Chemistry - Redox Reactions Question with Solution | TestHub

Let the atomic weight of$\text{X}\text{=}\text{y}$
No. of equivalents of$\text{KOH}$= No. of equivalents of${\text{H}}_{\text{2}}{\text{XO}}_{\text{4}}$
i.e.$\frac{\text{0}\text{.12544}}{\frac{\text{66}\text{+}\text{y}}{\text{2}}}\text{=}\text{0}\text{.12544}$
$\therefore \text{y}\text{=}\text{32}$