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

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

Solution:

Given that Henry’s law constant

k_H= 4.27 × 10⁵ mm Hg

 p = 760 mm Hg

If X is the molar fraction then

According to Henry’s law,

p = k_H X

760  = 4.27 × 10⁵ × X   

Divide by 4.27 × 10⁵  ,we get

Molar fraction, X 

= 178 × 10⁻⁵

So that, the molar fraction of methane in benzene = 178 × 10⁻⁵.