The time for a certain volume of a gas `A` to diffuse through a small hole is 2 minute If takes 5.65 minute for oxygen to diffuse under similar conditions Find the molecualr weight of `A` .

Explanation

To find the molecular weight of gas A, we can 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 molar mass. The formula can be expressed as:

\[ \frac{R_A}{R_O} = \sqrt{\frac{M_O}{M_A}} \]

Where:
- \( R_A \) = rate of diffusion of gas A
- \( R_O \) = rate of diffusion of oxygen (O₂)
- \( M_A \) = molar mass of gas A
- \( M_O \) = molar mass of oxygen (O₂), which is 32 g/mol

### Step-by-Step Solution:

1. **Identify the given data:**
- Time taken for gas A to diffuse, \( t_A = 2 \) minutes
- Time taken for oxygen to diffuse, \( t_O = 5.65 \) minutes

2. **Calculate the rates of diffusion:**
Since the rate of diffusion is inversely proportional to time, we can express the rates as:
\[ R_A = \frac{1}{t_A} = \frac{1}{2} \quad \text{and} \quad R_O = \frac{1}{t_O} = \frac{1}{5.65} \]

3. **Set up the ratio of rates:**
\[ \frac{R_A}{R_O} = \frac{\frac{1}{2}}{\frac{1}{5.65}} = \frac{5.65}{2} \]

4. **Substitute into Graham's law:**
\[ \frac{5.65}{2} = \sqrt{\frac{32}{M_A}} \]

5. **Square both sides to eliminate the square root:**
\[ \left(\frac{5.65}{2}\right)^2 = \frac{32}{M_A} \]
\[ \frac{31.9225}{4} = \frac{32}{M_A} \]
\[ 7.980625 = \frac{32}{M_A} \]

6. **Cross-multiply to solve for \( M_A \):**
\[ 7.980625 \cdot M_A = 32 \]
\[ M_A = \frac{32}{7.980625} \approx 4.01 \text{ g/mol} \]

7. **Round off the answer:**
The molecular weight of gas A is approximately \( 4 \text{ g/mol} \).

### Final Answer:
The molecular weight of gas A is approximately \( 4 \text{ g/mol} \).