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"Equivalent mass" =("Molecular mass/Atom...

"Equivalent mass" =("Molecular mass/Atomic mass" )/("n-factor")
n-factor is very important in redox as well as non-redox reactions.With the help of n-factor we can predict the molar ratio of the reactant species specis taking part in reactions. The reciprocal of n-factor's ratio of the reactions is the molar ratio of the reactants.
In general n-factor of acid/base is number of moles of `H^(+)//OH^(-)` furnished per mole of acid/base n-factor of a reactant is number of moles electrons lost or gained per mole of reactant.
Example 1:
(1)In acidic medium : `KMnO_(4) (n=5)to Mn^(2+)`
(2) In neutral medium : `KMnO_(4)(n=3) to Mn^(2+)`
(3) In basic medium : `KMnO_(4)(n=1) to Mn^(6+)`
Example 2 : `FeC_(2)O_(4)to Fe^(3+)+2CO_(2)`
Total number of moles `e^(-)` lost by 1 mole of `FeC_(2)O_(4)`
`=1+1xx2 implies 3`
For the reaction, `Fe_0.95O("molar mass=M") to Fe_(2)O_(3)` what is the eq. mass of `fe_(0.95)`O ?

A

`(M)/(0.85)`

B

`(M)/(0.95)`

C

`(M)/(0.8075)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
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"Equivalent mass" =("Molecular mass/Atomic mass" )/("n-factor") n-factor is very important in redox as well as non-redox reactions.With the help of n-factor we can predict the molar ratio of the reactant species specis taking part in reactions. The reciprocal of n-factor's ratio of the reactions is the molar ratio of the reactants. In general n-factor of acid/base is number of moles of H^(+)//OH^(-) furnished per mole of acid/base n-factor of a reactant is number of moles electrons lost or gained per mole of reactant. Example 1: (1)In acidic medium : KMnO_(4) (n=5)to Mn^(2+) (2) In neutral medium : KMnO_(4)(n=3) to Mn^(2+) (3) In basic medium : KMnO_(4)(n=1) to Mn^(6+) Example 2 : FeC_(2)O_(4)to Fe^(3+)+2CO_(2) Total number of moles e^(-) lost by 1 mole of FeC_(2)O_(4) =1+1xx2 implies 3 In the reaction, xVO+yFe_(2)O_(3) to FeO+V_(2)O_(5) what is the value of x and y respectively?

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