SOLUTION: 104.5 % labelled means 100 Oleum sample required 4.5 gm water to completely destroyed free SO3 present in 100 gm sample

100 gm Oleum sample require 4.5 gm water to destroyed all free SO3

Weight of SO

_{3 }in destroyed by 18 gm water is =18/18x80= 80 gm
Weight of H

_{2}SO_{4 }= 400-80= 320 gm present in 400 gm Oleum sample
Weight of H

_{2}SO_{4}newly formed is =18/18x98= 98 gm
Total Weight H

_{2}SO_{4 }=320+98=418 gm