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How many grams of KMnO4 would be required to prepare 2500 ml of 0.01 N solution, based on the following reaction? $MnO_4+Fe\rightarrow Mn+Fe$ [Mol. wt of $KMnO_4$ = 158]

1 Answer

0.79 g
Hence (B) is the correct answer.
answered Jun 9, 2014 by sreemathi.v
 

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