Volume V1 mL of 0.1 M K2Cr2O7 is needed ...

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 :

A

`(2)/(5)V_(1)`

B

`(5)/(2)V_(1)`

C

113`V_(1)`

D

can not be determined

Text Solution

Verified by Experts

The correct Answer is:

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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