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Standard XII

Chemistry

Concentration Terms

An equal volu...

Question

An equal volume of a reducing agent is titrated separately with $1 M$ $K M n O_{4}$ in acidic, neutral and alkaline medium. The volumes of $K M n O_{4}$ required are $20$ mL in acidic, $33.3$ mL in neutral and $100$ mL in alkaline medium. Find out the volume of $1$ M $K_{2} C r_{2} O_{7}$ consumed if the same volume of the reducing agent is titrated in acidic medium.

V = 16.7

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V = 33.4

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V = 50.1

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V = 22.6

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Solution

The correct option is C $V = 16.7$ mL
In acidic medium,
$1 m o l K M n O_{4} \equiv 5 m o l e^{-}$
$1 m o l K_{2} C r_{2} O_{7} \equiv 6 m o l e^{-}$
Thus, $5 m o l K_{2} C r_{2} O_{7} \equiv 6 m o l K M n O_{4}$
Thus, $\frac{5}{6} m o l K_{2} C r_{2} O_{7} \equiv 1 m o l K M n O_{4}$
Thus, $20$ mL $K M n O_{4} = \frac{5}{6} \times 20 = 16.67$ mL $K_{2} C r_{2} O_{7}$ .

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Similar questions

For titrating certain volume of a reducing substance by

1 M

K M n O_{4}

, it is found that

20

mL was used in acidic medium,

33.4

mL was used in alkaline medium and

100

mL was used in neutral medium.

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

is reduced to

M n^{2 +}

in acidic Medium.

To what oxidation state is

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

reduced to in alkaline medium?

Q. An equal volume of a reducing agent is treated separately with

1 M K M n O_{4}

in acid, neutral and alkaline media. The volume of

K M n O_{4}

required is

20 m L

in acid,

33.4 m L

in neutral and

100 m L

in alkaline medium. Find out the oxidation state of manganese in each reaction product. Give the balanced equations for all the three half reactions. Find out the volume of

1 M K_{2} C r_{2} O_{7}

consumed, if the same volume of the reducing agent is treated in acid medium.

Q. In acidic medium, 100 ml of 1 M

K M n O_{4}

solution oxidizes 100 ml of

H_{2} O_{2}

solution. The volume of 0.01 M

K M n O_{4}

, required to oxidize the same volume of

H_{2} O_{2}

in alkaline medium in ml is :

100

mL of

0.01 M

K M n O_{4}

oxidised

100

H_{2} O_{2}

in acidic medium. The volume of same

K M n O_{4}

required in strong alkaline medium to oxidise

100

mL of same

H_{2} O_{2}

is :

100

mL of

0.01 M

K M n O_{4}

oxidized

100

mL of

H_{2} O_{2}

in acidic medium. The volume of the same

K M n O_{4}

required in alkaline medium to oxidize

100

mL of the same

H_{2} O_{2}

will be :

(

M n O_{4}^{⊝}

changes to

M n^{2 +}

in acidic medium and to

M n O_{2}

in alkaline medium)

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The correct option is C V=16.7 mLIn acidic medium,1mol KMnO4≡5mole−1mol K2Cr2O7≡6mole−Thus, 5molK2Cr2O7≡6molKMnO4Thus, 56mol K2Cr2O7≡1molKMnO4Thus, 20 mL KMnO4=56×20=16.67 mL K2Cr2O7.