1.

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.

Answer»

SOLUTION :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)`.


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