Calculate the `%` of free `SO_(3)` in oleum ( a solution of `SO_(3)` in `H_(2)SO_(4)`) that is labelled `109% H_(2)SO_(4)` by weight.

Explanation

109% oleum represents that 100 g oleum on dilution produces 109 gm `H_(2)SO_(4)`.
`underset(80g)(SO_(3))(g)+H_(2)O(l) rarr underset(98 g)(H_(2)SO_(4))`
Let x gm `SO_(3)` was present in oleum then mass of `H_(2)SO_(4)` would be (100-x) gm.
Mass of `H_(2)SO_(4)` produced on dilution `= (98 xx x)/(80) gm`
Total mass of `H_(2)SO_(4) = (100- x) + (98x)/(80) = 109`
`(98x)/(80) - x = 9`
`18x = 720`
`x = 40`
`:.` % Composition of `SO_(3) =40`