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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

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

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प्रश्न

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

संख्यात्मक
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उत्तर

Solubility of H2S gas = 0.195 m

∴ Moles of H2S = 0.195 mol

Mass of water = 1000 g

No. of moles of water = `(1000  "g")/(18  "g mol"^(-1))`

= 55.55 mol

∴ Mole fraction of H2S gas in the solution (x) = `0.195/(0.195+55.56)`

= `0.195/55.745`

= 0.0035

Pressure at STP = 1 bar

According to Henry’s law,

`"p"_("H"_2"S") = "K"_"H" xx "x"_("H"_2"S")`

or, KH = `("p"_("H"_2"S"))/("x"_("H"_2"S"))`

= `(1  "bar")/(0.0035)`

= 285.7 bar

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अध्याय 1: Solutions - Intext Questions [पृष्ठ ९]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
अध्याय 1 Solutions
Intext Questions | Q 1.6 | पृष्ठ ९

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