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)`
`9.0 g` water is added into `100g` oleum sample labelled as `112%H_(2)SO_(4)` then the amount of free `SO_(3)` remaining in the solution is :
A. `14.93 L` at `STP`
B. `7.46 L` at `STP`
C. `3.78 L` at `STP`
D. `11.2 L` at `STP`

Answer 1

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