If the percent free `SO_(3)` in an oleum is 20% then label the sample of oleum in terms of percent `H_(2) SO_(4)`,

Explanation

Let `x g H_(2) O` combine with all the free `SO_(3)` in `100 g` of the oleum to give a total `(100 + x)g H_(2) SO_(4)`.
`18 g H_(2) O` combines with `80 g SO_(3)`
`x g H_(2) O` combines with `= (x)/(18) xx 80 = 20 g SO_(3)`
`:.x = (20 xx 18)/(80) = 4.5 g`
Labelling of oleum in terms of percent `H_(2) SO_(4)`
`= (100 + x) = (100 + 4.5) = 1.4.5% H_(2) SO_(4)`