Henry’s law constant for the molality of methane in benzene at 298 K is `4.27 xx 10^5` mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Explanation

Here `K_(H)=4.27xx10^(5)` mm Hg
p = 760 mm Hg
According to Henry.s law, `p=K_(H)xx x_(CH_(4))`
`x_(CH_(4))=p/(K_(H))=(760mmHg)/(4.27xx10^(5)mmHg)`
`=1.78xx10^(-3)`
Mole fraction of methane in benzene `x_(CH_(4))=1.78xx10^(-3)`