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if three different solutions of h202 in equAL volumes are mixed .if their volume strengths are v1 v2 v3 then which set of values of v1,v2,v3 are valid so that their 30 ml mixture easily be oxidised by 40 ml of .0 M KMno4

if three different solutions of h202 in equAL volumes are mixed .if their volume strengths are v1 v2 v3 then which set of values of v1,v2,v3 are valid so that their 30 ml mixture easily be oxidised by 40 ml of .0 M KMno4

Grade:11

1 Answers

Vikas TU
14149 Points
6 years ago
 
KMnO4 would oxidixed to Mn2+ and therfore
its valency factor would be => 7-2 => 5
Therdore
equivalents of H2O2 and the KMO4 should be equal.
Hence,
Equivalents of H2O2 needed = 40*10^-3*0.5*5 = 0.1
Now for H2O2=> M*0.03*2 = 0.1
M = 0.1/0.06 => 10M molarity strength H2O2 soln. would be more valid. 

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