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Equivalent Mass

5 moles of a ...

Question

$5$ moles of a mixture of $F e S O_{4}$ and $F e_{2} (S O_{4})_{3}$ required $150 m L$ of $1.0 M K M n O_{4}$ in acidic medium for the complete neutralisation. Calculate the mole fraction of $F e S O_{4}$ in the initial mixture.

0.2

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0.33

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1.5

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0.15

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Solution

The correct option is D $0.15$
As we know, $M n O_{4}^{-}$ will only oxidise $F e^{2 +}$ not $F e^{3 +}$ .
$+ 7 M n O_{4}^{-} + + 2 F e^{2 +}^{+} + 2 M n^{2 +} + + 3 F e^{3 +}$

$n_{f} = (| Change in O.S. |) \times Number of atoms$
$n_{f_{M n O_{4}^{-}}} = | + 2 - (+ 7) | \times 1 = 5$
$n_{f_{F e^{2 +}}} = | + 3 - (+ 2) | \times 1 = 1$

From the law of equivalence,
$Equivalents of F e^{2 +} = Equivalents of M n O_{4}^{-}$
$n_{F e^{2 +}} \times 1 = \frac{1.0 \times 150 \times 5}{1000}$
$∴ n_{F e^{2 +}} = 0.75 m o l$
$Mole fraction of FeSO4=Mole fraction of Fe2+=0.755=0.15$

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