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 1: Understand Henry's Law
Henry's law states that the amount of gas that dissolves in a liquid at a given temperature is directly proportional to the partial pressure of the gas above the liquid. The mathematical expression 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)
- \( x \) = solubility of the gas (in molality)

### Step 2: Rearrange the Equation
To find the solubility \( x \), we can rearrange the equation:

\[ x = \frac{P}{k_H} \]

### Step 3: Substitute the Known Values
From the problem, we know:
- \( P = 760 \, \text{mm Hg} \)
- \( k_H = 4.27 \times 10^5 \, \text{mm Hg} \)

Now we can substitute these values into the rearranged equation:

\[ x = \frac{760 \, \text{mm Hg}}{4.27 \times 10^5 \, \text{mm Hg}} \]

### Step 4: Perform the Calculation
Now, we perform the calculation:

\[ x = \frac{760}{4.27 \times 10^5} \]

Calculating this gives:

\[ x \approx 1.78 \times 10^{-3} \]

### Step 5: Conclusion
Thus, the solubility of methane in benzene at 298 K under a pressure of 760 mm Hg is:

\[ x \approx 1.78 \times 10^{-3} \, \text{molality} \]