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$3.92 \mathrm{~g}$ of ferrous ammonium sulphate react completely with $50 \mathrm{ml} \frac{\mathrm{N}}{10} \mathrm{KMnO}_{4}$ solution. The percentage purity of the sample is
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The correct answer is:
50
Eq of $\mathrm{KMnO}_{4}$ used $=\frac{50 \times 1}{1000 \times 10}=0.005$
$\therefore$ Eq of FAS reacted $=0.005$
$\therefore$ weight of FAS needed
$$
=0.005 \times 392=1.96 \mathrm{~g}
$$
Thus percentage purity of FAS is $50 \%$
$\therefore$ Eq of FAS reacted $=0.005$
$\therefore$ weight of FAS needed
$$
=0.005 \times 392=1.96 \mathrm{~g}
$$
Thus percentage purity of FAS is $50 \%$
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