Question

In order to prepare 1L normal solution of ${\text{KMn}}{{\text{O}}_{\text{4}}}$, how many grams of ${\text{KMn}}{{\text{O}}_{\text{4}}}$ are required if the solution is to be used in acidic medium for oxidation.
A. 158 g
B. 31.6 g
C. 62 g
D. 790 g

Answer 1

Normality can be identified by the product of molarity and the n-factor. Molarity is defined as the mole of solute dissolved in a given volume of solution.

Formula used: {\text{Normality = }}\,{\text{Molarity}}\,{\text{ \times }}\,{\text{n}} - {\text{factor}}

Complete step by step answer:
n-factor represents the valency of the compound such as electrons gained or lost which is also known as the oxidation state of the ion.
Determine the n-factor for the potassium permanganate as follows:
The normal solution of potassium permanganate will be used for oxidation in an acidic medium, so potassium permanganate will be reduced in the acidic medium.
During the reduction of potassium permanganate the change in the oxidation state of manganese in an acidic medium is shown as follows:
${\text{M}}{{\text{n}}^{{\text{7 + }}}}\, + 5{{\text{e}}^ - }\, \to \,{\text{M}}{{\text{n}}^{{\text{2 + }}}}\,$
During the reaction five electron exchange so, the n-factor for the potassium permanganate is 5.
Determine the normality of the solution as follows:
The formula of the molarity is as follows:
${\text{Molarity = }}\,\dfrac{{{\text{Gram amount/ Molar mass of solute}}}}{{{\text{Volume of solution in litres}}}}$
So, the formula of normality will be,
{\text{Normality = }}\,\dfrac{{{\text{Gram amount/ Molar mass of solute}}}}{{{\text{Volume of solution in litres}}}}\,{\text{ \times }}\,{\text{n}} - {\text{factor}}
Molecular mass of ${\text{KMn}}{{\text{O}}_{\text{4}}}$ = 158.034 g
Substitute 158.034 g for the molar mass 1 for normality, 1L for the volume of solution and 5 for n-factor.
{\text{1N}}\,{\text{ = }}\,\dfrac{{{\text{Gram amount/ }}158.034\,{\text{g}}}}{{\text{1}}}\,{\text{ \times }}\,5
$\Rightarrow \dfrac{{{\text{1}} \times {\text{1}}}}{5}\,{\text{ = }}\,\dfrac{{{\text{Gram amount }}}}{{158.034\,{\text{g}}}}\,$
$\Rightarrow {\text{Gram amount}} = \dfrac{{158.034\,{\text{g}}}}{5}\,\,$
$\Rightarrow {\text{Gram amount}} = 31.6\,{\text{g}}\,\,$
So, in order to prepare 1L normal solution of ${\text{KMn}}{{\text{O}}_{\text{4}}}$, $31.6\,{\text{g}}\,$of ${\text{KMn}}{{\text{O}}_{\text{4}}}$ are required.

Therefore, option (B) 31.6 g is correct.

Note: The n-factor for potassium permanganate will be different in the basic medium. Normality is also determined by dividing the equivalent weight by the volume of the solution. The equivalent weight is determined by dividing the molar mass by the n-factor. On changing temperature the normality changes.