Volume `V_1` mL of 0.1 M `K_2Cr_2O_7` is needed for complete oxidation of 0.678 g `N_2H_4` in acidic medium.The volume of 0.3 M `KMnO_4` needed for same oxidation in acidic medium will be :

Options

A. `(2)/(5)V_(1)`
B. `(5)/(2)V_(1)`
C. 113`V_(1)`
D. can not be determined

Answer

A

Explanation

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)`