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.

Options

A. (a)30.6%
B. (b)40.6%
C. (c )20.6%
D. (d)50%

Answer

c

Explanation

Meq. of `H_(2)SO_(4)+ " Meq. Of " SO_(3) =" Meq. Of " NaOH`
`:. ((0.5-x))/(98//2)xx1000+x/(80//2)xx1000=26.7xx0.4 :. X=0.103`
`:. % " of " SO_(3)=0.103/0.5xx100=20.6%`