Question

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

Answer 1

Correct Answer - 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`