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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 is


  1. 50.6%

  2. 30.6%

  3. 40.6%

  4. 20.6%


Solution

The correct option is D

20.6%


Oleum is nothing but SO3 dissolved in H2SO4

H2SO4+SO3H2S2O7

Clearly, SO3 is an acidic oxide, so it will be neutralized by NaOH.

2NaOH+SO3Na2SO4+H2O

From the above equation, EWSO3=MWSO32

Let the amount of SO3 in in grams be = x 

therefore, the amount of H2SO4=(0.5x)

At neutralisation stage, Meq of sample = Meq of NaOH

{xEWSO3+0.5xEWH2SO4}×1000=0.4×26.7 

{x40+0.5x49}×1000=0.4×26.7 

Solving, we get x=0.104g

% age of free SO3=0.1040.5×10020.6

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