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NCERT Class XII Chemistry
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Question : 35
Total: 53
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.
Solution:  
Here, KH=4.27×105 mm Hg, p = 760 mm Hg
Applying Henry’s law, p=KHx
∴x=
p
KH
=
760
4.27×105
 =1.78×10−3
i.e., solubility in terms of mole fraction of methane in benzene =1.78×10–3
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