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Byju's Answer

Standard XII

Chemistry

Molarity

A solution of...

Question

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 $0.158 g$ of $N a_{2} S_{2} O_{3}$ ? (in mL as nearest integer)

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Solution

The redox changes are as follows:
$(S^{2 +})_{2} ⟶ 2 S^{6 +} + 8 e$ $[E_{N a_{2} S_{2} O_{3}} = \frac{M}{8}]$
$M n^{7 +} + 3 e ⟶ M n^{4 +}$ ('n' factor for $K M n O_{4} = 3$ )
$∴ M e q . o f K M n O_{4} = M e q . o f S_{2} O_{3}^{2 -}$
$0.1 \times 3 \times V = \frac{0.158}{\frac{158}{8}} \times 1000$

$∴ V = 26.67 m L$
So, the nearest integer value is $27$ mL.

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Similar questions

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^{-}

Q. A solution of

K M n O_{4}

containing

3

g/L is titrated with a solution of

H_{2} O_{2}

containing

2

g/L.

The volume (in ml) of

K M n O_{4}

required to react with

20

H_{2} O_{2}

solution is: (as nearest integer)

150 g

of a solution

(d = 1.2 g / m l)

contains

63 g

H N O_{3}

. What volume of this

H N O_{3}

(in mL) solution will be required to react with a

N a O H

solution, containing

24 g

N a O H

?
Reaction:

H N O_{3} + N a O H \to N a N O_{3} + H_{2} O

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

required to react with

40 m L

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

is ______

m L

.
(Rounded off to the nearest integer)

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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A solution of 0.1MKMnO4 is used for the reaction, S2O2−3+2MnO−4+H2O⟶MnO2+SO2−4+OH−. What volume of solution in mL will be required to react with 0.158g of Na2S2O3? (in mL as nearest integer)

The redox changes are as follows:(S2+)2⟶2S6++8e [ENa2S2O3=M8]Mn7++3e⟶Mn4+ ('n' factor for KMnO4=3)∴Meq.ofKMnO4=Meq.ofS2O2−30.1×3×V=0.1581588×1000∴V=26.67mLSo, the nearest integer value is 27 mL.

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 $0.158 g$ of $N a_{2} S_{2} O_{3}$ ? (in mL as nearest integer)