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Question

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 weight of H2O then the total mass of H2SO4 obtained after dilution is known as percentage labelling of oleum. For example , a oleum lablled as 109% H2SO4 means 109g total mass of pure H2SO4 will be formed when 100g of oleum is diluted by 9 g of H2O which combined with all the free SO3 to form H2SO4 as SO3+H2OH2SO4.
1 g of oleum sample is diluted with water. The solution require 54 ml of 0.4 N NaOH for completed neutralization for the % of free SO3 in the sample is:

A
74
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B
26
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C
20
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D
None of these
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Solution

The correct option is B 26
equivalent of H2SO4 + equivalent of SO3= equivalent of NaOH

x98×2+(1x)×280=54×0.4×103

x=0.74

% of free SO3=10.741×100=26%

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