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What is the normality of \[{\text{KMn}}{{\text{O}}_{\text{4}}}\] solution against \[{\text{M}}{{\text{n}}^{2 + }}\]? The Molarity of \[{\text{KMn}}{{\text{O}}_{\text{4}}}\] solution is 0.025M.
A) \[{\text{0}}{\text{.125 N}}\]
B) \[{\text{0}}{\text{.1 N}}\]
C) \[{\text{0}}{\text{.01 N}}\]
D) \[{\text{0}}{\text{.156 N}}\]

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Last updated date: 13th Jun 2024
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Answer
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Hint: Determine the n factor by calculating the decrease in the oxidation number of manganese. Then use the formula \[{\text{Normality = Molarity }} \times {\text{ n factor}}\] to convert molarity into normality.

Complete step by step answer:
The decrease in the oxidation number is the reduction. In a redox reaction, an oxidizing agent oxidizes other substances. A species, whose oxidation number decreases during the reaction, is known as an oxidizing agent. An oxidizing agent itself undergoes reduction. For an oxidizing agent, the n factor gives the decrease in the oxidation number.
For manganese, determine the n factor from the decrease in the oxidation number.
The oxidation number of manganese in \[{\text{KMn}}{{\text{O}}_{\text{4}}}\] is +7.
The oxidation number of manganese in \[{\text{M}}{{\text{n}}^{2 + }}\] is +2.
When \[{\text{KMn}}{{\text{O}}_{\text{4}}}\] is reduced to \[{\text{M}}{{\text{n}}^{2 + }}\] ions, the oxidation number of manganese decreases from +7 to +5. The decrease in the oxidation number of manganese is 5. Hence, the n factor is 5.
The molarity of \[{\text{KMn}}{{\text{O}}_{\text{4}}}\] solution is \[{\text{0}}{\text{.025 M}}\].
Use the formula \[{\text{Normality = Molarity }} \times {\text{ n factor}}\] to convert molarity into normality.
\[{\text{Normality = Molarity }} \times {\text{ n factor}} \\
{\text{Normality = 0}}{\text{.025 M }} \times {\text{ 5}} \\
{\text{Normality = 0}}{\text{.125 N}} \\\]
Hence, the normality of \[{\text{KMn}}{{\text{O}}_{\text{4}}}\] solution is \[{\text{0}}{\text{.125 N}}\].

Hence, the option A )\[{\text{0}}{\text{.125 N}}\] is the correct option.

Additional information: n factor can also be defined as the ratio of the normality to the molarity.

Note: For an oxidizing agent, the equivalent weight is the ratio of the molecular weight to the number of electrons gained. For an oxidizing agent, the equivalent weight is the ratio of the molecular weight to the decrease in the oxidation number.
The number of gram equivalents is the ratio of the mass of oxidising agents to its equivalent weight. Normality is the number of gram equivalents present in one litre of solution.