1.

Henry's law constant for the molality of methane in benzene at 298 K is 4.27 xx10^5 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Answer»

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Solution :`K_H = 4.27 xx 10^5 MM, p = 760 mm`
Applying Henry.s law
`p = K_H x " or " x = (p)/(K_H)`
Substituting the VALUES, we have
` x=(760mm)/(4.27 xx 10^5 mm) =- 1.78 xx 10^(-3) `
i.e., MOLE fraction of methane in benzene = `1.78 xx 10^(-3)`


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