Calculate the composition of `10 9%` oleum

Explanation

Let the mass of `SO_(3)` in the sample be 'w' g, then the mass of `H_(2)SO_(4)` would be `(100 - w)g`. On dilution, `underset(80g)(SO_(3))+underset(18g)(H_(2)O) rarr H_(2)SO_(4)`
Moles of `SO_(3)` in oleum `=(w)/(80) =`
Moles of `H_(2)SO_(3)` formed after dilution.
`:.` Mass of `H_(2)SO_(4)` formed on dilution `= (98w)/(80)`
Total mass of `H_(2)SO_(4)` present in oleum after dilution
`=(98 w)/(80) +(100 -w) = 109, :. w = 40`
Thus oleum sample contains `40% SO_(3)` and `60% H_(2)SO_(4)`.