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H(2)S, a toxic gas with rotten egg like ...

`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 0.195m, calculate Henry's law constant.

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a) Calculation of mole fraction of `H_(2)S` 0.195 m means that 0.195 mole of `H_(2)S` are dissolves in 1000 g of water.
Number of moles of water in 1000g, `(n_(H_(2)O))=((1000"g"))/((18" g mol"^(-1)))=55.55" mol"`
Mole fraction `H_(2)S(x_(H_(2)S))=(n_(H_(2)S))/(n_(H_(2)S)+n_(H_(2)O))`
`=((0.195" mol"))/((0.195+55.55)" mol")=((0.195" mol"))/((55.745" mol"))=0.0035`
b) Calculation of Henry's law constant :
According to Henry's law
`x_(H_(2)S)=("Partial pressure of "H_(2)S)/(K_(H)" for "H_(2)S)" at STP"`
`K_(H)" for "H_(2)S=("Partial pressure of "H_(2)S)/(x_(H_(2)S))=((0.987"bar"))/((0.0035))=282" bar"`
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