The equivalent weight of $F e C_{2} O_{4}$ when it reacts with acidified $K M n O_{4}$ is:

144

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Solution

The correct option is C 48
Reaction of $F e C_{2} O_{4}$ with $K M n O_{4}$ is,

$5 F e C_{2} O_{4} + 6 K M n O_{4} + 24 H_{2} S O_{4} ⟶ 3 K_{2} S O_{4} + 6 M n S O_{4} + 5 F e + 3 (S O_{4})_{3} + 24 H_{2} O + 10 C O_{2}$

So, $F e$ is oxidized to $+ 2$ to $+ 3$
$C$ is oxidised from +3 to +4

So, $n - f a c t o r = 1 + 2 \times 1 = 3$

$∴$ Equivalent weight = $\frac{M o l e c u l a r w e i g h t}{n - f a c t o r}$
$= \frac{144}{3} = 48$

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