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.
For `109%` labelled oleum if the number of moles of `H_(2)SO_(4)` and free `SO_(3)` be `x` and `y` respectively, then what will be the value of `(x+y)/(x-y)` ?

Options

A. `18`
B. `18`
C. `(1)/(3)`
D. `9.9`

Answer

D

Explanation

`x=(60)/(98) " "y=(40)/(80) " "implies (x+y)/(x-y)=9.9`