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`25.0` litre of natural gas measured at `25^(@)C` and `740 mm` of `Hg` is bubbled through `Pb_((aq))^(2+)` to give `0.535 g` of solid residue. If natural gas contains `H_(2)S`, the only component responsible for the formation of solid residue, calculate the volume `%` of `H_(2)S`, in natural gas. |
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Answer» `underset("Solid residue")(Pb^(2+)+H_(2)srarr PbS+2H^(+))` Eq.of `H_(2)S=` Eq. of `PbS` `(w)/(32//2)=(0.535)/(240//2)` `:. w_(H_(2)S)=7.58xx10^(-2)` `:.` Moles of `S=(7.85xx10^(-2))/(34)=2.23x10^(-3)` `:. V_(H_(2)s)=(nRT)/(P)` `=(2.23xx10^(-3)xx0.0821xx298xx760)/(740)` `5.6xx10^(-2)`litre `:. % Vol=(5.6xx10^(2))/(25)xx100= 0.224` |
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