Question

At T(K), Henry's law constant for the molality of methane in benzene is \(4.27\times 10^5\) mm Hg. The solubility of methane in benzene at T(K) under a pressure of 2 atmospheres is:

• A. \(1.78 \times 10^{-3}\)
• B. \(4.56 \times 10^{-3}\)
• C. \(3.56 \times 10^{-3}\)
• D. \(5.34 \times 10^{-3}\)

Hint:

For Henry's law using molality-based constant: \(m = \frac{P}{K_H}\), ensure the pressure units match the units of \(K_H\).

Answer 1

The Correct Option is C

Step 1: Henry's Law Formula.
Henry's law relates solubility (\(m\)) to the partial pressure (\(P\)) using the molality-based constant (\(K_H\)): \[ P = K_H \cdot m \] where \(P\) is in consistent units and \(K_H\) is given.

Step 2: Express solubility.
Solubility \(m = \frac{P}{K_H}\)
Given: \(P = 2 \, \text{atm} = 2 \times 760 \, \text{mm Hg} = 1520 \, \text{mm Hg}\), and \(K_H = 4.27 \times 10^5 \, \text{mm Hg}\).

Step 3: Calculate solubility.
\[ m = \frac{1520}{4.27 \times 10^5} \approx 3.56 \times 10^{-3} \]

Step 4: Final conclusion.
Hence, the solubility of methane in benzene under 2 atm is: \[ \boxed{3.56 \times 10^{-3}} \]