Question
Obtain the units of Henry’s law constant.

Answer

By Henry's law, $S=K_H \times P$, where $S$ is solubility of the gas in mol $dm ^{-3}, P$ is the pressure of the gas in atmosphere (or in bar) and $K _{ H }$ is Henry's law constant.
$\therefore K _{ H }=\frac{S^{\left( moldm ^{-3}\right)}}{P_{( atm )}} mol dm atm ^{-1}$ or $mol dm ^{-3} bar ^{-1}$
Hence the units of Henry's law constant $K _{ H }$ are $mol dm { }^{-3} atm ^{-1}$.

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