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# Define normality.$6.3gms$ of oxalic acid is present in $500ml$ of solution its normality is:

Last updated date: 17th Jun 2024
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Hint: We must know about the word normality. Generally, normality is a quantitative measurement for chemical solutions. We can use the below formula to determine the normality of the solution.
$Normality{\text{ }}of{\text{ }}solution{\text{ }}N = \dfrac{{Weight}}{{equivalent{\text{ }}wt.}} \times \dfrac{{1000}}{{volume(ml)}}$

The concentration of any solution that is acknowledged in terms of gram equivalents is the normality of the solution.
We can use normality as the number of mole equivalents per liter of solution. Usually, the normality of the solution is represented by N or equivalent per liter ($eq/L$)units.
We know that the chemical formula of oxalic acid is ${C_2}{H_2}{O_4}.2{H_2}O$
Ans also the molecular mass of${C_2}{H_2}{O_4}.2{H_2}O$ =$126g$
Now, to calculate normality, we will use below formula
$Normality{\text{ }}of{\text{ }}oxalic{\text{ }}acid,N = \dfrac{{Wt.{\text{ }}of{\text{ }}oxalic{\text{ }}acid}}{{equivalent{\text{ }}wt.}} \times \dfrac{{1000}}{{volume(ml)}}$
To solve this, we must have a value of the equivalent weight of oxalic acid.
So, equivalent weight of Oxalic acid = $\dfrac{{molecular{\text{ }}mass}}{{Basicity}}$
∴ equivalent weight of Oxalic acid = $\dfrac{{126}}{2}$ =$63{\text{ }}gm/eq$.
Now, substituting the values, in the formula of normality equation, we get

Normality= $\dfrac{{6.3}}{{63}} \times \dfrac{{1000}}{{500}} = 0.1 \times 2{\text{ }} = 0.2{\text{ }}gm.mole{\text{ }}per{\text{ }}liter$
Hence, the Normality of oxalic acid is$0.2{\text{ }}gram{\text{ }}moles{\text{ }}per{\text{ }}litre$.
We can use normality instead of molarity because often $1{\text{ }}mole$of acid does not neutralize $1{\text{ }}mole$of base.
i.$Normality{\text{ }} = {\text{ }}Number{\text{ }}of{\text{ }}gram{\text{ }}equivalents{\text{ }} \times {\text{ }}{\left[ {volume{\text{ }}of{\text{ }}solution{\text{ }}in{\text{ }}litres} \right]^{ - 1}}$
ii.$Number{\text{ }}of{\text{ }}gram{\text{ }}equivalents{\text{ }} = {\text{ }}weight{\text{ }}of{\text{ }}solute\; \times {\text{ }}{\left[ {Equivalent{\text{ }}weight{\text{ }}of{\text{ }}solute} \right]^{ - 1}}$
iii.$N{\text{ }} = {\text{ }}Weight{\text{ }}of{\text{ }}Solute{\text{ }}\left( {gram} \right){\text{ }} \times {\text{ }}\left[ {Equivalent{\text{ }}weight{\text{ }} \times {\text{ }}Volume{\text{ }}\left( L \right)} \right]$
iv.$N{\text{ }} = {\text{ }}Molarity{\text{ }} \times {\text{ }}Molar{\text{ }}mass{\text{ }} \times {\text{ }}{\left[ {Equivalent{\text{ }}mass} \right]^{ - 1}}$
v.$N{\text{ }} = {\text{ }}Molarity\; \times {\text{ }}Basicity\; = {\text{ }}Molarity{\text{ }} \times {\text{ }}Acidity$