Hydrogen And Its Compounds Question 258
Question: 100 mL of 0.01 $ MKMnO_4 $ oxidises 100 mL $ H_2O_2 $ in acidic medium. Volume of the same $ KMnO_4 $ required in alkaline medium to oxidise 100 mL of the same $ H_2O_2 $ will be ( $ MnO_4^{-} $ changes to $ M{n^{2+}} $ in acidic medium and to $ MnO_2 $ in alkaline medium)
Options:
A) $ \frac{100}{3}mL $
B) $ \frac{500}{3}mL $
C) $ \frac{300}{5}mL $
D) None
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ MnO_4^{-}+5{e^{-}}\to M{n^{2+}} $ (acidic) $ MnO_4^{-}+3{e^{-}}\to MnO_2 $ (basic) $ 100mLH_2O_2\equiv 100\times 5N $
$ MnO_4^{-}\equiv V\times 3NMnO_2 $
$ N=\frac{500}{3}mL $
BETA
