The molecular weight of a gas which diffuses through a porous plug at `1//6^(th)` of the speed of hydrogen under identical condition is:

Options

A. 27
B. 72
C. 36
D. 48

Explanation

To solve the problem of finding the molecular weight of a gas that diffuses through a porous plug at \( \frac{1}{6} \) of the speed of hydrogen, we can use Graham's law of effusion. Graham's law states that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass.

### Step-by-Step Solution:

1. **Understand Graham's Law**: According to Graham's law, the relationship between the rates of diffusion of two gases is given by:
\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]
where \( r_1 \) and \( r_2 \) are the rates of diffusion of gas 1 and gas 2, and \( M_1 \) and \( M_2 \) are their respective molar masses.

2. **Identify the Gases**: Let gas 1 be hydrogen (H₂) and gas 2 be the unknown gas (X). The molar mass of hydrogen (H₂) is approximately 2 g/mol.

3. **Set Up the Ratio**: According to the problem, the unknown gas diffuses at \( \frac{1}{6} \) the speed of hydrogen. Thus, we can express this as:
\[ r_X = \frac{1}{6} r_{H_2} \]

4. **Substituting into Graham's Law**: Substitute the known values into Graham's law:
\[ \frac{r_{H_2}}{r_X} = \sqrt{\frac{M_X}{M_{H_2}}} \]
Replacing \( r_X \) with \( \frac{1}{6} r_{H_2} \):
\[ \frac{r_{H_2}}{\frac{1}{6} r_{H_2}} = \sqrt{\frac{M_X}{2}} \]

5. **Simplifying the Equation**: This simplifies to:
\[ 6 = \sqrt{\frac{M_X}{2}} \]

6. **Squaring Both Sides**: To eliminate the square root, square both sides:
\[ 6^2 = \frac{M_X}{2} \]
\[ 36 = \frac{M_X}{2} \]

7. **Solving for Molar Mass**: Multiply both sides by 2 to isolate \( M_X \):
\[ M_X = 36 \times 2 = 72 \text{ g/mol} \]

### Final Answer:
The molecular weight of the gas is **72 g/mol**.