${Hg}_{5} {({IO}_{6})}_{2}$ oxidizes KI to $I_{2}$ in acid medium and the other product containing iodine is $K_{2} {HgI}_{4}$ . If the $I_{2}$ liberated in the reaction requires 0.004 mole of ${Na}_{2} S_{2} O_{3}$ , the number of moles of ${Hg}_{5} {({IO}_{6})}_{2}$ that have reacted ·

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$10^{- 3}$

$2.5 \times 10^{- 2}$

$2.5 \times 10^{- 4}$

$10^{- 4}$

answer is C.

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Detailed Solution

Here,

$n_{1_{2}}$ Produced $= \frac{0.004}{2}$

$\begin{array}{l} {Hg}_{5} {({IO}_{6}^{+ 7})}_{2}^{- 10} ⟶ I^{-} \\ n . f . = 16 \\ n \times 16 = \frac{0.004}{2} \times 2 \Rightarrow n = 2.5 \times 10^{- 4} \end{array}$

Thus, number of moles required is $2.5 \times 10^{- 4}$

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