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Oleum is considered as a solution of SO(...

Oleum is considered as a solution of `SO_(3)` in `H_(2)SO_(4)`, which is obtained by passing `SO_(3)` in solution of `H_(2)SO_(4)` When 100 g sample of oleum is diluted with desired mass of `H_(2)O` then the total mass of `H_(2)SO_(4)` obtained after dilution is known is known as % labelling in oleum.
For example, a oleum bottle labelled as '`019% H_(2)SO_(4)`' means the 109 g total mass of pure `H_(2)SO_(4)` will be formed when 100 g of oleum is diluted by 9 g of `H_(2)O` which combines with all the free `SO_(3)` present in oleum to form `H_(2)SO_(4)` as `SO_(3)+H_(2)O to H_(2)SO_(4)`
What is the % of free `SO_(3)` in an oleum that is labelled as `'104.5% H_(2)SO_(4)'` ?

A

`10`

B

`20`

C

`40`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
`%` free `SO_(3) = (4.5)/(18) xx 80 = 20`
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