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Standard XII

Chemistry

Stoichiometry

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 the volume of dry $O_{2}$ evolved at $27^{\circ} C$ and $750$ mm of pressure (in mL as nearest integer).

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Solution

The redox changes are as follows:
$M n^{7 +} + 5 e ⟶ M n^{2 +}$

$(O^{-})_{2} ⟶ O_{2} + 2 e$ $[E_{O_{2}} = \frac{M}{2}]$
Eq. of $O_{2} =$ Eq. of $K M n O_{4}$

$\frac{w}{\frac{32}{2}} = \frac{0.316 \times 5}{158}$ $⟹ w_{O_{2}} = 0.16$ g

We know, $P V = \frac{w}{M} R T$ .

Substituting the values, we get

$\frac{750}{760} \times V = \frac{0.16}{32} \times 0.0821 \times 300$

$∴ V_{O_{2}} = 0.12479$ L $= 124.79$ mL
So, the nearest integer is $125$ mL.

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20

mL of a solution containing

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reacts with

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g of

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of a solution containing

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(i) Purity of

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(ii) Volume of dry

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

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reacts with

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in presence of

H_{2} S O_{4}

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H_{2} O_{2}

Q. A

1

g sample of

H_{2} O_{2}

containing

x

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

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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 the volume of dry O2 evolved at 27∘C and 750 mm of pressure (in mL as nearest integer).

The redox changes are as follows:Mn7++5e⟶Mn2+(O−)2⟶O2+2e [EO2=M2]Eq. of O2= Eq. of KMnO4w322=0.316×5158 ⟹wO2=0.16 gWe know, PV=wMRT.Substituting the values, we get750760×V=0.1632×0.0821×300∴VO2=0.12479 L =124.79 mLSo, the nearest integer is 125 mL.

$20$ 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).