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The volume of 0.2 M H_2O_2 that will be completely oxidised by 10ml of 0.3M KMnO_4 solutioninfaintly alkaline medium |
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Answer» SOLUTION :`M_1`=0.2 M, `V_1`=? `n_1=2 to ` for `H_2O_2` `M_2` =0.3 M , `V_2`=10 ml `n_2=3` (In slightlybasic MEDIUM `MnO_4^(-) to MnO_2` ) `M_1V_1n_1=M_2V_2n_2` `0.2xxv_1xx2=0.3xx10xx3` `V_1`=22.5 ml |
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