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Comprehension # 7 Oleum is considered ...

Comprehension # 7
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 `100g` sample of oleum is diluted with desired weight of `H_(2)O` then the total mass of `H_(2)SO_(4)` obtained after dilution is known as `%` labelling in oleum.
For example, a oleum bottle labelled as `109% H_(2)SO_(4)` means the `109g ` total mass of pure `H_(2)SO_(4)` will be formed when `100g` of oleum is diluted by `9g` of `H_(2)O` which combines combines with all the free `SO_(3)` present in oleum to form `H_(2)SO_(4)` as `SO_(3)+H_(2)OrarrH_(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 Na_(2)CO_(3)` then find the volume of `CO_(2)` evloved at `1atm` pressure and `300 K` temperature after the completion of the reaction

A

`2.46 L`

B

`24.6 L`

C

`1.23 L`

D

`12.3 L`

Text Solution

Verified by Experts

The correct Answer is:
C
`H_(2)SO_(4)+Na_(2)CO_(3)rarrNa_(2)SO_(4)+2H_(2)O+CO_(2)`
`{:((112)/(98),(5.3)/(106)," "0.05),(1.1428,0.05,):}`
`V_(co^(2)) = 0.05xx24.63 = 1.23l`
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Comprehension # 7 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 100g sample of oleum is diluted with desired weight of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known as % labelling in oleum. For example, a oleum bottle labelled as 109% H_(2)SO_(4) means the 109g total mass of pure H_(2)SO_(4) will be formed when 100g of oleum is diluted by 9g of H_(2)O which combines combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)OrarrH_(2)SO_(4) 9.0 g water is added into 100g oleum sample labelled as 112%H_(2)SO_(4) then the amount of free SO_(3) remaining in the solution is :

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) 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) 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 :

Find the mass of Free SO_(3) present in 100gm, 109% oleum sample

What is the percentage of free SO_(3) in an oleum sample that is labelled as '104.5% H_(2)SO_(4) ?

A sample of oleum is labelled 109% . The % of free SO_(3) in the sample is