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Byju's Answer

Standard XII

Chemistry

Introduction

A bottle of ...

Question

A bottle of $H_{2} O_{2}$ is labelled as $10$ vol $H_{2} O_{2}$ . $112$ mL of this solution of $H_{2} O_{2}$ is titrated against $0.04 M$ acidified solution of $K M n O_{4}$ . The volume of $K M n O_{4}$ (in L) is :

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Solution

The correct option is A $1$
$5.6 V H_{2} O_{2} = 1 N$
$10 V H_{2} O_{2} = \frac{10}{5.6} N$
Milliequivalents of $H_{2} O_{2} =$ Milliequivalents of $M n O_{4}^{⊝}$
$\frac{10}{5.6} \times 112 \equiv 0.04 \times 5$ ( $n$ factor) $\times V$
$V \equiv 1000$ mL $= 1$ L

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Similar questions

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)

Q. Decomposition of

H_{2} O_{2}

is a first order reaction. A solution of

H_{2} O_{2}

labelled as 20 volumes was left open. Due to this, some

H_{2} O_{2}

decomposed. To determine the new volume strength after 6 hours, 10 mL of this solution was diluted to 100 mL. 10 mL of this diluted solution was titrated against 25 mL of 0.025 N

K M n O_{4}

solution under acidic conditions. The rate constant for decomposition of

H_{2} O_{2}

is:

Q. 50 ml of

H_{2} O_{2}

solution when titrated with 0.1 M

K M n O_{4}

solution in acidic medium, volume required is 40 ml. Volume strength of

H_{2} O_{2}

solution is

Q. Decomposition of

H_{2} O_{2}

is a first order reaction. A solution of

H_{2} O_{2}

labelled as

20

volumes was left open. Due to this some

H_{2} O_{2}

decomposed. To determine the new volume strength after

6

hours,

10 m L

of this solution was diluted to

100 m L . 10 m L

of this diluted solution was titrated against

25 m L

0.025 M K M n O_{4}

acidified solution. Calculate the rate constant for decomposition of

H_{2} O_{2}

10

mL of a solution of

H_{2} O_{2}

10

volume strength decolourises

100

mL of

K M n O_{4}

solution acidified with dil.

H_{2} S O_{4}

. The amount of

K M n O_{4}

in the given solution is (

K = 39, M n = 55

) :

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A bottle of H2O2 is labelled as 10 vol H2O2. 112 mL of this solution of H2O2 is titrated against 0.04 M acidified solution of KMnO4. The volume of KMnO4 (in L) is :

The correct option is A 15.6 V H2O2=1 N10 V H2O2=105.6NMilliequivalents of H2O2= Milliequivalents of MnO⊝4105.6×112≡0.04×5 (n factor) ×VV≡1000 mL =1 L

A bottle of $H_{2} O_{2}$ is labelled as $10$ vol $H_{2} O_{2}$ . $112$ mL of this solution of $H_{2} O_{2}$ is titrated against $0.04 M$ acidified solution of $K M n O_{4}$ . The volume of $K M n O_{4}$ (in L) is :

The correct option is A $1$
$5.6 V H_{2} O_{2} = 1 N$
$10 V H_{2} O_{2} = \frac{10}{5.6} N$
Milliequivalents of $H_{2} O_{2} =$ Milliequivalents of $M n O_{4}^{⊝}$
$\frac{10}{5.6} \times 112 \equiv 0.04 \times 5$ ( $n$ factor) $\times V$
$V \equiv 1000$ mL $= 1$ L