Comprehension # 6
The percentage labelling of oleum is a unique process by means of which, the percentage composition of `H_(2)SO_(4), SO_(3)` (free) and `SO_(3)` (combined) is calculated.
Oleum is nothing but it is a mixture of `H_(2)SO_(4)` and `SO_(3)` i.e., `H_(2)S_(2)O_(7)`, which is obtained by passing. `SO_(3)` in solution of `H_(2)SO_(4)`. In order of dissolve free `SO_(3)` in oleum, dilution of oleum is done, in which oleum converts into pure `H_(2)SO_(4)`. It is shown by the reaction as under :
`H_(2)SO_(4)+SO_(3)+H_(2)Orarr2H_(2)SO_(4)("pure")`
or " "`SO_(3)+H_(2)OrarrH_(2)SO_(4)("pure")`
When `100g` sample of oleum is diluted with desired weight of `H_(2)O("in" g)`, then the total mass of pure `H_(2)SO_(4)` obtained after dilution is known as percentage labelling in oleum.
For example, if the oleum sample is labelled as `""109%H_(2)SO_(4)"` it means that `100 g` of oleum on dilution with `9m` of `H_(2)O` provides `109g` pure `H_(2)SO_(4)`, in which all free `SO_(2)` in `100g` of oleum is dissolved.
What volume of `10 M NaOH ("in" mL)` will be required to react completely with free `SO_(3)` in `118%` labelled oleum sample?

Options

A. `500 mL`
B. `200 mL`
C. `400 mL`
D. `600 mL`

Answer

B