Henry's law constan for the solubility of methane in benzene is `4.27 xx 10^(-5) mm^(-1)` Hg mol `dm^(-3)` at constant temperature .Calculate the solubility of methane at 760 mm Hg pressure at same temperature .

Explanation

To solve the problem, we will use Henry's law, which states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the liquid. The formula for Henry's law is:

\[ S = K \times P \]

Where:
- \( S \) = solubility of the gas (in mol/dm³)
- \( K \) = Henry's law constant (in mmHg⁻¹ mol dm⁻³)
- \( P \) = pressure of the gas (in mmHg)

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Henry's law constant, \( K = 4.27 \times 10^{-5} \) mmHg⁻¹ mol dm⁻³
- Pressure, \( P = 760 \) mmHg

2. **Substitute the Values into the Formula:**
Using the formula \( S = K \times P \):
\[ S = (4.27 \times 10^{-5} \, \text{mmHg}^{-1} \, \text{mol dm}^{-3}) \times (760 \, \text{mmHg}) \]

3. **Perform the Calculation:**
\[ S = 4.27 \times 10^{-5} \times 760 \]
\[ S = 3.2452 \times 10^{-2} \, \text{mol/dm}^3 \]

4. **Final Answer:**
The solubility of methane in benzene at 760 mmHg pressure is:
\[ S \approx 3.245 \times 10^{-2} \, \text{mol/dm}^3 \]