Henry's law constant for the molality of methane in benzene at 298 K is `4.27xx10^(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, "p=760 mm`
`"Applying Henry's law, "p=K_(H)x," we have "x=(p)/(K_(H))=("760 mm")/(4.27xx10^(5)" mm")=1.78xx10^(-3)`
i.e., mole fraction of methane in benzene `=1.78xx10^(-3)`.