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Standard XII

Chemistry

Molarity

A solution of...

Question

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 $20$ mL $H_{2} O_{2}$ solution is: (as nearest integer)

28

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20

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25

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Solution

The correct option is D $25$
The reaction is as follows:
$2 M n O_{4}^{-} + 5 H_{2} O_{2} + 6 H^{+} ⟶ 2 M n^{2 +} + 5 O_{2} + 8 H_{2} O$ .
As we know, $N_{1} V_{1} = N_{2} V_{2}$
$V = 24.8 \approx 25$ ml

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Similar questions

$A 20.0 m L$ solution containing $0.2 g$ impure $H_{2} O_{2}$ reacts completely with $0.316 g$ of $K M n O_{4}$ in acid solution. The purity of $H_{2} O_{2}$ (in%) is (Nearest integer) $(mol.wt.of H_{2} O_{2} = 34 mole.wt.of.KM n O_{4} = 158)$

Q. A

1

g sample of

H_{2} O_{2}

containing

x

percent

H_{2} O_{2}

by weight, requires

x

mL of

K M n O_{4}

solution for complete oxidation in acidic medium. Calculate the normality of

K M n O_{4}

solution as nearest integer.

20

mL of a solution containing

0.2

g of an impure sample of

H_{2} O_{2}

reacts with

0.316

g of

K M n O_{4}

(acidic medium). Calculate the volume of dry

O_{2}

evolved at

27^{\circ} C

and

750

mm of pressure (in mL as nearest integer).

20

mL of a solution containing

0.2

g of an impure sample of

H_{2} O_{2}

reacts with

0.316

g of

K M n O_{4}

(acidic medium). Calculate % purity of

H_{2} O_{2}

20

mL of a solution containing

0.2

g of impure sample of

H_{2} O_{2}

reacts with

0.316

g of

K M n O_{4}

in presence of

H_{2} S O_{4}

. What is the

%

purity of

H_{2} O_{2}

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A solution of KMnO4 containing 3 g/L is titrated with a solution of H2O2 containing 2 g/L.The volume (in ml) of KMnO4 required to react with 20 mL H2O2 solution is: (as nearest integer)

The correct option is D 25The reaction is as follows:2MnO−4+5H2O2+6H+⟶2Mn2++5O2+8H2O.As we know, N1V1=N2V2V=24.8≈25 ml

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 $20$ mL $H_{2} O_{2}$ solution is: (as nearest integer)

The correct option is D $25$
The reaction is as follows:
$2 M n O_{4}^{-} + 5 H_{2} O_{2} + 6 H^{+} ⟶ 2 M n^{2 +} + 5 O_{2} + 8 H_{2} O$ .
As we know, $N_{1} V_{1} = N_{2} V_{2}$
$V = 24.8 \approx 25$ ml