If 150cc of `CO` effused in 25 seconds, what volume of methane would effuse in the same time?

Answer

To solve the problem of how much volume of methane (CH₄) would effuse in the same time as 150 cc of carbon monoxide (CO) effused in 25 seconds, we can use Graham's law of effusion. According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Volume of CO effused (V_CO) = 150 cc
- Time for CO effusion (t) = 25 seconds
- Molar mass of CO (M_CO) = 28 g/mol (12 g/mol for C + 16 g/mol for O)
- Molar mass of CH₄ (M_CH₄) = 16 g/mol (12 g/mol for C + 4 g/mol for H)

2. **Write the Relationship from Graham's Law:**
\[ \frac{V_{CO}}{V_{CH₄}} = \frac{\sqrt{M_{CH₄}}}{\sqrt{M_{CO}}} \]
Rearranging gives:
\[ V_{CH₄} = V_{CO} \cdot \frac{\sqrt{M_{CO}}}{\sqrt{M_{CH₄}}} \]

3. **Substitute the Known Values:**
\[ V_{CH₄} = 150 \cdot \frac{\sqrt{28}}{\sqrt{16}} \]

4. **Calculate the Square Roots:**
- \(\sqrt{28} \approx 5.29\)
- \(\sqrt{16} = 4\)

5. **Substitute the Square Roots Back into the Equation:**
\[ V_{CH₄} = 150 \cdot \frac{5.29}{4} \]

6. **Perform the Calculation:**
\[ V_{CH₄} = 150 \cdot 1.3225 \approx 198.4 \text{ cc} \]

### Final Answer:
The volume of methane that would effuse in the same time is approximately **198.4 cc**.