A gas 'A' having molecular weight 4 diffuses thrice as fast as the gas B.The molecular weight of gas B is -

Options

A. 36
B. 12
C. 18
D. 24

Answer

To find the molecular weight of gas B given that gas A has a molecular weight of 4 and diffuses thrice as fast as gas B, we can use Graham's law of diffusion. Here’s a step-by-step solution:

### Step 1: Understand Graham's Law of Diffusion
Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (molecular weight). This can be expressed mathematically as:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
where:
- \( R_1 \) = rate of diffusion of gas A
- \( R_2 \) = rate of diffusion of gas B
- \( M_1 \) = molar mass of gas A
- \( M_2 \) = molar mass of gas B

### Step 2: Set Up the Known Values
From the problem, we know:
- \( M_1 = 4 \) g/mol (molecular weight of gas A)
- \( R_1 = 3R_2 \) (gas A diffuses thrice as fast as gas B)

### Step 3: Substitute the Known Values into Graham's Law
Using the relationship \( R_1 = 3R_2 \), we can express \( R_2 \) as:
\[ R_2 = \frac{R_1}{3} \]

Now, substituting \( R_1 \) and \( R_2 \) into Graham's law:
\[ \frac{R_1}{\frac{R_1}{3}} = \sqrt{\frac{M_2}{M_1}} \]

This simplifies to:
\[ 3 = \sqrt{\frac{M_2}{4}} \]

### Step 4: Square Both Sides to Eliminate the Square Root
Squaring both sides gives:
\[ 9 = \frac{M_2}{4} \]

### Step 5: Solve for \( M_2 \)
To find \( M_2 \), multiply both sides by 4:
\[ M_2 = 9 \times 4 = 36 \text{ g/mol} \]

### Conclusion
The molecular weight of gas B is 36 g/mol.