100cc of `CO_2` gas is diffused in 25 seconds through a porous membrane. How much time does the same volume fo sulphur dioxide take to diffuse ?  

Answer

To solve the problem of how much time it takes for the same volume of sulfur dioxide (SO₂) to diffuse through a porous membrane compared to carbon dioxide (CO₂), we can use Graham's law of effusion/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. **Identify Given Data:**
- Volume of CO₂ = 100 cc
- Time taken for CO₂ to diffuse (T_CO₂) = 25 seconds
- We need to find the time taken for SO₂ to diffuse (T_SO₂).

2. **Calculate Molar Masses:**
- Molar mass of CO₂:
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol
- CO₂ = 12 + (2 × 16) = 44 g/mol
- Molar mass of SO₂:
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol
- SO₂ = 32 + (2 × 16) = 64 g/mol

3. **Apply Graham's Law:**
According to Graham's law:
\[ \frac{T_{SO₂}}{T_{CO₂}} = \frac{\sqrt{M_{SO₂}}}{\sqrt{M_{CO₂}}} \]
Rearranging gives:
\[ T_{SO₂} = T_{CO₂} \times \frac{\sqrt{M_{SO₂}}}{\sqrt{M_{CO₂}}} \]

4. **Substitute the Values:**
- Substitute the known values:
\[ T_{SO₂} = 25 \, \text{seconds} \times \frac{\sqrt{64}}{\sqrt{44}} \]

5. **Calculate the Square Roots:**
- \(\sqrt{64} = 8\)
- \(\sqrt{44} \approx 6.63\)

6. **Calculate the Ratio:**
\[ \frac{\sqrt{64}}{\sqrt{44}} \approx \frac{8}{6.63} \approx 1.205 \]

7. **Calculate T_SO₂:**
\[ T_{SO₂} = 25 \, \text{seconds} \times 1.205 \approx 30.125 \, \text{seconds} \]

### Final Answer:
The time taken for the same volume of sulfur dioxide to diffuse is approximately **30.125 seconds**.