Home
Class 11
CHEMISTRY
Oleum is considered as a solution of SO(...

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 '`109% 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

`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`
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|3 Videos

  • IUPAC NOMENCLATURE

    ALLEN|Exercise Exercise - 05(B)|7 Videos

  • QUANTUM NUMBER & PERIODIC TABLE

    ALLEN|Exercise J-ADVANCED EXERCISE|24 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 109% 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 ' 109% 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 : (a)74 (b)26 (c)20 (d)None of these

What mass of H_(2)SO_(4) contains 32g oxygen ? .

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

H_(4)underline(S_(2))O_(6)+H_(2)O to H_(2)SO_(3)+H_(2)SO_(4)

H_(4)underline(S_(2))O_(6)+H_(2)O to H_(2)SO_(3)+H_(2)SO_(4)

Balance the following equation: C+H_(2)SO_(4)toCO_(2)+H_(2)O+SO_(2)

Oleum is mixture of H_(2)SO_(4) and SO_(3) i.e. H_(2)S_(2)O_(7) which is obtained by passing SO_(3) is solution of H_(2)SO_(4) . In order to dissolve SO_(3) in oleum, dilution of oleum is done by water in which oleum is converted into pure H_(2)SO as shown below: H_(2)SO_(4)+SO_(3)+H_(2)Oto2H_(2)SO_(4) (pure) When 100 gm oleum is diluted with water then total mass of diluted oleum is known as percentage labelling in oleum. For example: 109% H_(2)SO_(4) labelling of oleum sample means that 109 gm pure H_(2)SO_(4) is obtained on diluting 100 gm oleum with 9 gm H_(2)O which dissolves al free SO_(3) in oleum. If the number of moles of free SO_(3), H_(2)SO_(4) , and H_(2)O be x, y and z respectively in 118% H_(2)SO_(4) labelled oleum, the value of (x+y+z) is

The oxide that give H_(2)O_(2) on treatment with dilute H_(2)SO_(4) is