A solution of H2O2 is titrated against a solution of KMnO4-The reaction is- 2MnO4- - 5H2O2 - 6H- 2Mn2- - 5O2 - 8H2O-If it requires 46-9 mL of 0-145 M KMnO4 to oxidise 20 g of H2O2- the mass percentage of H2O2 in this solution is-

The correct option is C 2.9 Number of moles of KMnO_4 = MV1000 = 0.145× 46.91000 = 6.8× 10^ 3 Number of moles of H_2O_2 = 6.8× 10^ 3× 2.5 = ...

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Standard XII

Chemistry

Heat of Reaction

A solution of...

Question

A solution of $H_{2} O_{2}$ is titrated against a solution of $K M n O_{4}$ .
The reaction is, $2 {M n O_{4}}^{-} + 5 H_{2} O_{2} + 6 H^{+} \to 2 M n^{2 +} + 5 O_{2} + 8 H_{2} O$ .
If it requires $46.9 m L$ of $0.145 M K M n O_{4}$ to oxidise $20 g$ of $H_{2} O_{2}$ , the mass percentage of $H_{2} O_{2}$ in this solution is___________.

2.9

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Solution

The correct option is C $2.9$
Number of moles of $K M n O_{4} = \frac{M V}{1000} = \frac{0.145 \times 46.9}{1000}$
$= 6.8 \times 10^{- 3}$
Number of moles of $H_{2} O_{2} = 6.8 \times 10^{- 3} \times 2.5 = 0.017$
Mass of $H_{2} O_{2} = 0.017 \times 34 = 0.578$
Mass % of $H_{2} O_{2} = \frac{0.578}{20} \times 100 = 2.9$ .

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Similar questions

20

g sample of

H_{2} O_{2}

is oxidized by

46.9

mL of

0.145

K M n O_{4}

following the change:

2 M n O_{4}^{-} + 5 H_{2} O_{2} + 6 H^{+} ⟶ 2 M n^{2 +} + 5 O_{2} + 8 H_{2} O

The mass

%

H_{2} O_{2}

in the sample is :

Q. The reaction between

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and

K M n O_{4}

2 K M n O_{4} + 3 H_{2} S O_{4} + 5 H_{2} O_{2} \to K_{2} S O_{4} + 2 M n S O_{4} + 8 H_{2} O + 5 O_{2}

In a reaction excess of

H_{2} O_{2}

is added to 0.1 mole of acidified

K M n O_{4}

solution. Then the STP volume of

O_{2}

liberated is:

Q. A bottle of

H_{2} O_{2}

is labelled as

10

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112

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H_{2} O_{2}

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K M n O_{4}

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Q. A solution of

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3

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A solution of H2O2 is titrated against a solution of KMnO4.The reaction is, 2MnO4−+5H2O2+6H+→2Mn2++5O2+8H2O.If it requires 46.9 mL of 0.145 M KMnO4 to oxidise 20 g of H2O2, the mass percentage of H2O2 in this solution is___________.

The correct option is C 2.9Number of moles of KMnO4=MV1000=0.145×46.91000=6.8×10−3Number of moles of H2O2=6.8×10−3×2.5=0.017Mass of H2O2=0.017×34=0.578Mass % of H2O2=0.57820×100=2.9.