From a certain apparatus, the diffusion rate of hydrogen has an average value of `28.7 cm^(3) s^(-1)`. The diffusion of another gas under the same condition is measured to have an average rate of `7.2 cm^(3) s^(-1)`. Identify the gas.

Options

A. Nitrogen
B. Helium
C. Argon
D. Oxygen

Explanation

To identify the gas based on the given diffusion rates, we will 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 (molecular weight).

### Step-by-Step Solution:

1. **Write down the given data:**
- Rate of diffusion of hydrogen, \( R_1 = 28.7 \, \text{cm}^3/\text{s} \)
- Rate of diffusion of another gas, \( R_2 = 7.2 \, \text{cm}^3/\text{s} \)
- Molar mass of hydrogen, \( M_1 = 2 \, \text{g/mol} \)

2. **Apply Graham's Law of Effusion:**
According to Graham's Law:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
where \( M_2 \) is the molar mass of the unknown gas.

3. **Rearrange the equation to solve for \( M_2 \):**
\[ M_2 = M_1 \left( \frac{R_1}{R_2} \right)^2 \]

4. **Substitute the known values into the equation:**
\[ M_2 = 2 \left( \frac{28.7}{7.2} \right)^2 \]

5. **Calculate \( \frac{R_1}{R_2} \):**
\[ \frac{28.7}{7.2} \approx 3.98 \]

6. **Square the result:**
\[ (3.98)^2 \approx 15.84 \]

7. **Now calculate \( M_2 \):**
\[ M_2 = 2 \times 15.84 \approx 31.68 \, \text{g/mol} \]

8. **Identify the gas based on its molar mass:**
The molar mass of approximately \( 32 \, \text{g/mol} \) corresponds to oxygen (O₂).

### Conclusion:
The gas is identified as **oxygen (O₂)**.

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