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An oleum sample contains 10 g SO(3) and ...

An oleum sample contains `10 g SO_(3)` and `15 g H_(2) SO_(4)`
Answer the following questions on the basis of above information :
`%` labelling of oleum sample is .

A

`27.25%`

B

`106%`

C

`109%`

D

`118%`

Text Solution

Verified by Experts

The correct Answer is:
(C )
`%` labelling of oleum sample `= (100 + x) %`
`{:(SO_(3),+,H_(2)O,rarr,H_(2)SO_(4)),(10g,,,,),(1//8mol,,1//8mol,,):}`
`2.25g`
`:. %` labelling `= (100 + 9) % = 109 %` .
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Knowledge Check

  • An oleum sample contains 10 g SO_(3) and 15 g H_(2) SO_(4) Answer the following questions on the basis of above information : Find new % labeling of 0.45g of H_(2)O is added to the above oleum sample

    A
    `100%`
    B
    `102.83%`
    C
    `107.07%`
    D
    `109%`

  • An oleum sample has SO_(3) and H_(2)SO_(4) in 2:3 mass ratio. Select the correct statement(s).

    A
    % labelling of sample is 109%
    B
    % labelling of sample is 118%
    C
    If 9 gm `H_(2)O` is added to200 gm of above sample, new labelling would be 104.5%
    D
    If 9 gm `H_(2)O` is added to 200 gm of above sample, new labelling would be 104.3%

  • A 110% sample of oleum contains

    A
    44.4% of `SO_(3)`
    B
    55.6% of sulphuric acid
    C
    55.6% of `SO_(3)`
    D
    44.4% of sulphuric acid

    • Similar Questions

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

    A 50 gm oleum sample contains (400/49) gm of combined SO_(3). Find percent label of the oleum sample.

    A mixutre is prepared by mixing 10 gm H_(2)SO_(4) and 40 gm SO_(3) calculate, (a) mole fraction of H_(2)SO_(4) (b) % labelling of oleum

    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)' ?

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