Question

In order to prepare one liter 1N solution of \[KMn{O_4}\], how many grams of \[KMn{O_4}\] are required , if the solution is to be used in an acidic medium for oxidation?
(A) 128 g
(B) 41.75 g
(C) 31.60 g
(D) 62.34 g

Answer 1

Hint: The formula for Normality is, \[{\text{Normality = }}\frac{{{\text{Weight of Solute}}}}{{{\text{Equivalent weight of Solute }} \times {\text{ Volume of solution}}}}\]

Potassium permanganate will give the following reaction in acidic media.

\[\mathop {KMn{O_4}}\limits_{Mn = + 7} + 5{e^ - } \to M{n^{ + 2}}\]


Formula used: \[{\text{Normality = }}\frac{{{\text{Weight of Solute}}}}{{{\text{Equivalent weight of Solute }} \times {\text{ Volume of solution}}}}\]


Complete step by step answer:

Here, we are asked to find the weight of \[KMn{O_4}\] required to form its one litre 1N solution. We can use the formula of normality and can easily find the weight of \[KMn{O_4}\] required.
\[{\text{Normality = }}\frac{{{\text{Weight of Solute}}}}{{{\text{Equivalent weight of Solute }} \times {\text{ Volume of solution}}}}\]

Here we are given the normality of the resultant solution, which is 1.

We will need to find the equivalent weight of \[KMn{O_4}\] in order to use the formula of its normality.

-We know that \[KMn{O_4}\] is neither an acid nor a base but it will give a redox reaction in an acidic medium. Let’s see how it will react.
\[\mathop {KMn{O_4}}\limits_{Mn = + 7} + 5{e^ - } \to M{n^{ + 2}}\]

So, we can say that the number of electrons gained in this reaction by a Manganese atom is 5.
So, we know that \[{\text{Equivalent weight of Redox reagent = }}\frac{{{\text{Molecular weight of reagent}}}}{{{\text{Number of }}{{\text{e}}^ - }{\text{ lost or gained}}}}\]

We can find the molecular weight of \[KMn{O_4}\] by following the formula.
Molecular weight of \[KMn{O_4}\] = Atomic weight of K + Atomic weight of Mn + 4(Atomic weight of O),
Molecular weight of \[KMn{O_4}\] = 39 + 55 + 4(16),
Molecular weight of \[KMn{O_4}\] = 39 + 55 + 64
Molecular weight of \[KMn{O_4}\] = 158 \[gmmo{l^{ - 1}}\].

So, we can write the formula of equivalent weight of \[KMn{O_4}\] as,
\[{\text{Equivalent weight = }}\frac{{158}}{5}\],
\[{\text{Equivalent weight = 31}}{\text{.6}}\]\[gmo{l^{ - 1}}\].

So, we can write the Normality as,
\[{\text{Normality = }}\frac{{{\text{Weight of Solute}}}}{{{\text{Equivalent weight of Solute }} \times {\text{ Volume of solution}}}}\]

We know that the equivalent weight of solute is 31.4 \[{\text{gmol}^{ - 1}}\] and the volume of solution is given 1 liter. So,
\[1 = \frac{{{\text{Weight of KMn}}{{\text{O}}_4}}}{{31.6 \times 1}}\]
\[{\text{Weight of KMn}}{{\text{O}}_4} = 1 \times 31.6 \times 1\]
\[{\text{Weight of KMn}}{{\text{O}}_4} = 31.6gm\]

Hence we can say that we need to dissolve 31.6 g of potassium permanganate in order to prepare 1 liter 1N solution.

So, the correct answer is option (C) 31.60 g.

Note:

Do not get confused between Normality and Molarity, remember that Normality uses equivalent weight of solute and in molarity, molecular weight of solute is used. Remember that Potassium permanganate will gain 5 electrons in acidic medium and will get reduced while if the reaction is in neutral medium, then it will gain only 3 electrons in the process of reduction.