Question

$N_2{H}_4$ reduces $I{O}_3^{-}$ to $I^{+}$ in acidic medium:

$N_2{H}_4+I{O}_3^{-}+H^{+}⟶N_2+I^{+}+H_{2}O$

Equivalent mass of $N_2{H}_4$ (molar mass = 32g mol${}^{-1}$) is :

• A. 32
• B. 16
• C. 8
• D. 24

Answer 1

Answer: (C)

8

$N_2{H}_4⟶N_2$

$-2$ $0$

N is changing oxidation state from -2 to zero so total loss in electron is 4 for 2 N- atom of the molecule.

Equivalent mass $=\frac{Mw}{4}=\frac{32}{4}=8$.