Home
Class 11
CHEMISTRY
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)`
`1 g` of oleum sample is diluted with water. The solution required `54 mL` of `0.4 N NaOH` for complete neutralization. The `%` of free `SO_(3)` in the sample is :

A

`74`

B

`26`

C

`20`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
`H_(2)SO_(4)+2NaOH rarr Na_(2)SO_(4)+2H_(2)O`
`{:(,H_(2)SO_(4),+,2NaOHrarrNa_(2)SO_(4)+2H_(2)O),(,(0.4xx54)/(2xx1000),,0.4 N),(,,,54 ml):}`
`wt`. Of `H_(2)SO_(4) = 1.058 gm`
mole of `H_(2)O = (0.0584)/(18)`
`n_(SO_(3)) = 0.003244`
`wt`. of `SO_(3) = 0.26 gm`
`%` free `SO_(3) = 26`
Promotional Banner

Topper's Solved these Questions

  • MOLE CONCEPT

    ALLEN|Exercise Comprehension # 8|3 Videos

  • MOLE CONCEPT

    ALLEN|Exercise Comprehension # 9|4 Videos

  • MOLE CONCEPT

    ALLEN|Exercise Comprehension # 6|5 Videos

  • IUPAC NOMENCLATURE

    ALLEN|Exercise Exercise - 05(B)|8 Videos

  • QUANTUM NUMBER & PERIODIC TABLE

    ALLEN|Exercise J-ADVANCED EXERCISE|26 Videos

Similar Questions

Explore conceptually related problems

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 :

1 g of oleum sample is dilute with water. The solution required 54 mL of 0.4 N NaOH for complete neutralization. Find the percentage of free SO_(3) in the sample?

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) 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) 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 :

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