0.5 g of fuming H 2 S O 4 (Oleum) is diluted with water. This solution is completely neutralized by 26.7 mL…

0.5 g of fuming H2SO4 (Oleum) is diluted with water. This solution is completely neutralized by 26.7 mL of 0.4 N NaOH. The percentage of free SO3 in the sample is
  1. 30.6%
  2. 40.6%
  3. 20.6%
  4. 50.6%

Solution

2NaOH+ H 2 S O 4 →N a 2 S O 4 +2 H 2 O
2NaOH+S O 3 →N a 2 S O 4 +2 H 2 O
Meq of H2SO4+Meq of SO3=Meq of NaOH
(0.5-a)49×1000+a80/2×1000=26.7×0.4
So a=0.103
So % of SO3=0.1030.5×100=20.6% . ^

Asked in: JEE-TOPICTESTS-CHEMISTRY

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