NEET Chemistry Redox Reactions Questions Solved


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:

(a) 30.6% 

(b) 40.6%

(c) 20.6%

(d) 50%

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(c) Meq. of H2SO4 + Meq.Of SO3 = Meq. of NaOH

(0.5-a)/49*1000 +a/40*1000 = 26.7 x 0.4

a = 0.103

% of SO3 = 0.103/0.5*100 = 20.6%

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