Answer
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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.
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.
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.
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