Question

Henry’s law constant for the molality of methane in benzene at 298 K is \(4.27 \times 105\, mm\, Hg\). Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Answer 1

The correct answer is: \( 178 \times 10^{ - 5}\)
Here,
p = 760 mm Hg
\(k_H = 4.27 \times 10^ 5 mm Hg \)
According to Henry's law,
\(p = k_Hx\)
\(⇒x=\frac{p}{K_H}\)
\(=\frac{760mm Hg}{4.27\times10^5mmHg}\)
\(= 177.99 \times 10^{ - 5 }\)
\(= 178 \times 10^{ - 5}\) (approximately)
Hence, the mole fraction of methane in benzene is \( 178 \times 10^{ - 5}\)