Question

Calculate the amount of oxalic acid used to prepare $100{\text{ }}ml$ of $0.1N$ and $0.1M$ solution.

Answer 1

To solve this question, we will focus on the concept of morality and normality. Molarity $(M)$ indicates the number of moles of solute per litre of solution (moles/Litre). Normality $(N)$ is defined as the number of mole equivalents per litre of solution.

Complete answer:
First let’s calculate the molarity of oxalic acid. For it, we will need the molecular weight of oxalic acid. The molecular weight can be calculated by a molecular formula which is ${H_2}{C_2}{O_4}.2{H_2}O$ . We will multiply the molar mass of each element with the number of each element present in the molecule to obtain the molecular weight. Hence,
Molecular weight $= 1 \times 2 + 2 \times 12 + 4 \times 16 + 2 \times (1 \times 2 + 16)$
$= 2 + 24 + 64 + 2 \times (18)$
$= 2 + 24 + 64 + 36$
$= 126g/mol$
Since the $n$ factor is $2$ , the equivalent weight becomes $63g/mol$ .
Molarity $= \dfrac{{{\text{weight}} \times 1000}}{{{\text{molecular weight }} \times {\text{ volume}}}}$
$0.1 = \dfrac{{a \times 1000}}{{126 \times 100}}$
Here, $a \to$ the weight required for $0.1M$ solution.
On further solving,
$a = 1.26g$ (in $100mL$ )
Similarly,
Normality $= \dfrac{{{\text{weight}} \times 1000}}{{{\text{equivalent weight }} \times {\text{ volume}}}}$
$0.1 = \dfrac{{b \times 1000}}{{63 \times 100}}$
Here, $b \to$ the weight required for $0.1N$ solution.
On further solving,
$b = 0.63g$ (in $100mL$ )
Hence, the amount of oxalic acid required to prepare $100{\text{ }}ml$ solution of $0.1N$ is $0.63g$ and $0.1M$ is $1.26g$ .

Additional Information:
Sometimes, in place of morality, chemists prefer to use normality because often $1$ mole of acid does not completely neutralize $1$ mole of base. Hence, in order to have a one-to-one relationship between acids and bases, many chemists prefer to express the concentration of acids and bases in normality.

Note:
The normal concentration of a solution or normality is always equal to or greater than the molar concentration or molarity of a solution. The normal concentration can be directly calculated by multiplying the molar concentration by the number of equivalents per mole of solute.