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The number of $m o l e s$ of acidified $K M n O_{4}$ required to oxidise $o n e m o l e$ of ferrous oxalate $(F e C_{2} O_{4})$ is:
$3$
$5$
$0.6$
$1.5$

A

3

B

0.6

C

5

D

1.5

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Solution

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$∴ 5 m o l e F e C_{2} O_{4} \equiv 3 m o l e K M n O_{4}$
$1 m o l e F e C_{2} O_{4} \equiv 0.6 m o l e K M n O_{4}$ .

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