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Volume `V_(1)mL` of 0.1 M `K_(2)Cr_(2)O_(7)` is needed for complete oxidation of 0.678g `N_(2)H_(4)` in acidic medium. The volume of 0.3M `KMnO_(4)` needed for same oxidation in acidic medium will be :A. `(2)/(5)V_(1)`B. `(5)/(2)V_(1)`C. 113`V_(1)`D. can not be determined |
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Answer» Correct Answer - A Equivalent of `K_(2)Cr_(2)O_(7)` = equivalent of `N_(2)H_(4)` also equivalent of `KMnO_(4)` = equivalent of `N_(2)H_(4)` So, equivalent of `K_(2)Cr_(2)O_(7)` = equivalent of `KMnO_(4)` `0.1xx6xxV_(1)=0.3xx5xxV_(2) " " :. " so" V_(2)=2//5V_(1)` |
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