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Calculate the percent free SO(3) in an o...

Calculate the percent free `SO_(3)` in an oleum which is labelled `'118% H_(2) SO_(4)'`.

Text Solution

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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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