1. Tardigrade
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  4. In basic medium CrO42 - oxidizes S2O32 - to form SO42 - and itself changes into Cr(.OH.)4- . The volume of 0.154MCrO42 - required to react with 40mL of 0.25MS2O32 - is in mL . (Rounded-off to the nearest integer)

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Q. In basic medium $CrO_{4}^{2 -}$ oxidizes $S_{2}O_{3}^{2 -}$ to form $SO_{4}^{2 -}$ and itself changes into $Cr\left(\right.OH\left.\right)_{4}^{-}$ . The volume of $0.154MCrO_{4}^{2 -}$ required to react with $40mL$ of $0.25MS_{2}O_{3}^{2 -}$ is in $mL$ . (Rounded-off to the nearest integer)

NTA AbhyasNTA Abhyas 2022

Solution:

$17 H _{2} O +8 CrO _{4}^{2-}+3 S _{2} O _{3}^{2-} \longrightarrow 6 SO _{4}^{2-}+8 Cr \left( OH +2 OH ^{-}\right.$
Applying mole-mole analysis
$\frac{0 . 154 \times V}{8}=\frac{40 \times 0 . 25}{3}$
$V=173mL$