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Percentage Composition

1 g impure sa...

Question

$1$ g impure sample of $P_{4} O_{6}$ is treated with $20$ mL $a M K M n O_{4}$ in acidic medium to produce $H_{3} P O_{4}$ and $M n C l_{2}$ . $20$ mL of same $K M n O_{4}$ requires $10$ mL of $0.2 M$ $F e S O_{4}$ for complete reduction. What is percentage purity of $P_{4} O_{6}$ ?

3.5

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8.7

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5.5

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None of the above

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Solution

The correct option is C $5.5$
The redox changes are as follows:
$P_{4} O_{6} + K M n O_{4} ⟶ H_{3} P O_{4} + M n C l_{2}$
$(P^{3 +})_{4} ⟶ 4 (P^{5 +}) + 8 e$
$5 e + M n^{7 +} ⟶ M n^{2 +}$
Equivalents of $P_{4} O_{6} =$ Equivalents of $K M n O_{4}$ $= a \times 5 \times 20$
Also, milliequivalents of $K M n O_{4} =$ milliequivalents of $F e S O_{4} [\begin{matrix} M n^{7 +} + 5 e ⟶ M n^{2 +} F e^{2 +} ⟶ F e^{3 +} + I \end{matrix}]$
$a \times 5 \times 20 = 0.2 \times 1 \times 10$
$∴ a = 0.02$
$∴$ Milliequivalentsof $P_{4} O_{6} = 0.02 \times 5 \times 20 = 2.0$
$∴$ Mass of $P_{4} O_{6} = \frac{2 \times 220}{1000 \times 8}$ $[E_{P_{2} O_{5}} = \frac{M}{8}]$ $= 0.055$ g
Percentage of $P_{4} O_{6} = 5.5 %$

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1 g impure sample of P4O6 is treated with 20 mL aMKMnO4 in acidic medium to produce H3PO4 and MnCl2. 20 mL of same KMnO4 requires 10 mL of 0.2 M FeSO4 for complete reduction. What is percentage purity of P4O6?

$1$ g impure sample of $P_{4} O_{6}$ is treated with $20$ mL $a M K M n O_{4}$ in acidic medium to produce $H_{3} P O_{4}$ and $M n C l_{2}$ . $20$ mL of same $K M n O_{4}$ requires $10$ mL of $0.2 M$ $F e S O_{4}$ for complete reduction. What is percentage purity of $P_{4} O_{6}$ ?