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 `"118% H_(2)SO_(4)` labelled oleum, if the number of moles of free `SO_(3)`, number of moles of `H_(2)SO_(4)` and number of moles of `H_(2)O` be `x`, `y` and `z` respectively, then what will be value `x+y+z` ?A. `3.2`B. `3.2`C. `4.2`D. `2.2`

Comprehension # 6 The percentage labelling of oleum is a unique proces

1.	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 `"118% H_(2)SO_(4)` labelled oleum, if the number of moles of free `SO_(3)`, number of moles of `H_(2)SO_(4)` and number of moles of `H_(2)O` be `x`, `y` and `z` respectively, then what will be value `x+y+z` ?A. `3.2`B. `3.2`C. `4.2`D. `2.2`
Answer» Correct Answer - D

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