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Standard XII

Chemistry

Lewis Dot Structure

Volume of 0...

Question

Volume of $0.1 M$ oxalic acid, which can completely decolourise $20 m l$ of $0.1 M K M n O_{4}$ in acidic medium, is:

100 m l

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50 m l

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20 m l

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4 m l

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Solution

The correct option is B $50 m l$
Valency factor of oxalic acid $= 2$
Valency factor of $K M n O_{4} = 5$
Therefore, using stoichiometry:
$N V_{C O O H} = N V_{K M n O_{4}}$
$2 \times 0.1 \times V = 20 \times 0.1 \times 5$
$V = 50 m l$

So, the correct option is $B$ .

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Similar questions

K M n O_{4}

reacts with oxalic acid according to the equation,

2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O

Here, 20 mL of 0.1 M

K M n O_{4}

is equivalent to:

A known mass of oxalic acid dihydrate is dissolved in water to make 100 ml solution. 10 ml of this solution is diluted to 100 ml. 20 mL of the solution requires 20 mL of 0.1 N NaOH for complete neutralisation. (eq. mass of oxalic acid dihydrate = 63). The mass of oxalic acid in gram in the original solution is

K M n O_{4}

reacts with oxalic acid according to the equation:

2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} ⟶ 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O

Here, 20 mL of 0.1 M

K M n O_{4}

is equivalent to:

To neutralise completely 20 ml of 0.1 M of aqueous solution of phosphoric acid the volume of 0.1 M aqueous KOH solution required is ??

Q. To neutralise completely

20 m L

0.1 M

aqueous solution of phosphorous acid

H_{3} {P O}_{3}

, the volume of

0.1 M

K O H

solution required is:

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Volume of 0.1M oxalic acid, which can completely decolourise 20ml of 0.1M KMnO4 in acidic medium, is:

The correct option is B 50mlValency factor of oxalic acid =2Valency factor of KMnO4=5Therefore, using stoichiometry:NVCOOH=NVKMnO42×0.1×V=20×0.1×5V=50 ml So, the correct option is B.

Volume of $0.1 M$ oxalic acid, which can completely decolourise $20 m l$ of $0.1 M K M n O_{4}$ in acidic medium, is:

The correct option is B $50 m l$
Valency factor of oxalic acid $= 2$
Valency factor of $K M n O_{4} = 5$
Therefore, using stoichiometry:
$N V_{C O O H} = N V_{K M n O_{4}}$
$2 \times 0.1 \times V = 20 \times 0.1 \times 5$
$V = 50 m l$

So, the correct option is $B$ .