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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. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

$\begin{array}{1 1}1.78\times 10^{-3}\\2.78\times 10^{-3}\\3.78\times 10^{-3}\\4.78\times 10^{-3}\end{array} $

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Given :
$K_H=4.27\times 10^5$ mm
Applying Hentry's law
$\;\;\;=\large\frac{760m}{4.27\times 10^5mm}$
$\;\;\;=1.78\times 10^{-3}$
(ie) Mole fraction of methane in benzene =$1.78\times 10^{-3}$
answered Aug 6, 2014 by sreemathi.v
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