Find out the % labelling of new oleum sample obtained by mixing of 4.5 gm of water in 100 gm of 109% labelled oleum sample ?.

SOLUTION: Wt of SO3 in Original Oleum

The amount of free SO₃ destroyed by 4.5 gm water is added 

The amount of free SO₃ destroyed by 4.5 gm water is = x80=20 gm

Wt of left SO3 =40-20=20 gm

               Given Wt of (Free) SO₃= 20 % Find % labelling (X)

             % labeling(X) = 104.5 %