If rate of diffusion of A is 7 times that of B, what will be the density ratio of B and A?

Answer

To solve the problem, we need to 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 density.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Let the rate of diffusion of gas A be \( R_A \).
- Let the rate of diffusion of gas B be \( R_B \).
- According to the problem, \( R_A = 7 R_B \).

2. **Apply Graham's Law**:
- Graham's law states that:
\[ \frac{R_A}{R_B} = \sqrt{\frac{d_B}{d_A}} \]
- Here, \( d_A \) and \( d_B \) are the densities of gases A and B, respectively.

3. **Substitute the Known Values**:
- From the information given, we can substitute \( R_A \) in the equation:
\[ \frac{7 R_B}{R_B} = \sqrt{\frac{d_B}{d_A}} \]
- This simplifies to:
\[ 7 = \sqrt{\frac{d_B}{d_A}} \]

4. **Square Both Sides**:
- To eliminate the square root, square both sides of the equation:
\[ 7^2 = \frac{d_B}{d_A} \]
- This gives:
\[ 49 = \frac{d_B}{d_A} \]

5. **Find the Density Ratio**:
- Rearranging the equation gives us the density ratio:
\[ \frac{d_B}{d_A} = 49 \]
- This means that the density of gas B is 49 times the density of gas A.

### Conclusion:
The density ratio of B to A is \( 49:1 \).

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