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0.5 g of fuming sulphuric acid (H2SO4+SO...

0.5 g of fuming sulphuric acid `(H_2SO_4+SO_3)`, called oleum, is diluted with water. Thus solution completely neutralised 26.7 " mL of " 0.4 M `NaOH`. Find the percentage of free `SO_3` in the sample solution.

A

`30.6%`

B

`40.6%`

C

`20.6 %`

D

`50%`

Text Solution

Verified by Experts

The correct Answer is:
C
Meq. of `H_(2)SO_(4) + " Meq. of " SO_(3) = ` Meq.of NaOH
` :. ((0.5 -x))/(98//2) xx 1000 + x/(80//2) xx 1000 = 26.7 xx 0.4`
`:. x = 0.103`
`:. % " of " SO_(3) = (0.103)/(0.5) xx 100 = 20.6 %`
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0.5 gm of fuming H_(2)SO_(4) (Oleum) is diluted with water. This solution is completely neutralised by 26.7 ml of 0.4 M NaOH solution. Calculate the percentage of free SO_(3) in the given sample. Give your answer excluding the decimal places.

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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 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as ' 019% H_(2)SO_(4) ' means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) 1 g of oleum sample is diluted with water. The solution required 54 mL of 0.4 N NaOH for complete neutralization. The % free SO_(3) in the sample is :

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