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

`Ew(SO_3)=(80)/(2)=40[SO_3+H_2OtoH_2SO_4,(n=2)]`
Ew of `H_2SO_4=(98)/(2)=49`
Let x g of `SO_3` and `(0.5-x)g of H_2SO_4` be present
m" Eq of "`SO_3+m" Eq of "H_2SO_4=mEq` of NaOH
`((x)/(40)+(0.5-x)/(49))xx10^(3)mEq=26.7xx0.4xx1mEq`
`thereforex=0.103g`
`%` free `SO_3=(0.103)/(0.5)xx100=20.6%`