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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.
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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