At STP, 0.48 g of `O_(2)` diffused through a porous partition in 1200 seconds. What volume of `CO_(2)` will diffuse in the same time and under the same conditions?

Options

A. 286. 5 ml
B. 346.7 ml
C. 112.2 ml
D. 224.8ml

Answer

To solve the problem, 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.

### Step-by-Step Solution:

1. **Identify Given Data**:
- Mass of \( O_2 \) (oxygen) = 0.48 g
- Molar mass of \( O_2 \) = 32 g/mol
- Molar mass of \( CO_2 \) (carbon dioxide) = 44 g/mol
- Time for diffusion = 1200 seconds (same for both gases)

2. **Calculate Moles of \( O_2 \)**:
\[ \text{Number of moles of } O_2 (N_1) = \frac{\text{mass}}{\text{molar mass}} = \frac{0.48 \, \text{g}}{32 \, \text{g/mol}} = 0.015 \, \text{mol} \]

3. **Apply Graham's Law**:
According to Graham's law, the ratio of the rates of diffusion of two gases is given by:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
where \( R_1 \) and \( R_2 \) are the rates of diffusion of \( O_2 \) and \( CO_2 \) respectively, and \( M_1 \) and \( M_2 \) are their molar masses.

4. **Set Up the Equation**:
Since the time for diffusion is the same for both gases, we can express the rates in terms of moles:
\[ \frac{N_1/T}{N_2/T} = \sqrt{\frac{M_{CO_2}}{M_{O_2}}} \]
Simplifying gives:
\[ \frac{N_1}{N_2} = \sqrt{\frac{M_{CO_2}}{M_{O_2}}} \]

5. **Substituting Values**:
\[ \frac{0.015}{N_2} = \sqrt{\frac{44}{32}} \]
Calculating the square root:
\[ \sqrt{\frac{44}{32}} = \sqrt{1.375} \approx 1.17 \]
Thus, we have:
\[ \frac{0.015}{N_2} = 1.17 \]

6. **Solving for \( N_2 \)**:
\[ N_2 = \frac{0.015}{1.17} \approx 0.01282 \, \text{mol} \]

7. **Calculate Volume of \( CO_2 \)**:
At STP, 1 mole of gas occupies 22,400 mL. Therefore, the volume of \( CO_2 \) can be calculated as:
\[ \text{Volume of } CO_2 = N_2 \times 22400 \, \text{mL/mol} = 0.01282 \, \text{mol} \times 22400 \, \text{mL/mol} \approx 287.6 \, \text{mL} \]

### Final Answer:
The volume of \( CO_2 \) that will diffuse in the same time and under the same conditions is approximately **287.6 mL**.