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Q.

$H_{2} S$ gas is used in qualitative analysis of inorganic cations. Its solubility in water at STP is 0.195 mol Kg^-1. Thus, Henry's law constant $({in atm molal}^{- 1})$ for $H_{2} S$ is

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a

$2.628 \times 10^{- 4}$

b

5.128

c

0.185

d

$3.826 \times 10^{3}$

answer is B.

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Detailed Solution

By Henry's law, $p = K_{H} χ = K_{H} S$ (where, S = solubility)

$∴ K_{H} = \frac{p}{S} = \frac{1 atm}{0.195 {mol kg}^{- 1}}$

$= 5.128 {atm kg mol}^{- 1}$

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