Chemistry - Redox Reactions Question with Solution | TestHub

Equivalents are conserved: 0.46/(E + 8) = 0.92/(E + 35.5), using the equivalent weights of ${\mathrm{O}}^{2-}$ (8) and $\mathrm{C}{\mathrm{l}}^{-}$ (35.5). So 0.46E + 16.33 = 0.92E + 7.36 $\to$ E = 19.5. Atomic weight = valency $x$ 19.5, giving 19.5, 39, 58.5, ... ; hence A may be 39 (valency 2). Thus (b) and (c).