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Question

4.5 g water is added into an oleum sample labelled as 109% H2SO4. What is the amount of free SO3 remaining in the solution?

A
10 g
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B
12 g
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C
20 g
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D
24 g
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Solution

The correct option is C 20 g
109% H2SO4 refers to the total mass of pure H2SO4 i.e 109 g that 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
Moles of H2O present=918=0.5 mol
Moles of H2O added=4.518=0.25 mol

SO3+H2OH2SO4
Moles of SO3 Moles of H2O
Moles of free SO3 remaining =0.50.25=0.25 mol
Amount of free SO3 remaining=0.25×80=20 g

Theory:
Strength of Oleum :
Percentage Labelling of Oleum :
1. Oleum is a mixture of H2SO4 and SO3 i.e., H2S2O7
2. This is obtained by passing SO3 in solution of H2SO4
3. To dissolve free SO3, addition of water is done.
Dilution is continued till the entire SO3 gets converted into H2SO4
H2SO4(aq)+SO3(g)+H2O(l)2H2SO4(aq)
OleumH2SO4(aq)+SO3(g)+H2O(l)

Percentage labelling of Oleum :

When a 100 g sample of oleum is diluted with the desired weight of H2O (in g), then the total mass of H2SO4 obtained after dilution is known as the % labelling in oleum.


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