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

The correct option is B $1.78\times {10}^{-3}$

$K_H=4.27\times {10}^5$ mm Hg (at 298 K)

$p=760mm$

Applying Henry's law,

$p=K_{H}x$, where x= Mole fraction or solubility of methane.

$x=\frac{p}{K_H}=\frac{\left(760mm\right)}{(4.27\times {10}^{5}mm)}$

$=1.78\times {10}^{-3}$

Hence, the correct option is B.