1.

Calculate the normality of a solution containing 15.8 g of `KMnO_(4)` in 50 mL acidic solution.

Answer» Normality(N) `=(Wxx1000)/(ExxV)`
Here W=15.8g, V=50mL E = `("molar mass of KMnO"_(4))/("Valency factor")=158//5=31.6`
So, normality = 10 N


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