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Question

$$H_2S$$ a toxic gas with rottenegg like smell is used for the qualetative analysis, if the solubility of $$H_2S$$ in water,at STP is $$0.195 mole.kg^{-1}$$, the Henry's law constant is:


A
282 bar
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B
324.8 bar
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C
462.9 bar
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D
534.8 bar
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Solution

The correct option is D 282 bar
Solubility of $${ H }_{ 2 }S=0.195$$ moles/kg

$$1$$ kg of solvent contains $$=\dfrac { 1000 }{ 18 } =55.55$$ moles

$$\therefore$$   $${ x }_{ { H }_{ 2 }S }=\dfrac { { n }_{ { H }_{ 2 }S } }{ { n }_{ { H }_{ 2 }S }+{ n }_{ { H }_{ 2 }O } } =\dfrac { 0.195 }{ 0.195+55.55 } =\dfrac { 0.195 }{ 55.745 } =0.0035$$

Pressure at $$STP=0.987$$ bar $$=1$$ atm

$${ P }_{ { H }_{ 2 }S }={ K }_{ H }\times { x }_{ { H }_{ 2 }S }$$

$$\therefore$$   $${ K }_{ H }=\dfrac { 0.987\ bar }{ 0.0035 } =282$$ bar

Hence, the correct option is $$A$$

Chemistry

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