Find the % labelling of 100 gm oleum sample if it contains 20 gm `SO_(3)`

Find the % labeiling of 100gm oleum sample if it contains 20gm SO_(3) .

Find the % labelling of 100g oleum sample containing 10g SO_(3)

A 110% sample of oleum contains-

The percentage labelling (mixture of H_(2)SO_(4) and SO_(3)) refers to the total mass of pure H_(2)SO_(4) . The total amount of H_(2)SO_(4) found after adding calculated amount of water to 100 g oleum is the percentage labelling of oleum. The higher the percentage lebeling of oleum higher is the amount of free SO_(3) in the oleum sample. What is the amount of free SO_(3) in an oleum sample labelled as '118%'.

The percentage labelling (mixture of H_(2)SO_(4) and SO_(3)) refers to the total mass of pure H_(2)SO_(4) . The total amount of H_(2)SO_(4) found after adding calculated amount of water to 100 g oleum is the percentage labelling of oleum. Higher the percentage lebeling of oleum higher is the amount of free SO_(3) in the oleum sample. What is the amount of free SO_(3) in an oleum sample labelled as '118%'.

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 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 the number of moles of free SO_(3), H_(2)SO_(4) , and H_(2)O be x, y and z respectively in 118% H_(2)SO_(4) labelled oleum, the value of (x+y+z) is