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Calculate the volume of 0.05 M KMnO_(4) solution required to oxidise completely 2.70 of oxalic acid (H_(2)C_(2)O_(4)) in acidic medium |
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Answer» `2KMnO_(4)+5(COOH)_(2)+3H_(2)SO_(4)rarrK_(2)SO_(4)+2MnSO_(4)+10CO_(2)+8H_(2)O` No of moles of oxalic acid `=2.70//90 =0.03` MOLE From the balanced equation 5 moles of `(COOH)_(2)=2 "moles of" KmnO_(4)` `therefore 0.03 "mole of" (COOH)_(2)=2//5xx0.03=0.012 "mole of" KMnO_(4)` Now 0.05 mole of `KmnO_(4)` is present in solution =`(1000xx0.012)/(0.05)=240` mL |
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