MHT CET · Chemistry · Solutions
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 \(0.08 \mathrm{~mol} \mathrm{~L}^{-1}\) ?
- A \(0.24 \mathrm{bar}\)
- B \(1.6 \mathrm{bar}\)
- C \(0.5 \mathrm{bar}\)
- D \(4.0 \mathrm{bar}\)
Answer & Solution
Correct Answer
(C) \(0.5 \mathrm{bar}\)
Step-by-step Solution
Detailed explanation
According to Henry's Law
Solubility of gas \(=\mathrm{K}_{\mathrm{H}}\). \(\mathrm{P}_{\text {gas }}\)
\(
\begin{aligned}
& \mathrm{P}_{\text {gas }}=\text { Solubility } / \mathrm{K}_{\mathrm{H}} \\
& =\frac{0.08}{0.16} \\
& =0.5 \mathrm{bar}
\end{aligned}
\)
Solubility of gas \(=\mathrm{K}_{\mathrm{H}}\). \(\mathrm{P}_{\text {gas }}\)
\(
\begin{aligned}
& \mathrm{P}_{\text {gas }}=\text { Solubility } / \mathrm{K}_{\mathrm{H}} \\
& =\frac{0.08}{0.16} \\
& =0.5 \mathrm{bar}
\end{aligned}
\)
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