D And F Block Elements Question 41
Question: Equivalent weight of $ KMnO_4 $ acting as an oxidant in acidic medium is equal to
[CPMT 1990; MP PMT 1999]
Options:
A) Molecular weight of $ KMnO_4 $
B) $ \frac{1}{2} $ ´ Molecular weight of $ KMnO_4 $
C) $ \frac{1}{3} $ ´ Molecular weight of $ KMnO_4 $
D) $ \frac{1}{5} $ ´ Molecular weight of $ KMnO_4 $
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Answer:
Correct Answer: D
Solution:
$ \frac{1}{5}\times \text{ molecular }\text{weigth of }KMnO_4 $ as transfer of $ 5{{e}^{-}} $ takes place when $ KMnO_4 $ acts as oxidant in acidic medium. $ 2KMnO_4+3H_2SO_4\to K_2SO_4+2MnSO_4+3H_2O $ $ +5O $
BETA
