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 :

Options

A. `2/5V_1`
B. `5/2V_1`
C. `113 V_1`
D. can not be determined

Answer

A

Explanation

Equivalent of `K_2Cr_2O_7`=equivalent of `N_2H_4`
also equivalent of `KMnO_4`=equivalent of `N_2H_4`
So, equivalent of `K_2Cr_2O_7`=equivalent of `KMnO_4`
`0.1xx6xxV_1=0.3xx5xxV_2`
so `V_2=2//5V_1`