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If 1.27 g of iodine is liberated,when 10 ml solution of H2O2 is reacted with acidified KI solution.Then the volume strenghth of H2O2 is.?

Shriya Mehrotra , 10 Years ago
Grade 12
anser 4 Answers
Ramreddy

Last Activity: 10 Years ago

N1V1= N2V2
N1 X 10 = 2.54/254
= 0.001 N
Volume strength of H2O2 = 5.6 X Normality
= 5.6 X 0.001
= 0.0056 Vol.
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Shriya Mehrotra

Last Activity: 10 Years ago

i didnt get the ans .the ans is 2.8

Ramreddy

Last Activity: 10 Years ago

Meq of H2O2 = Meq of I2

(w/34)*1000 = [1.27/(254/2)]*1000

w=0.34 g

Thus

0.34 g of H2O2will give =(11.2/34)*.34 =0.028 litre =28 ml

Thus volume strength of H2O2is = 28/10=2.8
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Shriya Mehrotra

Last Activity: 10 Years ago

thanku sir

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