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A solution of `H_(2)O_(2)` is titrated against a solution of `KMnO_(4)`. The reaction is : `2MnO_(4)^(-)+5H_(2)O_(2)+6H^(+) to 2Mn^(2+)+5O_(2)+8H_(2)O` If it requires 46.9 mL of 0.145 M `KMnO_(4)` to oxidise 20 g of `H_(2)O_(2)`, the mass percentage of `H_(2)O_(2)` in this solution is :A. 2.9B. 29C. 21D. 4.9 |
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Answer» Correct Answer - A Number of moles of `KMnO_(4)=(MV)/(1000)=(0.145xx46.9)/(1000)` `=6.8xx10^(-3)` Number of moles of `H_(2)O_(2)=6.8xx10^(-3)xx2.5=0.017` Mass of `H_(2)O_(2)=0.017xx34=0.578` Mass % of `H_(2)O_(2)=(0.578)/(20)xx100=2.9` |
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