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Question

What is the % of free SO3 in an oleum that is labelled as 104.5% H2SO4 ?

A
10
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B
20
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C
40
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D
50
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Solution

The correct option is B 20
Oleum is mixture of H2SO4 and SO3 gas. When we add water to oleum, SO3 present in it reacts with water to from H2SO4 according to the following equation :
SO3+H2OH2SO4
104.5% labeled oleum means 100 g oleum reacts with H2O to form 104.5g H2SO4. Hence mass of water is 4.5 g.
According to stochiometry coefficient :
18 g water combines with 80 g SO3
4.5 g of H2O combines with 20 g of SO3
100 g of oleum contains 20 g of SO3
Hence 20% of SO3 was free.

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