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.
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Oleum consists of `SO_(3) " and " H_(2)SO_(4)`. Let the mass of `SO_(3)` in the given sample of oleum be=x g Mass of `H_(2)SO_(4)` in the given sample of oleum `=(0.5-x)g` Eq. mass of `SO_(3)=(80)/(2)=40` No. of g equivalents of `SO_(3)=(x)/(40)` `[2NaOH+SO_(3)to Na_(2)SO_(4)+H_(2)O` `2NaOH+H_(2)SO_(4) to Na_(2)SO_(4)+2H_(2)O]` Eq. mass of `H_(2)SO_(4)=(98)/(2)=49` No. of g equivalents of `H_(2)SO_(4)=((0.5-x))/(49)` Total no. of a equivalents `=(x)/(40)+((0.5-x))/(49)` 26.7 mL of 0.4 N NaOH contain no. of equivalents of NaOH `=(0.4)/(1000)xx26.7` At equivalence point, No. of g equivalents of `NaOH=(x)/(40)+((0.5-x))/(49)` So, `(0.4xx26.7)/(1000)=(49x+(40xx0.5-40x))/(40xx49)` `x=(0.9328)/(9)=0.1036` Hence, % of free `SO_(3)=(0.1036)/(0.5)xx100` =20.72
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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 :
