1.

H_(2)S, a toxis gas with rotten egg like small, is used for qualitative analysis. If the solubility of H_(2)S in water at STP is 0.195 m, calculate Henry's law constant.

Answer»

Solution :Solubility of `H_(2)S` gas = 0.195 m = 0.195 mole in 1 kg of the solvent (water )
`"1 kg of the solvent (water) = 1000 g "=(1000g)/("18 g mol"^(-1))="55.55 moles"`
`THEREFORE"Mole fraction of "H_(2)S" gas in the solution (x)"=(0.195)/(0.195+55.55)=(0.195)/(55.745)=0.0035`
Pressure at STP = 0.987 bar
`"Applying Henry's LAW, "p_(H_(2)S)=K_(H)xxx_(H_(2)S) or K_(H)=(p_(H_(2)S))/(x_(H_(2)S))=("0.987 bar")/("0.0035")="282 bar."`


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