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Charge and Current

Cl2g+S2O32-⟶ ...

Question

$C l_{2} (g) + S_{2} O_{3}^{2 -} ⟶ S O_{4}^{2 -} + C l^{-} + S$

If $50.0$ mL of $0.01 M N a_{2} S_{2} O_{3}$ solution and $5 \times 10^{- 4}$ moles of $C l_{2}$ react according to the given equation then what is the molarity of $N a_{2} S O_{4}$ in this solution?

A

0.080 M

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B

0.040 M

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C

0.020 M

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D

0.010 M

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Solution

The correct option is B $0.010 M$
$1$ mole of $C l_{2}$ (oxidation state $= 0$ ) changes to $2$ moles of $C l^{-}$ (oxidation state $= - 1$ ).

Net change in oxidation state $= 2 \times 1 - 2 \times 0 = 2$

$1$ mole of $N a_{2} S_{2} O_{3}$ (average oxidation state of $S$ is $2$ ) changes to $1$ mole of $N a S O_{4}$ (oxidation state of $S$ is $6$ ) and $1$ mole of $S$ .

Net change in oxidation state $= 1 \times 61 \times 0 - 2 \times 2 = 2$

Hence, $1$ mole of $C l_{2}$ oxidizes $1$ mole of $N a_{2} S_{2} O_{3}$ .

Hence, moles of sulphate formed $= \frac{5 \times 10^{- 4}}{1} = 5 \times 10^{- 4}$

Molarity of sulphate $= \frac{5 \times 10^{- 4}}{0.05} = 0.01 M$

Hence, the correct option is $(D)$

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Similar questions

C l_{2} (g) + S_{2} O_{3}^{2 -} ⟶ S O_{4}^{2 -} + C l^{-} + S

50.0

0.01 M N a_{2} S_{2} O_{3}

5 \times 10^{- 4}

C l_{2}

react according to given equation then how many moles of

S_{2} O_{3}^{2 -}

are in the above sample?

C l_{2} (g) + S_{2} O_{3}^{2 -} ⟶ S O_{4}^{2 -} + C l^{-} + S

50.0

0.01 M N a_{2} S_{2} O_{3}

5 \times 10^{- 4}

C l_{2}

react according to the given equation then how many millimoles of the reducing agent are left?

Q. The reaction :

C l_{2} (g) + S_{2} O_{3}^{2 -} ⟶ S O_{4}^{2 -} + C l^{-}

is to be carried out in basic medium. Starting with

0.15

C l_{2}

0.010

S_{2} O_{3}^{2 -}

0.30

O H^{-}

, how many moles of

O H^{-}

will be left in solution after the reaction is complete? Assume no other reaction occurs.

Q. A solution of

0.1 M K M n O_{4}

is used for the reaction,

S_{2} O_{3}^{2 -} + 2 M n O_{4}^{-} + H_{2} O ⟶ M n O_{2} + S O_{4}^{2 -} + O H^{-}

. What volume of solution in mL will be required to react with

0.158 g

N a_{2} S_{2} O_{3}

? (in mL as nearest integer)

0.1 M K M n O_{4}

is used for the following titration. What volume of the solution in

m L

will be required to react will

0.158 g

N a_{2} S_{2} O_{3}

?

{S_{2} O_{3}}^{2 -} + {M n O_{4}}^{-} + H_{2} O \to M n O_{2} (s) + {S O_{4}}^{2 -} + O H^{-}

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