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In basic medium ${CrO}_{4}^{2 -}$ oxidises $S_{2} O_{3}^{2 -}$ to form ${SO}_{4}^{2 -}$ and itself changes into $Cr {(OH)}_{4}^{-}$ . The volume of 0.154 M ${CrO}_{4}^{2 -}$ required to react with 40 mL of 0.25 M $S_{2} O_{3}^{2 -}$ is_______mL (Rounded off to the nearest integer)

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Detailed Solution

$\overset{+ 6}{C} r O_{4}^{2 -} + {\overset{+ 2}{S}}_{2} O_{3}^{2 -} \to \overset{+ 6}{S} O_{4}^{2 -} + \overset{+ 3}{C} r {(O H)}_{4}^{-}$

$g m e q u i . o f C r O_{4}^{2 -} = S_{2} O_{3}^{2 -}$

$\begin{array}{l} 0.14 \times 3 \times v = 0.25 \times 40 \times 8 \\ v = 173.16 = 173 ml \end{array}$

Hence answer is (173)

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