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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 '`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

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`
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