1.

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

Answer»

SOLUTION :`K_(H) = 4.27 XX 10^(5)` 1 Hg
P = 762 mm of Hg
Henry.s law`P = K_(H) xx x_(CH_(4)) `
`x_(CH_(4)) = (P)/(K_(H)) = (760 "mm Hg")/(4.27 xx 10^(5) mm "Hg")`
` = 1.78 xx 10^(-3)`
mole fraction of methane in benzene ,
`x_(CH_(4)) = 1.78 xx 10^(-3)` .


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