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Q.

The solubility of $H_{2} S$ gas in water at STP is 0.2 mol/kg. Thus Henry’s law constant $K_{H} ({in atm molal}^{- 1})$ for $H_{2} S$ is

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answer is 5.

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Detailed Solution

$K_{H} = \frac{P_{H_{2} S}}{solubitliy} = \frac{1atm}{0.2m} =5atm/molal$

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