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

Options

A. `2.56xx10^(-4)`
B. `1.78xx10^(-3)`
C. `3.78xx10^(-2)`
D. `4.13xx10^(-1)`

Answer

To solve the problem of calculating the solubility of methane in benzene at 298 K under a pressure of 760 mm Hg using Henry's law, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Henry's Law**:
Henry's law states that the solubility of a gas in a liquid at a constant temperature is directly proportional to the partial pressure of that gas above the liquid. The mathematical representation of Henry's law is:
\[ P = k_H \cdot x \]
where:
- \( P \) = partial pressure of the gas (in mm Hg)
- \( k_H \) = Henry's law constant (in mm Hg)
- \( x \) = mole fraction (solubility) of the gas in the liquid

2. **Identify Given Values**:
From the problem, we have:
- Henry's law constant, \( k_H = 4.27 \times 10^5 \) mm Hg
- Pressure, \( P = 760 \) mm Hg

3. **Rearrange the Formula to Solve for \( x \)**:
To find the mole fraction \( x \), we can rearrange the equation:
\[ x = \frac{P}{k_H} \]

4. **Substitute the Values**:
Now, substituting the values into the equation:
\[ x = \frac{760 \text{ mm Hg}}{4.27 \times 10^5 \text{ mm Hg}} \]

5. **Calculate \( x \)**:
Performing the calculation:
\[ x = \frac{760}{4.27 \times 10^5} \approx 1.78 \times 10^{-6} \]

6. **Conclusion**:
The solubility of methane in benzene at 298 K under a pressure of 760 mm Hg is approximately \( 1.78 \times 10^{-6} \).

### Final Answer:
The solubility of methane in benzene at 298 K under 760 mm Hg is \( 1.78 \times 10^{-6} \).

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