Question

What is molarity of a $Mn{O_4}^ - $ solution if $32.00mL$of the solution is required to titrate $40.00mL$ of $0.400N\,F{e^{2 + }}$ ?
(A) $0.1M$
(B) \[0.2M\]
(C) $0.3M$
(D) $0.4M$

Answer 1

Hint: In order to this question, to find the molarity of the $Mn{O_4}^ - $ solution, we will first find the n-factor of both the solutions, and then we will equate the milliequivalent of both the solutions, and we will find the normality of $Mn{O_4}^ - $ , now we can easily find the molarity of $Mn{O_4}^ - $ solution by the help of its normality and the n-factor.

Complete answer:
Given that-
Volume of $Mn{O_4}^ - $ is \[32mL\] .
Volume of $F{e^{2 + }}$ is $40mL$ .
And, Normality of $F{e^{2 + }}$ is $0.400N$ .
So, according to the given question-
The half-reaction is:
$Mn{O_4}^ - + F{e^{2 + }}$
In the above reaction, the n-factor of $Mn{O_4}^ - $ , $nf = 4 + 1 = 5$
Similarly, the n-factor of $F{e^{2 + }}$ , $nf = {2^ + } = 2$ .
As we know, n-factor acts a very important role in finding the molarity of any ionic compound.
Now, as we know that the milliequivalent of any compound is equal to the milliequivalent of any other compound:
$\because {M_{eq}}\,of\,Mn{O_4}^ - = {M_{eq}}\,of\,F{e^{2 + }}$
$ \Rightarrow N \times V(Mn{O_4}^ - ) = N \times V(F{e^{2 + }})$
here, $N$ is the normality (i.e.. useful to calculate the molarity)
$V$ is the given volume of both compounds.
$
   \Rightarrow N(Mn{O_4}^ - ) \times 32 = 0.400 \times 40 \\
   \Rightarrow N = \dfrac{{0.400 \times 40}}{{32}} = 0.5 \\
 $
Thus, the normality of $Mn{O_4}^ - $ is $0.5N$ .
Therefore, the required molarity of $Mn{O_4}^ - $ solution $ = \dfrac{{Normality}}{{nf}} = \dfrac{{0.5}}{5} = 0.1M$ .
Hence, the correct option is (A).

Note:
Molarity and normality have a very similar relationship. A multiple of molarity can be used to describe normality. Normality refers to the molar concentration of the acid component exclusively, whereas molarity relates to the concentration of a substance or ion in a solution.