In a reaction the ferrous $\left(\mathrm{Fe}^{+2}ight)$ ion is oxidised to ferric…
- Half of the atomic weight
- $1 / 5$ of the atomic weight
- The atomic veight
- Twice the atomic weight
Solution
Equivalent weight $=\frac{\text { Atomic weight }}{\text { No. of e lost or gained }}$
$\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-}$
$\therefore$ Equivalent weight $=$ Atomic weight undefined
Asked in: JEE-TOPICTESTS-CHEMISTRY
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