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.

Explanation

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