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Calculate the % of free `SO_(3)` in an oleum, that is labelled 118%. |
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Answer» Oleum `=H_(2)SO_(4)+SO_(3)=H_(2)S_(2)O_(7)` If initial weight of labelled `H_(2)S_(2)O_(7)=100 g` Weight of `H_(2)SO_(4)`, after dilution = 118 g Wt. of `H_(2)O=18 g` Moles of `H_(2)O` = moles of `SO_(3)=(18)/(18)=1` Wt. of `SO_(3)=80` `therefore %` of free `SO_(3)=("Wt. of "SO_(3)xx100)/(100)=80%` |
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