$H_{2} S$ , toxic gas with a rotten egg-like smell, is used for the qualitative analysis. If the solubility of $H_{2} S$ in the water at STP is $0.195 m$
Calculate Henry's law constant.

132 atm

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200 atm

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282 atm

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343 atm

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Solution

The correct option is B 282 atm
Molality $= 0.195 m$

Solvent mass $= 1 k g$

Moles of solute $= 0.195 m o l$

Number of moles of water $= \frac{1000 g}{18} = 55.56 m o l$

Molecular fraction of $H_{2} S = \frac{0.195}{0.195 + 55.5}$

$\Rightarrow x = 0.0035$

Henry's constant $= K_{H} = \frac{P}{x} = \frac{0.987}{0.0035}$

$[P = 0.987 a t m,$ at STP $]$

$K_{H} = 282$ atm

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H2S, toxic gas with a rotten egg-like smell, is used for the qualitative analysis. If the solubility of H2S in the water at STP is 0.195 mCalculate Henry's law constant.

The correct option is B 282 atmMolality =0.195mSolvent mass =1kgMoles of solute =0.195molNumber of moles of water =1000g18=55.56molMolecular fraction of H2S=0.1950.195+55.5⇒x=0.0035Henry's constant =KH=Px=0.9870.0035[P=0.987 atm, at STP ]KH=282 atm

$H_{2} S$ , toxic gas with a rotten egg-like smell, is used for the qualitative analysis. If the solubility of $H_{2} S$ in the water at STP is $0.195 m$
Calculate Henry's law constant.