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Last updated date: 10th Dec 2023
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# Calculate the amount of oxalic acid used to prepare $100{\text{ }}ml$ of $0.1N$ and $0.1M$ solution.

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Hint: 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.

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 $1.26g$ .

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.