MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Mole Concept

0.1 M KMnO4 i...

Question

$0.1 M K M n O_{4}$ is used for the following titration. What volume of the solution in $m L$ will be required to react will $0.158 g$ of $N a_{2} S_{2} O_{3}$ ?
${S_{2} O_{3}}^{2 -} + {M n O_{4}}^{-} + H_{2} O \to M n O_{2} (s) + {S O_{4}}^{2 -} + O H^{-}$

Open in App

Solution

$S_{2} O_{3}^{2 -} \to 2 S O_{4}^{2 -}$
Change in oxidation number of sulphur per molecule of $S_{2} O_{3}^{2 -} = 2 \times (6 - 2) = 8$
Change in oxidation number of $M n$ per molecule of $M n O_{4}^{-} = 7 - 4 = 3$
No. of moles in $0.158 g$ of $N a_{2} S_{2} O_{3} = \frac{0.158}{158} = 1 \times 10^{- 3}$
No. of equivalents $= 8 \times 10^{- 3}$
Normality of $0.1 M K M n O_{4}$ solution $= 0.1 \times 3 = 0.3$
Let $V m L$ of volume of $K M n O_{4}$ be required; then
$\frac{V}{1000} \times 0.3 = 8 \times 10^{- 3}$
or $V = \frac{8}{0.3} \times 10^{- 3} = 10^{3} = 26.7 m L$ .

Suggest Corrections

thumbs-up

0

Similar questions

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)

M n^{2 +} (a q)

can be determined by titration with

{M n O_{4}}^{-} (a q)

.

M n^{2 +} + {M n O_{4}}^{-} + {O H}^{-} \to M n O_{2} + H_{2} O (unbalanced)

25.0

M n^{2 +}

34.77

0.05876 M K M n O_{4}

(aq) for its titration. The molarity of

M n^{2 +}

y \times 10^{- 4} M

, then y is ________.

150 g

(d = 1.2 g / m l)

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

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

Q. Given an equation:

C r O_{4}^{2 -} + S_{2} O_{3}^{2 -} + ⊝ O H \to [C r (O H)_{4}]^{-} + S O_{4}^{2 -}

0.2 M

N a_{2} C r O_{4}

solution is required to just react with

30

0.2 M

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

100

1 M K M n O_{4}

100

H_{2} O_{2}

in acidic medium (when

{M n O_{4}}^{-}

M n^{2 +}

). Volume of same

K M n O_{4}

required to oxidise

100

H_{2} O_{2}

in neutral medium (when

{M n O_{4}}^{-}

M n O_{2}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Mole Concept

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Standard X Chemistry

Join BYJU'S Learning Program

CrossIcon