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Statement 109 % H(2)SO(4) represent a wa...

Statement `109 % H_(2)SO_(4)` represent a way to express concentration of industrial `H_(2)SO_(4)`.
Explanation It represents that `9 g` ` H_(2)O` reacts with `40 g` ` SO_(3)` to produce `49 g ` `H_(2)SO_(4)` in addition to `100 g` ` H_(2)SO_(4)`.

A

Statement -1 is True, Statement -2 is True, Statement -2 is a correct

B

Statement -1 is True, Statement -2 is True, statement -2 is NOT a correct explanation for Statement-10

C

Statement-1 is True, Statement-2 is False

D

Statement-1 is False, Statement-2 is True

Text Solution

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The correct Answer is:
B
`109%` oleum represents the amount of free `SO_(3)` present.
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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)' ?

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) If excess water is added into a bottle sample labelled as "112% H_(2)SO_(4) " and is reacted with 5.3 g NaCO_(3) then find the volume of CO_(2) evolved at 1 atm pressure and 300 K temperature after the completion of the reaction :

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) 9.0 g water is added into oleum sample lablled as "112%" H_(2)SO_(4) then the amount of free SO_(3) remaining in the solution is : (STP=1 atm and 273 K)

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) 1 g of oleum sample is diluted with water. The solution required 54 mL of 0.4 N NaOH for complete neutralization. The % free SO_(3) in the sample is :

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