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Henry’s law constant for the molality of methane in benzene at 298 K is $4.27 \times 10^5$ mm Hg. Determine the solubility of methane in benzene at 298 K under 760 mm Hg.

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$K_H=4.27\times 10^5mm$ Hg
P=760mm Hg
$X=\large\frac{P}{K_H}=\frac{760}{4.27\times 10^5}$$=178\times 10^{-5}$
Hence (A) is the correct answer.
answered Jun 9, 2014 by sreemathi.v
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