Equivalent mass of ferrous oxalate $F e C_{2} O_{4}$ in the following reaction is:
$F e C_{2} O_{4} \to F e^{3 +} + 2 C O_{2}$ .

\frac{Molecular mass}{1}

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\frac{Molecular mass}{2}

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\frac{Molecular mass}{3}

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\frac{Molecular mass}{4}

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Solution

The correct option is C $\frac{Molecular mass}{3}$
$F e^{2 +} \to F e^{3 +} + e^{-}$
$C_{2} O_{4}^{2 -} \to 2 C O_{2} + 2 e^{-} - ---------------------- -$
$F e C_{2} O_{4} \to F e^{3 +} + 2 C O_{2} + 3 e^{-} - -------------------------------- -$

$n - f a c t o r = 3$ .

$E q u i v a l e n t m a s s = \frac{M o l e c u l a r m a s s}{n - f a c t o r} = \frac{M o l e c u l a r m a s s}{3}$

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