A gas diffuses at a rate which is twice that of another gas B. The ratio of molecular weights of A to B is

Options

A. 1
B. 0.75
C. 0.5
D. 0.25

Answer

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. **Understanding the Given Information**:
- Let the rate of diffusion of gas A be \( R_A \) and that of gas B be \( R_B \).
- According to the problem, \( R_A = 2R_B \).

2. **Applying Graham's Law of Diffusion**:
- Graham's law states:
\[ \frac{R_A}{R_B} = \sqrt{\frac{M_B}{M_A}} \]
- Here, \( M_A \) is the molar mass of gas A and \( M_B \) is the molar mass of gas B.

3. **Substituting the Known Rate**:
- Substitute \( R_A = 2R_B \) into the equation:
\[ \frac{2R_B}{R_B} = \sqrt{\frac{M_B}{M_A}} \]
- This simplifies to:
\[ 2 = \sqrt{\frac{M_B}{M_A}} \]

4. **Squaring Both Sides**:
- To eliminate the square root, square both sides of the equation:
\[ 2^2 = \frac{M_B}{M_A} \]
- This gives:
\[ 4 = \frac{M_B}{M_A} \]

5. **Finding the Ratio of Molecular Weights**:
- Rearranging the equation gives:
\[ \frac{M_A}{M_B} = \frac{1}{4} \]
- Therefore, the ratio of the molecular weights of A to B is:
\[ M_A : M_B = 1 : 4 \]

### Final Answer:
The ratio of the molecular weights of gas A to gas B is \( 1 : 4 \).