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one litre of a mixture of o2 and o3 at ntp was allowed to react with an excess of acidified solution of kI.the iodine liberated required 40 ml of M/10 sodium thiosulphate solution for titration.what is the percent age of ozone in mixture

one litre of a mixture of o2 and o3 at ntp was allowed to react with an excess of acidified solution of kI.the iodine liberated required 40 ml of  M/10 sodium thiosulphate solution for titration.what is the percent age of ozone in mixture

Grade:12

2 Answers

Akriti bhatt
19 Points
2 years ago
O3--> O2 +[O]
Ki +H2O +O3 --> 2KOH +I2
------------------------------------------
Ki +H2O +O3-->2KOH +I2 +O2
-----------------------------------------------
2Na2S2O3 +I2 --> Na2S4O6 +2NaI
^
¦
M/10
(n*1000)/40=1\10--> n=1/250
n(O3) = n(I2) = 1/2*n(hypo)=1/500
Volume of O3 at Ntp = no. Of moles *22.4 = 0.0448
Volume of O2 = 1-0.0448=0.9552
T
Now the question is similar to other questions % of ozone can easily be calculated...... 
 
 
 
Akriti bhatt
19 Points
2 years ago
% of ozone (by weight) =(48*0.0448*100)/(48*0.0448+0.9552*32)=0.0657*100=6.57%
 The whole answer is given  in the previous content. If you need to calculate the ozone % by volume in the mixture then simply divide the volume of ozone by total volume. The answer will be 4.48%
6.57% is the final % of ozone in the mixture. 

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