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H_(2)S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H_(2)S in water at STP is 0.195 m, calculate Henry's law constant. |
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Answer» Solution :It is given that the solubility of `H_(2)S` in water at STP is 0.195 m, i.e., 0.195 mol of `H_(2)S` is dissolved in 1000 G of water. Moles of water `= (1000)/(18)=55.56` mol `therefore` Mole fraction of `G_(2)S, X_(H_(2)O)` `= ("Moles of "H_(2)S)/("Moles of "H_(2)S+"Moles of water")` `= (0.195)/(0.195+55.56)=0.0035` At STP, pressure (p) = 0.987 BAR ACCORDING to Henry.s law : `p=K_(H).x` `K_(H)=(p)/(x)=(0.987)/(0.0035)=282` bar. |
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