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Volume V(1)mL of 0.1 M K(2)Cr(2)O(7) is ...

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

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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Knowledge Check

  • The mililitres of 0.2M KMnO_(2) required for the complete oxidation of 0.1 mol Fe^(2+ ) in acidic medium is -

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    200ml
    B
    100ml
    C
    400ml
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  • What volume of 0.05 M Cr_(2)O_(7)^(2-) in acid medium is needed for complete oxidation of 200 mL of 0.6 M FeC_(2)O_(4) solution ?

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    0.6 L
    B
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    D
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