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