Two gases `A` and `B` having the same volume diffuse through a porous partition in `20` and `10` seconds respectively. The molar mass of `A` is `49 u`. Molar mass of `B` will be

Options

A. 12.25 u
B. 6.50 u
C. 25.00 u
D. 50.00 u

Explanation

To solve the problem, we will use Graham's law of diffusion, 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:**
- Molar mass of gas A, \( M_A = 49 \, u \)
- Time taken for gas A to diffuse, \( t_A = 20 \, s \)
- Time taken for gas B to diffuse, \( t_B = 10 \, s \)

2. **Determine the Rates of Diffusion:**
- The rate of diffusion is inversely proportional to the time taken. Therefore:
\[ R_A = \frac{V}{t_A} = \frac{V}{20} \]
\[ R_B = \frac{V}{t_B} = \frac{V}{10} \]

3. **Set Up the Ratio of Rates of Diffusion:**
- According to Graham's law:
\[ \frac{R_A}{R_B} = \sqrt{\frac{M_B}{M_A}} \]
- Substituting the expressions for \( R_A \) and \( R_B \):
\[ \frac{\frac{V}{20}}{\frac{V}{10}} = \sqrt{\frac{M_B}{49}} \]
- The \( V \) cancels out:
\[ \frac{1}{20} \cdot 10 = \sqrt{\frac{M_B}{49}} \]
- Simplifying the left side:
\[ \frac{1}{2} = \sqrt{\frac{M_B}{49}} \]

4. **Square Both Sides to Eliminate the Square Root:**
\[ \left(\frac{1}{2}\right)^2 = \frac{M_B}{49} \]
\[ \frac{1}{4} = \frac{M_B}{49} \]

5. **Solve for Molar Mass of Gas B:**
- Cross-multiplying gives:
\[ M_B = \frac{49}{4} \]
- Calculating \( M_B \):
\[ M_B = 12.25 \, u \]

### Final Answer:
The molar mass of gas B is \( 12.25 \, u \).