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Standard VIII

Chemistry

Ways to Define Concentration

20 mL of a so...

Question

$20$ 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 $H_{2} O_{2}$ .

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Solution

The redox changes are:
$M n^{7 +} + 5 e ⟶ M n^{2 +}$
$(O^{-})_{2} ⟶ O_{2}^{0} + 2 e$ $[E_{O_{2}} = \frac{M}{2}]$
$∵ M e q . o f H_{2} O_{2} = M e q . o f K M n O_{4}$
$\frac{w \times 1000}{\frac{34}{2}} = \frac{0.316}{\frac{158}{5}} \times 1000$
$[∵ E_{K M n O_{4}} = \frac{158}{5}]$
$∴ w_{H_{2} O_{2}} = 0.17 g$
$∴ % p u r i t y o f s a m p l e H_{2} O_{2} = \frac{0.17 \times 100}{0.2} = 85 %$

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Similar questions

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}

$A 20.0 m L$ 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) $(mol.wt.of H_{2} O_{2} = 34 mole.wt.of.KM n O_{4} = 158)$

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 m L

of a solution containing

0.2 g

of impure sample of

H_{2} O_{2}

reacts with

0.316 g

K M n O_{4}

(acidic). Calculate:
(i) Purity of

H_{2} O_{2}

.
(ii) Volume of dry

O_{2}

evolved at

27^{\circ} C

and

750 m m P

Q. The purity of

H_{2} O_{2}

in a given sample is

85 %

. Calculate the weight of impure sample of

H_{2} O_{2}

which requires

10

mL of

\frac{M}{5} K M n O_{4}

solution in a titration in acidic medium.

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20 mL of a solution containing 0.2 g of an impure sample of H2O2 reacts with 0.316 g of KMnO4 (acidic medium). Calculate % purity of H2O2.

The redox changes are:Mn7++5e⟶Mn2+(O−)2⟶O02+2e [EO2=M2] ∵Meq.ofH2O2=Meq.ofKMnO4w×1000342=0.3161585×1000[∵EKMnO4=1585]∴wH2O2=0.17g∴%purityofsampleH2O2=0.17×1000.2=85%

$20$ 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 $H_{2} O_{2}$ .