Oleum is mixture of `H_(2)SO_(4)` and `SO_(3)` i.e. `H_(2)S_(2)O_(7)` which is obtained by passing `SO_(3)` is solution of `H_(2)SO_(4)`. In order to dissolve `SO_(3)` in oleum, dilution of oleum is done by water in which oleum is converted into pure `H_(2)SO` as shown below:
`H_(2)SO_(4)+SO_(3)+H_(2)Oto2H_(2)SO_(4)` (pure)
When 100 gm oleum is diluted with water then total mass of diluted oleum is known as percentage labelling in oleum.
For example: `109% H_(2)SO_(4)` labelling of oleum sample means that 109 gm pure `H_(2)SO_(4)` is obtained on diluting 100 gm oleum with 9 gm `H_(2)O` which dissolves al free `SO_(3)` in oleum.
If 109% `H_(2)SO_(4)` labelled oleum, the percent of free `SO_(3)` and `H_(2)SO_(4)` are

Options

A. `30%,70%`
B. `40%,60%`
C. `60%,40%`
D. `15%,85%`

Answer

B

Explanation

109% `H_(2)SO_(4)` labelled oleum
100g oleum + `9g H_(2)O`
`80g-18gimplies40g-9g`
`therefore 100g` oleum contain `40g SO_(3) and 60g H_(2)SO_(4)`
`%SO_(3)=(40)/(100)xx100=40`
`%H_(2)SO_(4)=(60)/(100)xx100=60`