What is the percentage of free SO(3) in...

What is the percentage of free `SO_(3)` in an oleum sample that is labelled as '104.5% `H_(2)SO_(4)`?

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Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as ' 019% H_(2)SO_(4) ' means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) What is the % of free SO_(3) in an oleum that is labelled as '104.5% H_(2)SO_(4)' ?

Calculate the % of free SO_(3) in an oleum, that is labelled 118%.

Calculate the percent free SO_(3) in an oleum which is labelled '118% H_(2) SO_(4)' .

The percentage labelling (mixture of H_(2)SO_(4) and SO_(3)) refers to the total mass of pure H_(2)SO_(4) . The total amount of H_(2)SO_(4) found after adding calculated amount of water to 100 g oleum is the percentage labelling of oleum. The higher the percentage lebeling of oleum higher is the amount of free SO_(3) in the oleum sample. What is the amount of free SO_(3) in an oleum sample labelled as '118%'.

Calculate the % of free SO_(3) in oleum ( a solution of SO_(3) in H_(2)SO_(4) ) that is labelled 109% H_(2)SO_(4) by weight.

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