The solubility of a gas in water is $0.001 M$ at $S T P$ . Determine its Henry's law constant.

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Solution

$C = K_{H} \times P$
$0.001 = K_{H} \times 1$ (mole ${a t m}^{- 1}$ will be the unit of $K_{H}$ )
$∴ K_{H} = 0.001 m o l e / a t m$

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The solubility of a gas in water is 0.001M at STP. Determine its Henry's law constant.

C=KH×P0.001=KH×1 (mole atm−1 will be the unit of KH )∴KH=0.001mole/atm

The solubility of a gas in water is $0.001 M$ at $S T P$ . Determine its Henry's law constant.

$C = K_{H} \times P$
$0.001 = K_{H} \times 1$ (mole ${a t m}^{- 1}$ will be the unit of $K_{H}$ )
$∴ K_{H} = 0.001 m o l e / a t m$