Comprehension # 6 The percentage labelling of oleum is a unique process by means of which,

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. In the above question number `1`, what will be the percentage of combined `SO_(3)` in the given oleum sample?