Home
Class 12
CHEMISTRY
A mixture having 0.5 mole KMnO(4) and 0...

A mixture having 0.5 mole `KMnO_(4)` and 0.5 mole `K_(2)Cr_(2)O_(7)` (in acidic medium) was completely reduced by 0.5 moles of a reducing agent `(M^(@) = 132)`. Then equivalent mass of the reducing agent is :

A

`13.20`

B

`16.50`

C

`11.45`

D

`12.00`

Text Solution

Verified by Experts

The correct Answer is:
D
`Sigmax_(O.A) = Sigmax_(R.A)`
`5n + 6n = n (n_(f)) therefore n_(f)= 11 rArr E^(@) = (M^(@))/(n_(f)) = (132)/(11)=12`
Promotional Banner

Similar Questions

Explore conceptually related problems

For a given reductant , ratio of volumes of 0.2 M KMnO_(4) and 1M K_(2)Cr_(2)O_(7) in acidic medium will be

Calculate the number of moles of Sn^(2+) ion oxidise by 1 mole of K_(2)Cr_(2)O_(7) in acidic medium.

A certain amount of reducing agent reduces x mole of KMnO_(4) & y mole of K_(2)Cr_(2)O_(7) in difference experiments in acidic medium. If the change in oxidation state of reducing agent is same in both experiments then x:y is

How many moles of KMnO_(4) are reduced by 1 mole of ferrous oxalate in acidic medium :-

How many moles of K_(2)Cr_(2) O_(7) are reduced by 1 mole of formic acid -

One mole of FeC_(2)O_(4) is oxidised by KMnO_(4) in acidic medium. Number of moles of KMnO_(4) used are