50 mL of a gas A diffuse through a membrane in the same time as for the diffusion of 40 mL of a gas B under identical pressure temperature conditions. If the molecular weight of `A = 64` that of B would be

Options

A. 100
B. 250
C. 200
D. 80

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. Let's go through the steps to find the molecular weight of gas B.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Volume of gas A (R1) = 50 mL
- Volume of gas B (R2) = 40 mL
- Molecular weight of gas A (M1) = 64 g/mol
- Molecular weight of gas B (M2) = ?

2. **Apply Graham's Law of Diffusion:**
According to Graham's law:
\[ \frac{R1}{R2} = \sqrt{\frac{M2}{M1}} \]

3. **Substitute the Known Values:**
Substitute R1, R2, and M1 into the equation:
\[ \frac{50}{40} = \sqrt{\frac{M2}{64}} \]

4. **Simplify the Left Side:**
Simplifying the left side gives:
\[ \frac{5}{4} = \sqrt{\frac{M2}{64}} \]

5. **Square Both Sides:**
To eliminate the square root, square both sides:
\[ \left(\frac{5}{4}\right)^2 = \frac{M2}{64} \]
\[ \frac{25}{16} = \frac{M2}{64} \]

6. **Cross Multiply to Solve for M2:**
Cross multiplying gives:
\[ 25 \times 64 = 16 \times M2 \]
\[ 1600 = 16 \times M2 \]

7. **Divide Both Sides by 16:**
To find M2, divide both sides by 16:
\[ M2 = \frac{1600}{16} = 100 \]

8. **Conclusion:**
Therefore, the molecular weight of gas B is 100 g/mol.

### Final Answer:
The molecular weight of gas B is **100 g/mol**.

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