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  4. Weight of KMnO4 required to prepare 250 ml of its (N/10) . solution if equivalent weight of KMnO4 is 31.6, will be:

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Q. Weight of $ KMn{{O}_{4}} $ required to prepare 250 ml of its $ \frac{N}{10} $ . solution if equivalent weight of $ KMn{{O}_{4}} $ is 31.6, will be:

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Solution:

Milli equivalents of $ KMn{{O}_{4}} $ solution $ =\frac{1}{10}\times 250=25 $ Equivalents of $ KMn{{O}_{4}}=\frac{25}{1000}=0.025 $ Thus, required weight = number of equivalents $ \times $ equivalent weight. $ =0.025\times 31.6 $ $ =0.79g $