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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 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`
n-factor of `Ba(MNO_(4))_(2)` in acidic medium is :

A

2

B

6

C

10

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
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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 For the reaction, O("molar mass=M") to Fe_(2)O_(3) what is the eq. mass of fe_(0.95) O ?

"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 Consider the following reaction. H_(3)PO_(2)+NaOH to NaH_(2)PO_(2)+H_(2)O What is the equivalent mass of H_(3)PO_(2) ?(mol.Wt.is M)

Knowledge Check

  • Decreasing the concentration of an electrophile in a typical S_(N^2) reaction by a factor of 3 will cause the reaction ratio to :

    A
    increase by a factor of 3
    B
    increase by a factor of `3^(2)`
    C
    decrease by a factor of 3
    D
    remain about the same

  • Increasing the concentration of an electrophile in a typical S_(N^2) reaction by a factor of 3 and the concentration of the nucleophile by a factor of 3 will change the reaction rate to :

    A
    increase by a factor of 6
    B
    increase by a factor of 9
    C
    decrease by a factor of 3
    D
    remain about the same

    • Similar Questions

    Explore conceptually related problems

    "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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