IIT JEE 1980 question

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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 % of free SO₃ in the solution is

(A) 20.6%
(B) 26%
(C) 30%
(D) 19%

#Chemistry #Physical Chemistry

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2 Answers

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Abhishek Das ·2011-06-25 04:51:30

26.7ml of 0.4N NaOH = 10.68 meq acid
If x is the precentage of SO3 in oleum then
eq of SO3 = 0.5x/(40*100) (eq wt of SO3 = 40)
eq of H2SO4 = 0.5*(100-x)/(49*100) (eq wt of H2SO4 = 49)
So 0.5(x/4000+ (100-x)/4900)= 10.68/1000
Solve this to get x = 21%

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Abhishek Das ·2011-06-25 04:51:57

ans. (A) is correct..

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