IIT JEE 1980 question

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%

2 Answers

1
Abhishek Das ·

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%

1
Abhishek Das ·

ans. (A) is correct..

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