0.5 g of fuming H2SO4 (oleum) is diluted with water. This solution is completely neutralised by 26.7 ml of 0.4 N NaOH. The percentage of free SO3 in the sample is

Meq.  of  H2SO4+Meq.  of  SO3 =  Meq.  of  NaOH    WE×1000     + WE×1000   =  NV∴  0.5−x98/2 ×  1000  +  x80/2  ×  1000=26.7 × 0.4  ∴  x=0.103∴  %  of  SO3 =  0.1030.5 × 100=20.6%