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`.

Options

A. `1.78xx10^(-3)`
B. 17.43
C. 0.114
D. 2.814

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 partial pressure of that gas above the liquid. The formula for Henry's Law is given by:

\[ p = k_H \cdot x \]

Where:
- \( p \) = partial pressure of the gas (in mmHg)
- \( k_H \) = Henry's law constant (in mmHg)
- \( x \) = solubility of the gas in the solvent (in molality)

We are given:
- \( k_H = 4.27 \times 10^5 \, \text{mmHg} \)
- \( p = 760 \, \text{mmHg} \)

We need to find the solubility \( x \).

### Step 1: Rearranging Henry's Law
We can rearrange the formula to solve for \( x \):

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

### Step 2: Substituting the values
Now, substitute the values of \( p \) and \( k_H \) into the equation:

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

### Step 3: Performing the calculation
Now, we perform the division:

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

Calculating this gives:

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

### Final Answer
Thus, the solubility of methane in benzene at 298 K under 760 mmHg is:

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