Question

Henry's law constant for methane in benzene at $298$ K is $4.27\times 10^{5}$ mm Hg. Calculate the solubility of methane in benzene at $298$ K under $760$ mm Hg.

• A. $2.56\times 10^{- 4}$
• B. $1.78\times 10^{- 3}$
• C. $3.78\times 10^{- 2}$
• D. $4.13\times 10^{- 1}$

Answer 1

Answer: (B)

K_H = 4.27 × 10⁵ mm Hg (at 298 K)
p = 760 mm
Applying Henry's law
p = K_H x [x = Mole fraction / solubility of methane]
$\text{x} = \frac{\text{p}}{\left(\text{K}\right)_{\text{H}}} = \frac{\left(\text{760 mm}\right)}{\left(\text{4.27} \times \left(\text{10}\right)^{5} \text{mm}\right)} = \text{178} \times \left(\text{10}\right)^{- 5} = \text{1.78} \times \left(\text{10}\right)^{- 3}$