n factor of MnO2
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Answer:
Hello mate ✌️..
Explanation:
Molecular mass of KMnO4 = 158 gm
In acidic medium, no. of electrons involved n is:
(MnO4)^- + 8H^+ +5e^- ---> Mn^+2 + 4H2O
So, equivalent mass = molar mass/ electrons involved
= 158/ 5 = 31.6 gm/ equivalent
In strongly basic medium,
(MnO4)^- + 1e^- ---> (MnO4)^-2
So, n = 1 , equivalent weight = 158/1 = 158gm/ equivalent
In neutral medium (MnO4)^- + 4H2O +3e^- ---> MnO2 + 4OH^-
Here, n = 3 so, equivalent weight = 158/3 = 52.66 gm/equivalent
For, MnO2 (molar mass = 87) , here n = 3
So, equivalent weight = molar mass/ 3 = 87/3 = 29 gm/equivalent.
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