At identical temperature and pressure the rate of diffusion of hydrogen gas is `3sqrt3` 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 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. **Understand the relationship between rates of diffusion and molar masses**:
According to Graham's law, the rate of diffusion of gas A (hydrogen in this case) to gas B (the hydrocarbon) can be expressed as:
\[ \frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}} \]
where \( r_A \) and \( r_B \) are the rates of diffusion of gases A and B, and \( M_A \) and \( M_B \) are their molar masses.

2. **Identify the gases and their rates**:
Here, we have:
- \( r_A \) (rate of diffusion of hydrogen) = \( 3\sqrt{3} \times r_B \) (rate of diffusion of the hydrocarbon).
- Let \( M_A \) be the molar mass of hydrogen (H₂) = 2 g/mol.
- Let \( M_B \) be the molar mass of the hydrocarbon \( C_nH_{2n-n} \).

3. **Set up the equation using Graham's law**:
Plugging in the values into Graham's law gives:
\[ 3\sqrt{3} = \sqrt{\frac{M_B}{M_A}} = \sqrt{\frac{M_B}{2}} \]

4. **Square both sides**:
Squaring both sides to eliminate the square root:
\[ (3\sqrt{3})^2 = \frac{M_B}{2} \]
\[ 27 = \frac{M_B}{2} \]

5. **Solve for M_B**:
Multiply both sides by 2:
\[ M_B = 54 \text{ g/mol} \]

6. **Determine the molar mass of the hydrocarbon**:
The molar mass of the hydrocarbon \( C_nH_{2n-n} \) can be expressed as:
\[ M_B = 12n + (2n - n) = 12n + n = 13n \]

7. **Set up the equation**:
Now, we have:
\[ 13n = 54 \]

8. **Solve for n**:
Dividing both sides by 13:
\[ n = \frac{54}{13} = 4.15 \]

9. **Conclusion**:
Since \( n \) must be a whole number, we round \( n \) to the nearest whole number, which gives us \( n = 4 \).

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