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)

Question

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) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

17 H 2 O +8 CrO 42-+3 S 2 O 32- longrightarrow 6 SO 42-+8 Cr ( OH +2 OH -. Applying mole-mole analysis (0 . 154 × V/8)=(40 × 0 . 25/3) V=173mL

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

$17 H_{2} O + 8 C r O_{4}^{2 -} + 3 S_{2} O_{3}^{2 -} ⟶ 6 S O_{4}^{2 -} + 8 C r (O H + 2 O H^{-}$
Applying mole-mole analysis
$\frac{0.154 \times V}{8} = \frac{40 \times 0.25}{3}$
$V = 173 m L$

Q. In basic medium CrO42−​ oxidizes S2​O32−​ to form SO42−​ and itself changes into Cr(OH)4−​ . The volume of 0.154MCrO42−​ required to react with 40mL of 0.25MS2​O32−​ is in mL . (Rounded-off to the nearest integer)

17H2​O+8CrO42−​+3S2​O32−​⟶6SO42−​+8Cr(OH+2OH− Applying mole-mole analysis 80.154×V​=340×0.25​ V=173mL

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

$17 H_{2} O + 8 C r O_{4}^{2 -} + 3 S_{2} O_{3}^{2 -} ⟶ 6 S O_{4}^{2 -} + 8 C r (O H + 2 O H^{-}$
Applying mole-mole analysis
$\frac{0.154 \times V}{8} = \frac{40 \times 0.25}{3}$
$V = 173 m L$