Three different solutions of oxidizing agent ${KMnO}_{4}, K_{2} {Cr}_{2} O_{7}$ and $I_{2}$ is titrated separately with $0.158 gm$ of ${Na}_{2} S_{2} O_{3}$ .If molarity of each oxidizing agent is $0.1 M$ and reactions are:
I ${MnO}_{4}^{-} + S_{2} O_{3}^{2 -} ⟶ {MnO}_{2} + {SO}_{4}^{2 -}$
II ${CrO}_{7}^{2 -} + S_{2} O_{3}^{2 -} ⟶ {Cr}^{3 +} + {SO}_{4}^{2 -}$
III $I_{2} + S_{2} O_{3}^{2 -} ⟶ S_{4} O_{6}^{2 -} + I^{-}$
Then:

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gram equivalent of ${Na}_{2} S_{2} O_{3}$ are same in all the reactions.

volume of ${KMnO}_{4}$ used is maximum

volume of iodine used is maximum

weight of $I_{2}$ used in titration is maximum

answer is A, B.

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Detailed Solution

Three different solutions of oxidizing agent ${KMnO}_{4}, K_{2} {Cr}_{2} O_{7}$ and $I_{2}$ is titrated separately with $0.158 gm$ of ${Na}_{2} S_{2} O_{3}$

No. of moles of ${Na}_{2} S_{2} O_{3}$ $= \frac{0.158}{158} = 10^{- 3}$

I. $0.1 \times V_{1} \times 3 = 10^{- 3} \times 8$

II. $0.1 \times V_{2} \times 6 = 10^{- 3} \times 8$

III. $0.1 \times V_{3} \times 2 = 10^{- 3} \times 1$

Hence, the correct option is a and b.

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