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Grade: 12th pass
        
KHC2O4 solution is 4N when treated with NaOH. What is the normality of this solution when oxidised by KMnO4 ?
Please explain in detail....
 
 
8 months ago

Answers : (2)

Vikas TU
7767 Points
							
Equivalents of KMnO4 =  Equivalents of KHC2O4
KMnO4 oxidezed into Mn2+ and KHC2O4 into CO2.
Valence factor of KMnO4 therefore,= 7 – 5 = +5
Valence factor of KHC2O4 = 2
4*2 = N*5
N = 8/5 => 1.6.
7 months ago
Yasir Rather
31 Points
							
Please Vikas TU sir explain..............
eq of KMnO4 = Eq of KHC2O4
NV = NV
M *n-factor *V = M *n-factor *V
no. of moles* n-factor= no. of moles* n-factor
7 months ago
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