Oleum is

A

castor Oil

B

oil of vitriol

C

fuming `H_(2)SO_(4)`

D

none of them

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**Step-by-Step Solution:** 1. **Understanding Oleum**: Oleum is a solution of sulfur trioxide (SO3) in sulfuric acid (H2SO4). It is often referred to as "fuming sulfuric acid" because it releases fumes of sulfur trioxide when exposed to air. 2. **Identifying the Components**: The main components of oleum are concentrated sulfuric acid and sulfur trioxide. When sulfur trioxide is dissolved in concentrated sulfuric acid, it forms oleum. 3. **Analyzing the Options**: - **Option A: Castor Oil** - This is a type of vegetable oil and is not related to oleum. - **Option B: Oil of Vitriol** - This term is often used to refer to sulfuric acid, but it does not specifically denote oleum. - **Option C: Fuming H2SO4** - This is indeed another name for oleum, as it describes the fuming nature of the solution when SO3 is present. - **Option D: None of them** - Since we have identified option C as correct, this option is incorrect. 4. **Conclusion**: The correct answer is **Option C: Fuming H2SO4**. Oleum is essentially a mixture of sulfur trioxide in concentrated sulfuric acid, which gives it the fuming characteristic.

**Step-by-Step Solution:** 1. **Understanding Oleum**: Oleum is a solution of sulfur trioxide (SO3) in sulfuric acid (H2SO4). It is often referred to as "fuming sulfuric acid" because it releases fumes of sulfur trioxide when exposed to air. 2. **Identifying the Components**: The main components of oleum are concentrated sulfuric acid and sulfur trioxide. When sulfur trioxide is dissolved in concentrated sulfuric acid, it forms oleum. 3. **Analyzing the Options**: - **Option A: Castor Oil** - This is a type of vegetable oil and is not related to oleum. ...
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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) .

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%'.

DISHA PUBLICATION-THE P-BLOCK ELEMENTS (GROUP 15, 16, 17 & 18)-Exercise
  1. it is possible to obtain oxygen from air by fractional distillation be...

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  2. The number of S-S bonds in SO(3), S(2)O(3)^(2-), S(2)O(6)^(2-) and S(2...

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  3. Oleum is

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  4. By passing H(2)S gas in acidified KMnO(4) solution, we get

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  5. Oxidation of thiosulphate by iodine gives

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  6. When PbO(2) reacts with conc. HNO(3) the gas evolved is

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  7. The compound which gives off oxygen on moderate heating is:

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  8. Oxidation state exhibited by sulphur is

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  9. On electrolysis of dilute sulphuric acid using platinum electrodes, th...

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  10. A gas that cannot be collected over water is.

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  11. Sodium thiosulphate is prepared by

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  12. Which of the following is not oxidized by O(3)?

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  13. Which one of the following compounds is a peroxide?

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  14. Sulphur trioxide can be obtained by which of the following reactions:

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  15. Which of the following does not give oxygen on heating?

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  16. Acidity of diprotic acids in aqueous solutions increases in the order

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  17. In the reaction: 3Br(2)+6OH^(Theta)rarr 5Br^(Theta)+BrO(3)^(Theta)+3H...

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  18. In the case of alkali metals, the covalent character decreases in the ...

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  19. Potassium chlorate on heating with cone H2SO4 gives

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  20. In the manufacture of bromine from sea water the mother liquor contain...

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