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If 10gm of water4 is added to 150gm of oleum (104.5%), then the find solution:A. Mass of `SO_(3)` left is 10gmB. Mass of `H_(2)SO_(4)` is 156.75gmC. No water will be leftD. Labelling of new solution is 102.25% |
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Answer» 100gm convert into maximum 104.5gm `H_(2)SO_(4)` 150gm convert into pure `(104.5)/(100)xx150=156.75gm H_(2)SO_(4)` and `w_(H_(2)O)` requried `=(4.5)/(100)xx150=6.75gm` we add water `gt6.75gm` thus no `SO_(3)` remain after 10gm water addition labelling of new solution `=100%` as no `SO_(3)` left in it. |
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