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 expression.
The law connects solubility m to partial pressure P via P = K_H · m, with K_H given.

Step 2: Isolating solubility.
m = P / K_H. Given P = 2 atm = 2 × 760 = 1520 mm Hg, and K_H = 4.27 × 10⁵ mm Hg.

Step 3: Numerical computation.
m = 1520 / (4.27 × 10⁵) ≈ 3.56 × 10⁻³.

Step 4: Final answer.
Thus, methane's solubility in benzene under 2 atm is approximately 3.56 × 10⁻³.