Answer
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Hint: A certain amount of milliequivalents of $KMn{{O}_{4}}$ will titrate the same amount of milliequivalents of oxalic acid. Also, remember that the n-factor of $KMn{{O}_{4}}$ varies in acidic and basic medium.
Step-by-Step Solution:
Let us first go through the properties of Potassium Permanganate ($KMn{{O}_{4}}$) as a compound.
Potassium permanganate is an inorganic compound with the chemical formula $KMnO_4$ and composed of K+ and $MnO^{4-}$. It is a purplish-black crystalline solid that dissolves in water to give intensely pink or purple solutions.
Potassium permanganate is widely used in the chemical industry and laboratories as a strong oxidizing agent, and also as a medication for dermatitis, for cleaning wounds, and general disinfection.
Having established the properties and usage of $KMn{{O}_{4}}$, let us now move on to the particulars of this question.
Now, we know that Oxalic Acid’s chemical formula is $COOH-COOH$ and due to its presence in the initial titration, the medium would thus be acidic.
However, the later titration is done against alkaline ${{H}_{2}}{{O}_{2}}$ where the medium would be alkaline in nature.
With all this information now at our disposal along with the hint, let us now proceed to the step-by-step solution:
\[\begin{align}
& Milliequivalents\text{ }of\text{ }oxalic\text{ }acid~=50\times 0.1\times 2=10 \\
& \because Initial\text{ }medium\text{ }is\text{ }acidic~(\Rightarrow n=5~for~KMnO4). \\
& \therefore ~Molarity\text{ }of~KMn{{O}_{4}}=\dfrac{N}{5}=\dfrac{\dfrac{10}{100}}{5}=0.02M \\
& In\text{ }alkaline\text{ }medium,~\text{ }n=3~for~KMn{{O}_{4}}~ \\
& Hence,\text{ }normality~=0.02\times 3~=0.06N \\
\end{align}\]
Therefore, by our calculations, we can safely conclude that the required normality of $KMn{{O}_{4}}$ is 0.06N making the answer to this question c) 0.06N.
Note: Absolute caution must be taken during the titrations of any acidic or alkaline substance, regardless of its strength, so as to avoid any form of injury or damage. Also make sure that no mistakes are made in the n- factor calculation of Potassium Permanganate w.r.t change in mediums.
Step-by-Step Solution:
Let us first go through the properties of Potassium Permanganate ($KMn{{O}_{4}}$) as a compound.
Potassium permanganate is an inorganic compound with the chemical formula $KMnO_4$ and composed of K+ and $MnO^{4-}$. It is a purplish-black crystalline solid that dissolves in water to give intensely pink or purple solutions.
Potassium permanganate is widely used in the chemical industry and laboratories as a strong oxidizing agent, and also as a medication for dermatitis, for cleaning wounds, and general disinfection.
Having established the properties and usage of $KMn{{O}_{4}}$, let us now move on to the particulars of this question.
Now, we know that Oxalic Acid’s chemical formula is $COOH-COOH$ and due to its presence in the initial titration, the medium would thus be acidic.
However, the later titration is done against alkaline ${{H}_{2}}{{O}_{2}}$ where the medium would be alkaline in nature.
With all this information now at our disposal along with the hint, let us now proceed to the step-by-step solution:
\[\begin{align}
& Milliequivalents\text{ }of\text{ }oxalic\text{ }acid~=50\times 0.1\times 2=10 \\
& \because Initial\text{ }medium\text{ }is\text{ }acidic~(\Rightarrow n=5~for~KMnO4). \\
& \therefore ~Molarity\text{ }of~KMn{{O}_{4}}=\dfrac{N}{5}=\dfrac{\dfrac{10}{100}}{5}=0.02M \\
& In\text{ }alkaline\text{ }medium,~\text{ }n=3~for~KMn{{O}_{4}}~ \\
& Hence,\text{ }normality~=0.02\times 3~=0.06N \\
\end{align}\]
Therefore, by our calculations, we can safely conclude that the required normality of $KMn{{O}_{4}}$ is 0.06N making the answer to this question c) 0.06N.
Note: Absolute caution must be taken during the titrations of any acidic or alkaline substance, regardless of its strength, so as to avoid any form of injury or damage. Also make sure that no mistakes are made in the n- factor calculation of Potassium Permanganate w.r.t change in mediums.
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