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Standard XII

Chemistry

Normality

How many mill...

Question

How many millilitres of 0.02 M $K M n O_{4}$ solution would be required to exactly titrate 25.0 mL of 0.2 M $F e (N O_{3})_{2}$ solution?

50.00 m L

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70.00 m L

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8 0.00 m L

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None of these

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Solution

The correct option is A $50.00 m L$
n factor of $K M n O_{4}$ ,

Equivalents of $K M n O_{4} = 5 \times$ moles of $K M n O_{4}$
Normality $K M n O_{4} = 5 \times$ molarity of $K M n O_{4}$

For $F e (N O_{3})_{2},$ moles and equivalents are equal, as n factor is 1.

At the equivalence point,

Equivalents of $K M n O_{4}$ = Equivalents of $F e (N O_{3})_{2}$

or $V_{K M n O_{4}} \times$ Normality of $K M n O_{4}$
$= V_{F e (N O_{3})_{2}} \times$ Normality of $F e (N O_{3})_{2}$

For 0.02 M $K M n O_{4}$ solution.

Normality of $K M n O_{4} = (5) (0.02000) = 0.1$ N

and for 0.20 M $F e (N O_{3})_{2}$ solution
Normality of $F e (N O_{3})_{2} = 0.20$ N

putting the values,
$V_{K M n O_{4}} = (25.00 m L) (\frac{0.2000}{0.1000}) = 50.00 m L$

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Similar questions

Q. How many millilitres of 0.02 M

K M n O_{4}

solution would be required to exactly titrate 25.0 mL of 0.2 M

F e (N O_{3})_{2}

solution?

Q. A

0.46

g sample of

A s_{2} O_{3}

required

25.0

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solution for its titration. The molarity of

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Q. 25 ml of the given

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solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this

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How many millilitres of 0.1 M $H_{2} S O_{4}$ must be added to 50 mL of 0.1 M $N a O H$ to give a solution that has a concentration of 0.05 M in $H_{2} S O_{4}$ ?

The following titration method is given to determine the total content of the species with variable oxidation states. Answer the question given at the end of it.

A quantity of

25.0

mL of solution containing both

F e^{2 +}

and

F e^{3 +}

ions is titrated with

25.0

mL of

0.02 M

K M n O_{4}

(in dilute

H_{2} S O_{4}

). As a result, all of the

F e^{2 +}

ions are oxidised to

F e^{3 +}

ions. Next

25

mL of the original solution is treated with

Z n

metal. Finally, the solution requires

40.0

mL of the same

K M n O_{4}

solution for oxidation to

F e^{3 +}

{M n O_{4}}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

0.02 M K_{2} C r_{2} O_{7}

is used instead of

0.02 M K M n O_{4}

, its volume required in these titrations will be respectively :

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Normality

Standard XII Chemistry

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How many millilitres of 0.02 M KMnO4 solution would be required to exactly titrate 25.0 mL of 0.2 M Fe(NO3)2 solution?

How many millilitres of 0.02 M $K M n O_{4}$ solution would be required to exactly titrate 25.0 mL of 0.2 M $F e (N O_{3})_{2}$ solution?