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Equivalent mass \( =\frac{\text { Molecular mass } / \text { Atomic mass }}{n \text {-factor }} \)
\( \mathrm{P} \)
\( n \)-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
W part in reactions. The reciprocal of \( n \)-factor's ratio of the reactants is the molar ratio of the reactants.
In general \( n \)-factor of acid/base is number of moles of \( \mathrm{H}^{+} / \mathrm{OH}^{-} \)furnished per mole of acid/base. \( n \)-factor of a reactant is number of moles of electrons lost or gained per mole of reactant.
Example 1 :
1. In acidic medium : \( \mathrm{KMnO}_{4}(n=5) \longrightarrow \mathrm{Mn}^{2+} \)
2. In neutral medium : \( \mathrm{KMnO}_{4}(n=3) \longrightarrow \mathrm{Mn}^{2+} \)
3. In basic medium : \( \quad \mathrm{KMnO}_{4}(n=1) \longrightarrow \mathrm{Mn}^{6+} \)
Example 2 : \( \mathrm{FeC}_{2} \mathrm{O}_{4} \longrightarrow \mathrm{Fe}^{3+}+2 \mathrm{CO}_{2} \)
Total number of moles of \( e^{-} \)lost by \( 1 \mathrm{~mole}^{\text {of } \mathrm{FeC}_{2} \mathrm{O}_{4}} \)
\( =1+1 \times 2 \Rightarrow 3 \)
\( \therefore \quad n \)-factor of \( \mathrm{FeC}_{2} \mathrm{O}_{4}=3 \)
In the reaction, \( x \mathrm{VO}+y \mathrm{Fe}_{2} \mathrm{O}_{3} \longrightarrow \mathrm{FeO}+\mathrm{V}_{2} \mathrm{O}_{5} \), what is the value of \( x \) and \( y \) respectively?
(a) 1,1
(b) 2,3
(c) 3,2
(d) None of these
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