Oleum is a mixture of

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 is the percentage of free SO_(3) and H_(2)SO_(4) in the oleum simple respectively ?

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?

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) ?

A recemic mixture is a mixture of

The fusion mixture is a mixture of

Similar to % labelling of oleum ,a mixture of H_(3)PO_(3) and P_(4)O_(6) is labelled as (100 +x)% where x is maximum mass of water which can reacts with P_(4)O_(6) present in 100 gm mixture of H_(3)PO_(3) and P_(4)O_(6) P_(4)O_(6) + H_(2)O rarr H_(3)PO_(3) For such a mixture of P_(4)O_(6) and H_(3)PO_(3) labelled as (100 +x)% . Value of x can lie in range of (maximum and minimum) :

Freezing mixture is a mixture