In basic medium ${C r O}_{4}^{2 -}$ oxidises $S_{2} O_{3}^{2 -}$ to form ${S O}_{4}^{2 -}$ and itself changes into $C r (O H)_{4}^{-}$ . The volume of $0.154 M {C r O}_{4}^{2 -}$ required to react with $40 m L$ of $0.25 M S_{2} O_{3}^{2 -}$ is $m L$ (Rounded off to the nearest integer)

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answer is 173.

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Detailed Solution

Given, Molarity of ${C r O}_{4}^{- 2} = 0.154 M$

Volume of $S_{2} O_{3}^{- 2} = = 40 m l$

Molarity of $S_{2} O_{3}^{- 2} = = 0.25 M$

Let the volume of ${C r O}_{4}^{- 2} = V$

Normality of ${C r O}_{4}^{- 2} = 30.154$

Normality of $S_{2} O_{3}^{- 2} = = 80.25$

$N_{1} V_{1} = N_{2} V_{2}$

$V \times 0.154 \times 3 = 0.25 \times 8 \times 40$

V = 173ml

Therefore, volume of ${C r O}_{4}^{- 2}$ is 173ml.

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