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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 labelling of oleum. For example , a oleum is diluted by 9 g of H2O which combined with all the free SO3 to form H2SO4 as SO3+H2OH2SO4.


In excess water is added into a 100 g bottle sample labelled as 112 % H2SO4 and is reacted with 5.3 g Na2CO3, then find the volume of CO2 evolved at 1 atm pressure and 300 K temperature after the completion of the reaction. [R=0.0821Latmmol1k1].

H2SO4+Na2CO3Na2SO4+H2O+CO2

A
2.46 L
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B
24.6 L
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C
1.23 L
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D
12.3 L
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Solution

The correct option is C 1.23 L
H2SO4+Na2CO3Na2SO4+H2O+CO2

Moles of CO2 formed =moles of Na2CO3 reacted ( is is limiting reagent )= 5.3106=0.05


Volumes of CO2 formed at 1 atm pressure and 300K=0.05×24.63=1.23L

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