A substance · 'A' undergoes disproportionation reaction to form $A^{3 +}$ and $A^{2 -}$ obeying first order kinetics. The rate · constant is obtained by measuring volume of an oxidizing agent capable of oxidizing A, $A^{3 +}$ and $A^{2 -}$ $A^{2 -}$ to $A^{2 +}, A^{4 +}, A^{1 +}$ respectively. From the following data, calculate moles of $A^{2 -}$ 6 minutes after start of the reaction if initially 10 moles of A was taken.

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answer is 5.

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Detailed Solution

$A ⟶ A^{3 +} + 3 e^{-} \times 2$ Using ${KMnO}_{4}$

$\frac{A + 2 e^{-} ⟶ A^{2 -} \times 3}{5 A ⟶ 2 A^{3 +} + 3 A^{2 -}}$ $\begin{array}{l} A ⟶ A^{2 +} \\ n_{f} = 2 \\ A^{3 +} ⟶ A^{4 +} \end{array}$

$t = t 10 - 5 x 2 x 3 x$ $\begin{array}{l} n_{f} = 1 \\ A^{2 -} ⟶ A^{1 +} \\ n_{f} = 3 \end{array}$

At $t = 4 \min$

$\begin{array}{l} (10 - 5 x) 2 + 2 x \times 1 + 3 x \times 3 = N \times 21.5 \\ 20 - 10 x + 11 x = N \times 21.5 \\ \frac{20 + x}{20} = \frac{N \times 21.5}{N \times 20} \\ 1 + \frac{x}{20} = \frac{215}{200} = \frac{43}{40} \\ \frac{x}{20} = \frac{3}{40} \\ x = 1.5 mole \end{array}$