Question
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)

Answer

$gm$ equi. of $CrO _{4}^{2-}= S _{2} O _{3}^{2-}$

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

$v =173.16=173 \,ml$

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