1.

H_2S , a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H_2S in water at STP is 0.195 m, calculate Henry's law constant

Answer»

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Solution :Solubility of `H_2S`gas = 0.195 m i.e., 0.195 MOLE in 1 kg of the solvent (water)
1 kg of the solvent (water) = 1000 g = 1000/18g = 55.55 moles
` therefore ` Mole fraction of `H_2S`gas in the solution (X) = `(0.195)/(0.195 + 55.55) = 0.0035`
Pressure at STP = 0.987 bar
Applying Henry.s law,
`p_(H_2S) = K_H xx x_(H_2S) " or" K_H = (p_(H_2S))/(x_(H_2S)) = (0.987 bar)/(0.0035) = 282 bar `


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