A 50 gm oleum sample contains `(400/49)` gm of combined `SO_(3).` Find percent label of the oleum sample.

25 gm of an oleum sample contains 15 gm H_(2)SO_(4) . What is % labellling of this oleum sample-

A 110% sample of oleum contains-

Find the mass of Free SO_(3) present in 100gm, 109% oleum sample

Find the mass of Free SO_(3) present in 100gm, 109% oleum sample

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

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

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. In the above question number 1 , what will be the percentage of combined SO_(3) in the given oleum sample?