Henry's law constant for $\mathrm{CH}_3 \mathrm{Br}$ is $0.16 \mathrm{~mol} \mathrm{~L}^{-1}…
Henry's law constant for $\mathrm{CH}_3 \mathrm{Br}$ is $0.16 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{bar}^{-1}$ at $298 \mathrm{~K}$. What pressure is required to have solubility of $0.08 \mathrm{~mol} \mathrm{~L}^{-1}$ ?
$0.24 \mathrm{bar}$
$1.6 \mathrm{bar}$
$0.5 \mathrm{bar}$
$4.0 \mathrm{bar}$
Solution
According to Henry's Law
Solubility of gas $=\mathrm{K}_{\mathrm{H}}$. $\mathrm{P}_{\text {gas }}$
$\begin{aligned}
& \mathrm{P}_{\text {gas }}=\text { Solubility } / \mathrm{K}_{\mathrm{H}} \\
& =\frac{0.08}{0.16} \\
& =0.5 \mathrm{bar}
\end{aligned}$