Home
Class 12
CHEMISTRY
0.1 M solution of KMnO(4) (in acidic me...

0.1 M solution of `KMnO_(4)` (in acidic medium) may oxidies

A

`0.25 M CO_(2)O_(4)^(-2)`

B

`0.8 M Fe^(+2)`

C

`0.1 M FeC_(2)O_(4)`

D

`0.6 M Cr_(2)O_(7)^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`MN_(4)^(-)toMn^(-2),v.f.=5`
`"no. of eq" =0.1xx5xx1=0.5`
Option
(1) `C_(2)O_(4)^(2-)toCO_(2),v.f.=1xx2=2`
`"no.f of eq"=0.25xx2xx1=0.5`
(2) `Fe^(2+)toFe^(3+),v.f.=1`
`"no.f of eq" = 0.8xx1xx1=0.8`
(3) `FeC_(2)O_(4)toFe^(3+)+CO_(2)`,
` v.f.=3,"no. of eq" =0.1xx3xx1=0.3`
(4) `Cr_(2)O_(7)^(2-)toCr^(=3)rArrv.f.=6`
`rArr"no. of eq" 0.6xx6xx1=3.6`
Promotional Banner

Topper's Solved these Questions

  • CHEMISTRY AT A GLANCE

    ALLEN|Exercise INORGANIC CHEMISTRY|300 Videos

  • CHEMISTRY AT A GLANCE

    ALLEN|Exercise ORGANIC CHEMISTRY|483 Videos

  • Chemical Equilibrium

    ALLEN|Exercise All Questions|24 Videos

  • ELECTROCHEMISTRY

    ALLEN|Exercise EXERCISE -05 [B]|38 Videos

Similar Questions

Explore conceptually related problems

On oxidation with KMnO_(4) in acidic medium , SO_(2) is oxidised to

The weight of oxalic acid required to reduce 80 ml of 0.4 M KMnO_(4) in acidic medium is

STATEMENT-1: The equivalent mass of of KMnO_(4) in acidic medium is (M)/(5) where M=molecular mass of KMnO_(4) STATEMENT-2: Equivalent mass is equal to product of molecular mass and change in oxidation number.

3 mole of FeSO_(4) are oxidised by a mole of KMnO_(4) in acidic medium whereas 3 moles of FeC_(2)O_(4) are oxidised by b mole of KMnO_(4) in acidic medium, the ratio of a and b is:

The molarity (m) of KMnO_(4) in the acidic medium is (densityh of KMnO_(4) solution =1.58gmL^(-1)Mw(KMnO_(4))=158gmol^(-1))

Molecular weight of KMnO_(4) in acidic medium and neutral medium will be respecitvely -