In basic medium- CrO2-4 reacts with S2O2-3 resulting in the formation of CrOH-4 and SO2-4- How many mL of 0-1 M Na2CrO4 is required to react with 40 mL of 0-25 M Na2S2O3-

The correct option is D 266.67 mLThe redox changes are given below: S_2O^2 _3→ 2SO^2 _4+8e^ 3e^ +CrO^2 _4→ [Cr(OH)_4]^⊝ CrO^2 _4≡ S_2O^2 _ ...

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Standard XII

Chemistry

Introduction to Oxidation and Reduction

In basic medi...

Question

In basic medium, $C r O_{4}^{2 -}$ reacts with $S_{2} O_{3}^{2 -}$ resulting in the formation of $C r (O H)_{4}^{⊝}$ and $S O_{4}^{2 -}$ . How many mL of $0.1 M$ $N a_{2} C r O_{4}$ is required to react with $40$ mL of $0.25 M$ $N a_{2} S_{2} O_{3}$ ?

240.2

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24.02

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266.67

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26.67

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Solution

The correct option is D $266.67$ mL
The redox changes are given below:
$S_{2} O_{3}^{2 -} \to 2 S O_{4}^{2 -} + 8 e^{-}$
$3 e^{-} + C r O_{4}^{2 -} \to [C r (O H)_{4}]^{⊝}$
$C r O_{4}^{2 -} \equiv S_{2} O_{3}^{2 -}$
mEq. of $C r O_{4}^{2 -}$ $=$ mEq. of $S_{2} O_{3}^{2 -}$
$V \times 0.1 \times 3 \equiv 40 \times 0.25 \times 8$
$V = 266.67$ mL

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In basic medium, CrO2−4 reacts with S2O2−3 resulting in the formation of Cr(OH)⊝4 and SO2−4. How many mL of 0.1 M Na2CrO4 is required to react with 40 mL of 0.25 M Na2S2O3?

The correct option is D 266.67 mLThe redox changes are given below:S2O2−3→2SO2−4+8e−3e−+CrO2−4→[Cr(OH)4]⊝CrO2−4≡S2O2−3mEq. of CrO2−4 = mEq. of S2O2−3V×0.1×3≡40×0.25×8V=266.67 mL

In basic medium, $C r O_{4}^{2 -}$ reacts with $S_{2} O_{3}^{2 -}$ resulting in the formation of $C r (O H)_{4}^{⊝}$ and $S O_{4}^{2 -}$ . How many mL of $0.1 M$ $N a_{2} C r O_{4}$ is required to react with $40$ mL of $0.25 M$ $N a_{2} S_{2} O_{3}$ ?