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<div>For decolourization of one mole of KMnO4, the required litre of 30 volume H2O2 in acidified medium?</div> <div>1. 0.94</div> <div>2. 9.4</div> <div>3. 0.094</div> <div>4. 3.01</div>
Asked by Mamali Nanda | 15 Jul, 2016, 10:15: PM
1 mole of KMnO4 requires 2.5 moles of H2O2 for decolourisation.
1 mole H2O2 = 34 g
Hence 2.5 moles H2O2 = 85 g
Also,
2H2O2 → 2H2O + O2
From the above equation,
2x34 g of H2O2 gives 22.4 L O2
Hence 30 L O2 can be obtained from = 30x2x34/22.4 = 91.07 g H2O2
1 L of 30 volumes of H2O2 contains 91.07 g H2O2
So, 85 g H2O2 corresponds to 0.933 L H2O2
Hence, correct ans is 0.94 (option 1)
Answered by Prachi Sawant | 17 Jul, 2016, 12:17: AM

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