`0.5` g of fuming `H_(2)SO_(4)` (oleum ) is diluted with water . The solution is completely neutralised by `26.7` mL of 0.4 N naOH .Find the percentage of free `SO_(3)` in the sample of oleum.

Explanation

Reactions involved are
`SO_(3) + 2NaOH to Na_(2)SO_(4) +H_(2)O `
`H_(2)SO_(4) +2NaOH to Na_(2)SO_(4) +2H_(2)O `
Eq. wt of `SO_(3) = 80/2 = 40 `
Eq. wt of `H_(2)SO_(4) = 98/2 = 49 `
Now m.e of `SO_(3)`+ m.e of `H_(2)SO_(4)` = m.e of NaOH
` :. " eq of " SO_(3) xx 1000 eq . of H_(2)SO_(4) xx 1000 = 0.4 xx 26.7 `
Suppose the wt. of `SO_(3) ` is x g .
` :. ` wt of `H_(2)SO_(4) = (0.5 - x) g `
` :. x/40 +((0.5 - x))/49 = 0.4 xx 26.7 x = 0.1036 g `
` :. " " % of SO_(3) = (0.1036)/(0.5) xx 100 = 20.67 % `