The volume, in mL, of $0.02 M K_{2} C r_{2} O_{7}$ solution required to react with 0.576g of ferrous oxalate in acidic medium is ______. (molar mass of $F e = 56 g m o l^{- 1}$ )

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answer is 10.

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Detailed Solution

$M . e q . o f K_{2} C r_{2} O_{7} = M e q . o f F e C_{2} O_{4}$
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n-factor of $K_{2} C r_{2} O_{7} = 2 (6 - 3) = 6$
n-factor of $F e C_{2} O_{4} = 1 (3 - 2) + 2 (4 - 3) = 3$
By law of equivalence,
n-factor of $K_{2} C r_{2} O_{7} \times$ no.of moles
=n-factor of $F e C_{2} O_{4} \times$ no.of moles
$\Rightarrow \frac{V}{1000} \times 0.2 \times 6 = \frac{0.576}{144} \times 3$
$\frac{V}{1000} \times 0.2 \times 2 = \frac{0.576}{144}$
$V = \frac{576}{144 \times 0.4} = 10 m l$