Cl2g-S2O32- SO42-Cl-SIf 50-0 mL of 0-01 M Na2S2O3 solution and 5- 10-4 moles of Cl2 react according to given equation then how many moles of S2O32- are in the above sample-

The correct option is A 0.00050 Given,Molarity of Na_2S_2O_3 is 0.01 M Volume of #160; Na_2S_2O_3 is 50 mL No of mole of Cl_2 ...

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Byju's Answer

Standard XII

Chemistry

Order of Reaction

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 given equation then how many moles of $S_{2} O_{3}^{2 -}$ are in the above sample?

0.00050

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Solution

The correct option is A $0.00050$
Given,
Molarity of $N a_{2} S_{2} O_{3}$ is $0.01 M$
Volume of $N a_{2} S_{2} O_{3}$ is $50 m L$
No of mole of $C l_{2} = 5 \times 10^{- 4}$
The number of moles of $S_{2} O_{3}^{2 -} = Volume (in L) \times molarity = \frac{50.0}{1000} \times 0.01 = 0.0005$ moles.

Hence, the correct option is $(A)$ .

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Similar questions

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

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 how many millimoles of the reducing agent are left?

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

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?

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

moles of

C l_{2}

0.010

moles of

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

and

0.30

moles of

O H^{-}

, how many moles of

O H^{-}

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

Q. In basic medium,

C r O_{4}^{2 -}

oxidises

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

to form

C r (O H)_{4}^{-}

and

S O_{4}^{2 -}

. How many mL(nearest integer) of

0.154 M

C r O_{4}^{2 -}

are required to react with

40.0 mL

0.246 M

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

[Hint : 0.0154 M = 0.154 \times 3 N C r O_{4}^{2 -} and 0.246 M = 0.246 \times 8 N S_{2} O_{3}^{2 -}]

Q. In basic solution,

{C r O_{4}}^{2 -}

oxidises

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

to form

{C r (O H)_{4}}^{-}

and

{S_{2} O_{4}}^{2 -}

. How many mL (nearest integer) of

0.154

{C r O_{4}}^{2 -}

are required to react with

40.0

mL of

0.246

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

?
[Hint :

0.0154

= 0.154 \times 3

{C r O_{4}}^{2 -}

and

0.246

= 0.246 \times 8

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

]

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Cl2(g)+S2O2−3⟶SO2−4+Cl−+SIf 50.0 mL of 0.01 M Na2S2O3 solution and 5×10−4 moles of Cl2 react according to given equation then how many moles of S2O2−3 are in the above sample?

The correct option is A 0.00050Given,Molarity of Na2S2O3 is 0.01MVolume of Na2S2O3 is 50 mLNo of mole of Cl2=5×10−4The number of moles of S2O2−3=Volume (in L)×molarity=50.01000×0.01=0.0005 moles.Hence, the correct option is (A).

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

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 given equation then how many moles of $S_{2} O_{3}^{2 -}$ are in the above sample?