The mass precent of oxgen in `109%` oleum is .

The mass precent of oxYgen in 109% oleum is .

The mass percent of oxgen in 109% oleum is .

Calculate the mass precent of a solution obtine by mixing 100 g of 10% urea solution with 150 g of 15% urea solution.

Calculate the mass precent of different elements present in sodium sulphate (Na_(2)SO_(4)) .

Calculate the mass precent of different elements present in sodium sulphate (Na_(2)SO_(4)) .

If the percent free SO_(3) in an oleum is 20% then label the sample of oleum in terms of percent H_(2) SO_(4) ,

If the percent free SO_(3) in an oleum is 20% then label the sample of oleum in terms of percent H_(2) SO_(4) ,

If the percent free SO_(3) in an oleum is 20% then label the sample of oleum in terms of percent H_(2) SO_(4) ,

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) ?

What is the percentage of free SO_(3) in an oleum sample that is labelled as '104.5% H_(2)SO_(4) ?