100 gm of 120% labelled Oleum is diluted with 15 gm of water. determined the new % labelling of Oleum ?.

SOLUTION: :   Wt of SO₃ in Original Oleum

The amount of free SO3 destroyed by 15 gm water is added

                  % labelling(X) = 105 %

The amount of free SO3 destroyed by 4.5 gm water=15/18 x 80=66.66 gm

Wt of left SO3 =88.88-66.66=22.22 gm

Given Wt of (Free) SO₃= 22.22 % Find % labelling (X)

% labelling(X) = 105 %