Question

The molarity of a \[0.2{\text{N}}\] \[{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}\] Solution will be:
A. \[0.05\] M
B. \[0.2\] M
C. \[0.1\] M
D. \[0.4\] M

Answer 1

Hint: Normality is defined as the number of grams / moles equivalent of solute present in one solution. Equivalents are the moles of the active molecule in the molecule. Normality is equal to Molarity times the \[n\] factor. \[n\] factor is the equivalent.

Complete step by step answer:
Let’s start with discussing the normality for better understanding of this question. Normality is defined as the number of grams / moles equivalent of solute present in one solution. Equivalents are the moles of the active molecule in the molecule.
Molarity is the number of moles per litre of the solution. So, what is the relation between the molarity and normality, well the first thing common between the two is that both of them are used for describing concentrations. And the relation between them is Normality is equal to Molarity times the \[n\] factor. \[n\] factor is the equivalent. So, molarity is equal to Normality divided by the \[n\] factor.
In question we are given \[0.2{\text{N}}\] \[{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}\] Solution, and we need to find its molarity. So, first find out the n factor.
\[{\text{N}}\] factor of the \[{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}\] is \[2\].
\[{\text{N}}\] factor is calculated by calculating the number of electrons that are transferred. Since, two electrons are transferred n factor is \[2\].
Put the values in the formula below, we get
${\text{Molarity = }}\dfrac{{{\text{Normality}}}}{{{\text{n factor}}}}$
Molarity = $\dfrac{{0.2{\text{ N}}}}{2} = 0.1M$
Hence, the answer to this question is option C. \[0.1\] M

$\therefore $ Option C) \[0.1\] M is correct .

Note:
We must observe that Normality is mostly used in precipitating reactions for measuring the number of ions which are likely to precipitate during reaction. It is also used in redox reactions to determine the number of electrons in reducing and oxidizing ends.