we know:
\[ \frac{R_{H_2}}{R_{C_nH_{2n-2}}} = 3\sqrt{3} \]
Therefore, we can write:
\[ 3\sqrt{3} = \sqrt{\frac{M_{C_nH_{2n-2}}}{M_{H_2}}} \]

3. **Square Both Sides**: To eliminate the square root, we square both sides:
\[ (3\sqrt{3})^2 = \frac{M_{C_nH_{2n-2}}}{M_{H_2}} \]
This simplifies to:
\[ 27 = \frac{M_{C_nH_{2n-2}}}{M_{H_2}} \]

4. **Calculate Molar Mass of Hydrogen**: The molar mass of hydrogen gas (\( H_2 \)) is:
\[ M_{H_2} = 2 \text{ g/mol} \]

5. **Calculate Molar Mass of Hydrocarbon**: Substitute \( M_{H_2} \) into the equation:
\[ 27 = \frac{M_{C_nH_{2n-2}}}{2} \]
Rearranging gives:
\[ M_{C_nH_{2n-2}} = 27 \times 2 = 54 \text{ g/mol} \]

6. **Express Molar Mass of Hydrocarbon**: The molar mass of the hydrocarbon \( C_nH_{2n-2} \) can be expressed as:
\[ M_{C_nH_{2n-2}} = 12n + (2n - 2) = 14n - 2 \]

7. **Set Up the Equation**: Now we can set up the equation:
\[ 14n - 2 = 54 \]

8. **Solve for \( n \)**: Rearranging gives:
\[ 14n = 54 + 2 = 56 \]
Thus,
\[ n = \frac{56}{14} = 4 \]

9. **Conclusion**: The value of \( n \) is \( 4 \).

### Final Answer:
The value of \( n \) is \( 4 \).