Question

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 $760 \,mm \,Hg$.

• A. $ 1.78\times 10^{-4} $
• B. $ 1.78\times 10^{-3} $
• C. $ 1.78\times 10^{4} $
• D. $ 1.78\times 10^{-6} $

Answer 1

Answer: (B)

$K _{ H }=4.27 \times 10^{5} mm\, Hg ($ at $298 \,K )$
$p =760\, mm$
Applying Henry's law,
$p = K _{ H x }, $ where $x =$ Mole fraction or solubility of methane.
$x =\frac{ p }{ K _{ H }}=\frac{(760\, mm )}{\left(4.27 \times 10^{5} mm \right)}$
$=1.78 \times 10^{-3}$