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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.
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उत्तर
Given: 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 + 55.55)`
= `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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