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. `1.1`
B. 5
C. `0.5`
D. `10.1`

Answer

D

Explanation

`109%` labeled oleum will contain `9g H_(2)O, 40g` free `SO_(3),60g H_(2)SO_(4)`
`40g` (free) `SO_(3) = (1)/(2)` mole `SO_(3)` (free) `=y (0.5)`
`60g(H_(2)SO_(4)) = (60)/(98) =` mole `H_(2)SO_(4) = x (0.61)`
`rArr (x+y)/(x-Y) = 10.09`