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Q)

$2.68 \times 10^{-3} $ moles of a solution containing an ion $A^{n+}$ acqired. $1.81 \times 10^{-3}$ moles of $MnO_4^{-}$ for the oxidation of $An^{+} +0 AO_3^{-}$ in acid medium. What is the value of n

$(a)\;3 \\ (b)\;2 \\(c)\;2.5 \\(d)\;3.5 $

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A)
The change in oxidation state of A when,
$A^{n+}$ is oxidised to $AO_3^{-}$ in acidic medium is form $+n$ to $+n \;+0 \; (+5)$ ie the change is equal to $(5-n)$
The number of electrons added to reduced
$1.61 \times 10^{-3}$ moles of $MnO^{-}_4$ to $Mn^{2+}$
(o.s of Mn changes from $+7$ to $+2=1.61 \times 10^{-3} \times 5)$
$\therefore 1.61 \times 10^{-3} \times 5 =2.68 \times 10^{-3}(5-n)$
or $(5-n) =\large\frac{1.61 \times 10^{-3} \times 5}{2.68 \times 10^{-3}}$
or $5-n=3=>n=2$
Hence b is the correct answer.
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