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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)`
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)

A

`14.93 L` at `STP`

B

`7.46 L` at `STP`

C

`3.78 L` at `STP`

D

`11.2 L` at `STP`

Text Solution

Verified by Experts

The correct Answer is:
C
`%` free `SO_(3) = (212)/(18) xx 80 = (160)/(3)`
`SO_(3)+H_(2)O rarr H_(2)SO_(4)`
`(160)/(3) " "(9)/(18)`
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