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Calculate the % of free SO(3) in oleum (...

Calculate the `%` of free `SO_(3)` in oleum ( a solution of `SO_(3)` in `H_(2)SO_(4)`) that is labelled `109% H_(2)SO_(4)` by weight.

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109 % `H_(2)SO_(4)` refers to the total mass of pure `H_(2)SO_(4)` i.e ., 109 g that will be formed when 100 g of oleum is diluted by 9 g of `H_(2)O ` which `(H_(2)O)` combines with all the free `SO_(3)` present in oleum to form `H_(2)SO_(4)`
`H_(2)O+SO_(3) to H_(2)SO_(4)`
1 mole of `H_(2)O` combines with 1 mole of `SO_(3)`
or 18 g of `H_(2)O ` combines with 80 g of `SO_(3)`
or 9 g of `H_(2)O ` combines with 40 g of `SO_(3)`
Thus , 100 g of oleum contains 40 g of `SO_(3)` or oleum 40 % of free `SO_(3)`
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Calculate the % of free SO_(3) in an oleum, that is labelled 118%.

Calculate the percent free SO_(3) in an oleum which is labelled '118% H_(2) SO_(4)' .

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

Calculate the normality of solution of 4.9 % (w/v) of H_(2)SO_(4) .

Oleum or fuming sulphuric acid contains SO_(3) dissolved in sulphuric acid and has the molecular formula H_(2)S_(2)O_(7) , It is formed by passing SO_(3) in H_(2)SO_(4) . When water is added to oleum, SO_(3) reacts with water to form H_(2)SO_(4) . SO_(3)(g) + H_(2)O(l) to H_(2)SO_(4)(aq) As a result, mass of H_(2)SO_(4) increases. When 100 g sample of oleum is diluted with desired amount of water (in gram) then the total mass of pure H_(2)SO_(4) obtained after dilution is known as percentage labelling of oleum. % Labelling of oleum = Total mass of H_(2)SO_(4) present in oleum after dilution or = Mass of H_(2)SO_(4) initially present + Mass of H_(2)SO_(4) produced after dilution From this, the percentage composition of H_(2)SO_(4) and SO_(3) (free) and SO_(3) (combined) can be calculated. The percentage of SO_(3) in 109% H_(2)SO_(4) is

Oleum or fuming sulphuric acid contains SO_(3) dissolved in sulphuric acid and has the molecular formula H_(2)S_(2)O_(7) , It is formed by passing SO_(3) in H_(2)SO_(4) . When water is added to oleum, SO_(3) reacts with water to form H_(2)SO_(4) . SO_(3)(g) + H_(2)O(l) to H_(2)SO_(4)(aq) As a result, mass of H_(2)SO_(4) increases. When 100 g sample of oleum is diluted with desired amount of water (in gram) then the total mass of pure H_(2)SO_(4) obtained after dilution is known as percentage labelling of oleum. % Labelling of oleum = Total mass of H_(2)SO_(4) present in oleum after dilution or = Mass of H_(2)SO_(4) initially present + Mass of H_(2)SO_(4) produced after dilution From this, the percentage composition of H_(2)SO_(4) and SO_(3) (free) and SO_(3) (combined) can be calculated. The percentage of free SO_(3) and H_(2)SO_(4) in 112% H_(2)SO_(4) 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. What is the amount of free SO_(3) in an oleum sample labelled as '118%'.

RC MUKHERJEE-VOLUMETRIC CALCULATIONS -OBJECTIVE PROBLEMS
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  2. A solution is defined as a :

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  3. In the reaction, 2S(2)O(3)^(2-)+I(2)rarrS(4)O(6)^(2-)+2I^(-). The eq. ...

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  4. In the reaction VO+Fe(2)O(3) rarr FeO+V(2)O(5), the eq.wt. of V(2)O(5)...

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  5. In the reaction Na(2)CO(3)+HCl toNaHCO(3)+NaCl the eq.wt of Na(2)CO(3)...

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  6. The eq.wt of iodine inI(2)+2S(2)O(3)^(2-) to 2I^(-)+S(4)O(6)^(2-) is e...

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  7. The eq.wt of K(2)CrO(4) as an oxidasing agent in acid medium is

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  8. In alkaline medium , KMnO(4) reacts as follows 2KMnO(4)+2KOH rarr 2K...

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  9. 0.126 g of acid required 20 mL of 0.1 N NaOH for complete neutralisati...

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  10. H(3)PO(4) is a tribasic acid and one of its salts of NaH(2)PO(4). What...

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  11. 2 gm of a base whose equiv wt. is 40 reacts with 3 gm of an acid. The ...

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  12. In a reaction, 4 mole of electrons are transferred to 1 mole of HNO(3)...

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  13. Normality of 1 % H(2)SO(4) solution is nearly

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  14. What volume of 0.1 N HNO(3) solution can be prepared from 6.3 " g of...

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  15. The volume of water to be added to 200 mL of seminormal HCl solution t...

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  16. 0.2 g of a sample of H(2)O(2) required 10 mL of 1 N KMnO(4) in a tit...

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  17. 100 mL of 0.5 N NaOH solution is added to 10 mL of 3 N H(2)SO(4) solut...

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  18. Which of the following has the highest normality ?

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  19. Eq. wt of a metal , x g of which reacts with 1 eq. of an acid is

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  20. The molarity of 98 % H(2)SO(4) (d = 1.8 "g/ml") by wt . is :

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  21. 0.7 g of Na(2)CO(3).xH(2)O is dissolved in 100 mL of water , 20 mL of ...

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