300 mL of nitrogen gas diffuse through a a porous membrane in `100` seconds. How long will it take for `400 mL` of `CO_(2)` to diffuse through the same membrane under similar conditions of temperature and pressure ?

Explanation

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.

### Step-by-Step Solution:

1. **Determine the rate of diffusion for nitrogen (N₂):**
- Given that 300 mL of nitrogen gas diffuses in 100 seconds, we can calculate the rate of diffusion (R₁) for nitrogen.
\[ R_1 = \frac{\text{Volume of N₂}}{\text{Time}} = \frac{300 \, \text{mL}}{100 \, \text{s}} = 3 \, \text{mL/s} \]

2. **Identify the molar masses of the gases:**
- Molar mass of nitrogen (N₂) = 28 g/mol (since N = 14 g/mol, and there are 2 nitrogen atoms).
- Molar mass of carbon dioxide (CO₂) = 44 g/mol (C = 12 g/mol + 2 × O = 16 g/mol).

3. **Set up the relationship using Graham's law of diffusion:**
- According to Graham's law:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
where \( R_1 \) is the rate of diffusion of nitrogen, \( R_2 \) is the rate of diffusion of carbon dioxide, \( M_1 \) is the molar mass of nitrogen, and \( M_2 \) is the molar mass of carbon dioxide.

4. **Express the rate of diffusion for carbon dioxide (R₂):**
- We know the volume of CO₂ that needs to diffuse is 400 mL. Thus, we can express the rate of diffusion for CO₂ as:
\[ R_2 = \frac{400 \, \text{mL}}{t} \]
where \( t \) is the time we need to find.

5. **Substitute the known values into Graham's law:**
\[ \frac{3 \, \text{mL/s}}{\frac{400 \, \text{mL}}{t}} = \sqrt{\frac{44}{28}} \]

6. **Cross-multiply to solve for t:**
\[ 3 \cdot t = 400 \cdot \sqrt{\frac{44}{28}} \]

7. **Calculate the square root and simplify:**
\[ \sqrt{\frac{44}{28}} = \sqrt{1.5714} \approx 1.25 \]
Now substitute this back:
\[ 3t = 400 \cdot 1.25 \]
\[ 3t = 500 \]

8. **Solve for t:**
\[ t = \frac{500}{3} \approx 166.67 \, \text{seconds} \]

### Final Answer:
It will take approximately **166.67 seconds** for 400 mL of CO₂ to diffuse through the same membrane under similar conditions.