Barium permangnate solution $(20 mL, 0 .1 M)$ is mixed with $0.1 {NI}^{-}$ giving precipitate of ${IO}_{3}^{-}$ and ${MnO}_{2}$ . Resulting solution is filtered and titrated against ${Mo}^{3 +}$ , giving ${MoO}_{2}^{2 +}$ and ${Mn}^{2 +}$ , which required 0.5 M, 10 mL acidified ${Mo}^{3 +}$ . Select the correct option(s)

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Volume of 1- solution taken is 30 mL

Per mole ${Mn}^{2 +}$ formed, 4 moles of H⁺ are consumed

Volume of 1- solution taken is 50 mL

Per mole ${IO}_{3}^{-}$ formed, 2 moles of ${MnO}_{4}^{-}$ are consumed

answer is A, D.

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Detailed Solution

Here,

$4 H^{+} + 3 {MnO}_{4}^{-} + 5 {Mo}^{3 +} ⟶ 3 {Mn}^{2 +} + \overset{+ 6}{5 {MoO}_{2}^{2 +}}$

$\begin{aligned} 20 ml \\ 0.2 M \\ (n f = 5) \end{aligned}$ $\begin{matrix} n_{f} = 3 \\ 0.5 M \\ 10 ml \end{matrix}$

$H_{2} O + \overset{+ 7}{2 {MnO}_{4}^{-}} + \overset{- 1}{I^{-}} ⟶ \overset{+ 5}{{IO}_{3}^{-}} + 2 {MnO}_{2} + 2 {OH}^{-}$

$\begin{array}{cc} 0.1 N \\ 1 mmol & 0 .5 mmol \\ (n_{f} = 3) & n_{f} = 6 \\ V = 30 ml \end{array}$

Thus, correct options are A and D.

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