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Oleum is considered as a solution of SO3 in H2SO4, which is obtained by passing SO3 in solution of H2SO4. When 100 g sample of oleum is diluted with desired mass of H2O then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum.
For example, a oleum bottle labelled as '109 % H2SO4' means the 109 g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of H2O which combines with all the free SO3 present in oleum to form H2SO4 as SO3+H2OH2SO4.

What is the % of free SO3 in  an oleum that is labelled as '104.5 % H2SO4'?
  1. 10
  2. 20
  3. 40
  4. None of the above


Solution

The correct option is B 20
For the reaction,
SO3+H2OH2SO4
from stoichiometry,
18 g water combines with 80 g SO3
4.5 g water combines with 20 g SO3
100 g oleum contains 20 g SO3
or 20% free SO3

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