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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 %

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