Three different solutions of oxidants $K_{2} C r_{2} O_{7}, I_{2} & K M n O_{4}$ is titrated separately with $0.19 g r K_{2} S_{2} O_{3} .$ The molarity of each oxidant is $0.1 M$ and the reactions are
$(i) C r_{2} O_{7}^{2 -} + S_{2} O_{3}^{2 -} \to C r^{3 +} + S O_{4}^{2 -}$ $(i i) I_{2} + S_{2} O_{3}^{2 -} \to I^{-} + S_{4} O_{6}^{2 -}$
$(i i i) M n O_{4}^{-} + S_{2} O_{3}^{2 -} \to M n O_{2} + S O_{4}^{2 -}$
Which of the following statements are correct

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Volume of $K M n O_{4}$ used in the titration is maximum

Volume of $K_{2} C r_{2} O_{7}$ used is maximum

Volume of $I_{2}$ used is minimum

All three oxidants can acts as self-indicators

answer is B, D.

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Detailed Solution

All three are self indicators
$n_{K_{2} S_{2} O_{3}} = \frac{0.19}{190} = 10^{- 3} = 1 m m o l e$
$i) m . e q o f S_{2} O_{3}^{2 -} = m . e q o f C r_{2} O_{7}^{2 -}$
$1 \times 8 = 0.1 \times 6 \times V$
$V = \frac{80}{6}$
$i i) m . e q o f S_{2} O_{3}^{2 -} = m . e q o f I_{2}$
$1 \times 1 = 0.1 \times 2 \times V \Rightarrow V_{I_{2}} = 5 m l$
$i i i) m . e q o f S_{2} O_{3}^{2 -} = m . e q o f M n O_{4}^{-}$
$1 \times 8 = 0.1 \times 3 \times V \Rightarrow V = \frac{80}{3}$
and also wt of $K_{2} C r_{2} O_{7} = 8 \times \frac{294}{6} \times 10^{- 3} = 0.392 g r$
wt of $I_{2} = 1 \times \frac{254}{2} \times 10^{- 3} = 0.127 g r$
wt of $K M n O_{4} = 8 \times \frac{158}{3} \times 10^{- 3} = 1.264 g r$