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Valency and Oxidation States

In the equati...

Question

In the equation potassium permanganate react with oxalic acid and hair 20 ml of 0.1 m Potassium permagnate is equivalent to:

A

20 ml of 0.5

H_{2} C_{2} O_{4}

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B

50 ml of 0.1 M

H_{2} C_{2} O_{4}

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C

50 ml of 0.5 M

H_{2} C_{2} O_{4}

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D

20 ml of 0.1 M

H_{2} C_{2} O_{4}

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Solution

The correct option is A 20 ml of 0.5 $H_{2} C_{2} O_{4}$
Normality = molarity x change in oxidation number
Normality of $0.1$ M $K M n O_{4}$ = $0.1 \times 5$ = $0.5$ N
Number of milli equivalents (meq)= Normality x volume in ml
Therefore
20ml of0.1M $K M n O_{4}$ =20ml 0.5N $K M n O_{4}$

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Similar questions

Q. Oxalic acid,

H_{2} C_{2} O_{4}

, reacts with paramagnet ion according to the balanced equation

5 H_{2} C_{2} O_{4} (a q) + 2 M n O_{4}^{-} (a q) ⇌ 2 M n^{2 +} (a q) + 10 C O_{2} (g) + 8 H_{2} O (l)

. The volume in mL of

0.0162

K M n O_{4}

solution required to react with

25.0

0.022

H_{2} C_{2} O_{4}

M n O_{4}^{-}

reacts with oxalate 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

20 m L

0.1 M M n O_{4}^{-}

is equivalent to -

Q. A mixture of

H_{2} C_{2} O_{4}

(oxalic acid) and

N a H C_{2} O_{4}

2.02 g

was dissolved in water and the solution made upto

1

10

mL of the solution required

3.0

0.1 N

sodium hydroxide solution for complete neutralisation. In another experiment,

10.0

mL of the same solution in hot dilute sulphuric acid medium required

4.0

0.1

N potassium permanganate solution for complete reaction. Calculate the amount of

H_{2} C_{2} O_{4}

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:

(H_{2} C_{2} O_{4})

is present in many plants and vegetables. If

24.0

0.01

K M n O_{4}

solution is needed to titrate

1.0

H_{2} C_{2} O_{4}

to the equivalence-point, the mass percentage (as nearest integer) of

H_{2} C_{2} O_{4}

in the sample is :

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