Question

# 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 is50.6% 30.6% 40.6% 20.6%

Solution

## The correct option is D 20.6% Oleum is nothing but SO3 dissolved in H2SO4 H2SO4+SO3⟶H2S2O7 Clearly, SO3 is an acidic oxide, so it will be neutralized by NaOH. 2NaOH+SO3⟶Na2SO4+H2O From the above equation, EWSO3=MWSO32 Let the amount of SO3 in in grams be = x  therefore, the amount of H2SO4=(0.5−x) At neutralisation stage, Meq of sample = Meq of NaOH ⇒{xEWSO3+0.5−xEWH2SO4}×1000=0.4×26.7  ⇒{x40+0.5−x49}×1000=0.4×26.7  Solving, we get x=0.104g % age of free SO3=0.1040.5×100≈20.6

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