Guys, please help me out: Calculate the normality of KMnO₄ in acidic medium. Where molar mass of KMnO₄ is 158 g/mol and it gets reduced from +7 oxidation state to +2 oxidation state
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so, eq wt = 158/5 = 31.6 g. Hence, eq wt = 158/3 = 52.67g. In many cases though if it's written alkaline, it mostly means the neutral one; for the highly alkaline thing, it'll especially mention it.
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