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How many mL of a 0.05 M `KMnO_(4)` solution are required to oxidise 2.0 g of `FeSO_(4)` in a dilute solution (acidic) ? |
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Answer» `underset(10xx151.8)(10 FeSO_(4))+underset(2xx158)(2KMnO_(4))+8H_(2)SO_(4) to K_(2)SO_(4)+2MnSO_(4)+5Fe_(2)(SO_(4))_(3)+8H_(2)O` `10xx151.8g " of" FeSO_(4)` require `KMnO_(4)=2xx158` g 2 g of `FeSO_(4)` will require `KMnO_(4)=(2xx158xx2)/(10xx151.8)g` Suppose, V mL of `KMnO_(4)` solution (0.05 M) is required. Amoount of `KMnO_(4)` in this solution `=(158xx0.05)/(1000)xxV` Thus, `(158xx0.05xx V)/(1000)=(2xx158xx2)/(10xx151.8)` V=52.7 mL |
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