Oleum is mixture of H_(2)SO_(4) and SO_(3) i.e. H_(2)S_(2)O_(7) which is obtained by passing SO_(3) is solution of H_(2)SO_(4).In order to dissolve SO_(3) in oleum, dilution of oleum is done by water in which oleum is converted into pure H_(2)SO as shown below: H_(2)SO_(4)+SO_(3)+H_(2)Oto2H_(2)SO_(4) (pure) When 100 gm oleum is diluted with water then total mass of diluted oleum is known as percentage labelling in oleum. For example: 109% H_(2)SO_(4) labelling of oleum sample means that 109 gm pure H_(2)SO_(4) is obtained on diluting 100 gm oleum with 9 gm H_(2)O which dissolves al free SO_(3) in oleum. If 109% H_(2)SO_(4) labelled oleum, the percent of free SO_(3) and H_(2)SO_(4) are

Oleum is mixture of H_(2)SO_(4) and SO_(3) i.e. H_(2)S_(2)O_(7) which

1.	Oleum is mixture of H_(2)SO_(4) and SO_(3) i.e. H_(2)S_(2)O_(7) which is obtained by passing SO_(3) is solution of H_(2)SO_(4).In order to dissolve SO_(3) in oleum, dilution of oleum is done by water in which oleum is converted into pure H_(2)SO as shown below: H_(2)SO_(4)+SO_(3)+H_(2)Oto2H_(2)SO_(4) (pure) When 100 gm oleum is diluted with water then total mass of diluted oleum is known as percentage labelling in oleum. For example: 109% H_(2)SO_(4) labelling of oleum sample means that 109 gm pure H_(2)SO_(4) is obtained on diluting 100 gm oleum with 9 gm H_(2)O which dissolves al free SO_(3) in oleum. If 109% H_(2)SO_(4) labelled oleum, the percent of free SO_(3) and H_(2)SO_(4) are
Answer» `30%,70%` `40%,60%` `60%,40%` `15%,85%` Solution :109% `H_(2)SO_(4)` LABELLED OLEUM 100g oleum + `9G H_(2)O` `80g-18gimplies40g-9g` `therefore 100g` oleum contain `40G SO_(3) and 60g H_(2)SO_(4)` `%SO_(3)=(40)/(100)xx100=40` `%H_(2)SO_(4)=(60)/(100)xx100=60`