Question

In an oxidation-reduction reaction, $ MnO_{4}^{-} $ ion is converted to $ M{{n}^{2+}} $ . What is the number of equivalents of $ KMn{{O}_{4}} $ (mol. wt. =158) present in $ 250\,\,mL $ of $ 0.04\,\,M\,\,KMn{{O}_{4}} $ solution?

• A. $ 0.02 $
• B. $ 0.05 $
• C. $ 0.04 $
• D. $ 0.07 $

Answer 1

Answer: (B)

Change in oxidation number $ (+7)-(+2)=5 $ $ \therefore $ Normality of solution $ =5\times 0.04=0.20\,\,N $ Volume $ =250\,\,mL $ $ \therefore $ Number of equivalents of $ KMn{{O}_{4}} $ $ =0.20\times \frac{250}{1000}=0.05 $