Oleum is

A

Castor oil

B

Oil of vitriol

C

Fuming of `H_2SO_4`

D

None of them

Text Solution

Verified by Experts

The correct Answer is:
C
Oleum is obtained by dissolving `SO_3` in `H_2SO_4`
`SO_3 +H_2SO_4 (conc.) to underset("Oleum")(H_2S_2O_7)`
Oleum is also called fuming sulphuric acid because it fumes in moist air due to `SO_3`.
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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. 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) ?

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 # 7 Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) . When 100g sample of oleum is diluted with desired weight of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known as % labelling in oleum. For example, a oleum bottle labelled as 109% H_(2)SO_(4) means the 109g total mass of pure H_(2)SO_(4) will be formed when 100g of oleum is diluted by 9g of H_(2)O which combines combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)OrarrH_(2)SO_(4) 9.0 g water is added into 100g oleum sample labelled as 112%H_(2)SO_(4) then the amount of free SO_(3) remaining in the solution is :

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. The percent free SO_(3) is an oleum is 20%. Label the sample of oleum in terms of percent H_(2) SO_(4) .