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.

Answer

3.84

Explanation

`N_(1)=1, V_(1)=? , N_(2)=26.7, V_(2)=0.4`
`N_(1)V_(1)=N_(2)V_(2)`
`I xx V_(1) = 26.7 xx 0.4`
`V_(1) =(26.7 xx 0.4)/1 = 10.68`
49 g (`therefore` eq. wt of `H_(2)SO_(4) = 49`) of `H_(2)SO_(4)` will be neutralised by 1 N 1000 mL of NaOH.
`SO_() + 2NaOH to Na_(2)SO+_(4) + H_(2)O`
`therefore` Eq. wt of `SO_(3) =("mol wt")/2 = 80/2 =40`
Wt. of `SO_(3)` in 0.48 ml of 1 M solution.
`=40/1000 xx 0.48 = 0.0192 g`
% of `SO_(3) = 0.0192/0.5 xx 100 = 3.84%`