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Standard XII

Mathematics

Right Hand Limit

20 g sample o...

Question

$20$ g sample of $H_{2} O_{2}$ is oxidized by $46.9$ mL of $0.145$ M $K M n O_{4}$ 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 $%$ of $H_{2} O_{2}$ in the sample is :

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2.9

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1.95

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4.93

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Solution

The correct option is B 2.9
$\begin{matrix} meq of {K M n O}_{4} = 0.145 \times 5 \times 46.9 = 34 meq of H_{2} O_{2} = 34 \begin{matrix} {(0^{- 1})}_{2}^{- 1} \to O_{2}^{\circ} + 2 e \frac{ω}{17} & \times 1000 = 34 ω_{H_{2} O_{2}} = 0.578 % H_{2} O_{2} = \frac{0.578}{20} \times 100 = 2.89 \end{matrix} \end{matrix}$

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Similar questions

Q. 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

0.145 M K M n O_{4}

to oxidise

20 g

H_{2} O_{2}

, the mass percentage of

H_{2} O_{2}

in this solution is___________.

Q. In the following reaction,

O^{18}

is in what?

2 M n O_{4}^{-} + 5 H_{2} O_{2}^{18} + 6 H^{+} \to 2 M n^{2 +} + 8 H_{2} O + 5 O_{2}

3.4

g sample of

H_{2} O_{2}

solution containing

x %

H_{2} O_{2}

by weight requires

x

mL of

K M n O_{4}

solution for complete oxidation under acidic condition. The normality of

K M n O_{4}

solution is :

Q. A

3.4 g

sample of

H_{2} O_{2}

solution that has

x % H_{2} O_{2}

by mass requires

x mL

K M n O_{4}

solution for complete oxidation under acidic conditions. The molarity of the

K M n O_{4}

solution is :

Q. The reaction between

H_{2} O_{2}

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:

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20 g sample of H2O2 is oxidized by 46.9 mL of 0.145 M KMnO4 following the change: 2MnO−4+5H2O2+6H+⟶2Mn2++5O2+8H2OThe mass % of H2O2 in the sample is :

The correct option is B 2.9 meq of KMnO4=0.145×5×46.9=34 meq of H2O2=34(0−1)2→O∘2+2eω17×1000=34ωH2O2=0.578%H2O2=0.57820×100=2.89

$20$ g sample of $H_{2} O_{2}$ is oxidized by $46.9$ mL of $0.145$ M $K M n O_{4}$ 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 $%$ of $H_{2} O_{2}$ in the sample is :