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 SO3 in destroyed by 18 gm water is =18/18x80= 80 gm
Weight of H2SO4 = 400-80= 320 gm present in 400 gm Oleum sample
Weight of H2SO4 newly formed is =18/18x98= 98 gm
Total Weight H2SO4 =320+98=418 gm