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 $760 \, mm \, Hg$.
(N/A) According to Henry's law,the partial pressure of a gas is proportional to its mole fraction in the solution: $p = k_{H} x$.
Given:
$p = 760 \, mm \, Hg$
$k_{H} = 4.27 \times 10^{5} \, mm \, Hg$
Rearranging the formula for solubility $(x)$:
$x = \frac{p}{k_{H}}$
Substituting the values:
$x = \frac{760 \, mm \, Hg}{4.27 \times 10^{5} \, mm \, Hg}$
$x = 177.9859 \times 10^{-5}$
Rounding to appropriate significant figures:
$x \approx 1.78 \times 10^{-3}$ or $178 \times 10^{-5}$.
Given:
$p = 760 \, mm \, Hg$
$k_{H} = 4.27 \times 10^{5} \, mm \, Hg$
Rearranging the formula for solubility $(x)$:
$x = \frac{p}{k_{H}}$
Substituting the values:
$x = \frac{760 \, mm \, Hg}{4.27 \times 10^{5} \, mm \, Hg}$
$x = 177.9859 \times 10^{-5}$
Rounding to appropriate significant figures:
$x \approx 1.78 \times 10^{-3}$ or $178 \times 10^{-5}$.
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MediumAIIMS 2007
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