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Question
Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
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Solution
Given: p = 760 mm Hg
KH = 4.27 × 105 mm Hg
According to Henry’s law,
p = KHχ
⇒ χ = `p/K_H`
= `(760 mm Hg)/(4.27 xx 10^(5) mm Hg)`
= 1.78 × 10−3
∴ Solubility in terms of mole fraction of methane in benzene is 1.78 × 10−3.
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