Henry's law constant for the molality of methane in benzene at `298K `is `4.27xx10^(5)mm Hg`. Calculate the solubility of methane in benzene at `298 K` under `760 mm Hg`.

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 can be expressed as:

\[ P = K_H \times S \]

Where:
- \( P \) is the pressure of the gas (in mm Hg),
- \( K_H \) is Henry's law constant (in mm Hg),
- \( S \) is the solubility of the gas (in molality or mol/L).

### Step-by-step solution:

1. **Identify the given values**:
- Henry's law constant \( K_H = 4.27 \times 10^5 \) mm Hg
- Pressure \( P = 760 \) mm Hg

2. **Rearrange Henry's Law to find solubility**:
\[ S = \frac{P}{K_H} \]

3. **Substitute the values into the equation**:
\[ S = \frac{760 \, \text{mm Hg}}{4.27 \times 10^5 \, \text{mm Hg}} \]

4. **Perform the calculation**:
\[ S = \frac{760}{4.27 \times 10^5} \]
\[ S \approx 1.78 \times 10^{-3} \, \text{molality} \]

5. **Final answer**:
The solubility of methane in benzene at 298 K under 760 mm Hg is approximately \( 1.78 \times 10^{-3} \) molality.