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)'` ?

Options

A. 10
B. 20
C. 40
D. None of these

Answer

B

Explanation

`SO_3+H_2OtoH_2SO_4` ,
18 g water combines with 80 g `SO_3`
`:.` 4.5 g of `H_2O` combines with 20 g of `SO_3`
`:.` 100 g of oleum contains 20 g of `SO_3` or 20 % free `SO_3`