Question

Match List-I with equivalent weight given in List-II.

The compound whose equivalent weight is needed is underlined.

Answer 1

Equivalent weight $=\frac{Mol.wt.}{Changeinoxidationstate}$
So,
For, $\underset{–}{P_2{H}_4}\to P{H}_3+P_4{H}_2$ ; $E=\frac{5M}{6}$
$\underset{–}{I_2}\to I^{-}+I{O}_3^{-}$ ; $E=\frac{3M}{5}$
$Mn{O}_4^{-}+M{n}^{2+}+H_{2}O\to \underset{–}{M{n}_3{O}_4}+H^{+}$ ; $E=\frac{15M}{26}$
$\underset{–}{H_{3}P{O}_2}\to P{H}_3+H_{3}P{O}_3$ ; $E=\frac{3M}{4}$