Question

Amount of oxalic acid required to prepare $250{\text{mL}}$of $\dfrac{{\text{N}}}{{10}}$solution (Molecular mass of oxalic acid is $126$) is:
A. $1.5759{\text{g}}$
B. $3.15{\text{g}}$
C. $15.75{\text{g}}$
D. $63.0{\text{g}}$

Answer 1

Hint : Normality is the number of equivalents of solute in one litre of solution. The strength of the solution is determined using normality. Equivalent weight can be calculated by dividing molecular weight by valency.
Given data:
Normality of oxalic acid, ${\text{N}} = \dfrac{{\text{N}}}{{10}} = \dfrac{1}{{10}}{\text{N}} = 0.1{\text{N}}$
Volume of oxalic acid, ${\text{V}} = 250{\text{mL}}$

Complete step by step solution :
Oxalic acid is having the formula${\text{COOH}} - {\text{COOH}}.2{{\text{H}}_2}{\text{O}}$. It is a dibasic acid since it can donate two protons. Therefore its n-factor or valency is 2.
Normality is expressed as the number of gram equivalents/one litre of solution. Its unit is generally expressed as N. Number of gram equivalents indicates the number of grams of reactive species in a compound.
Given that normality ${\text{N}} = \dfrac{{\text{N}}}{{10}} = \dfrac{1}{{10}}{\text{N}} = 0.1{\text{N}}$
Volume of acid, ${\text{V}} = 250{\text{mL}}$
Molecular mass of oxalic acid, ${\text{M}} = 126$
Normality=Number of gram equivalent in volume (litre)
Equivalent weight of oxalic acid, ${\text{E}} = \dfrac{{\text{M}}}{{\text{n}}}$, where ${\text{M}}$is the molecular weight of oxalic acid.
${\text{n}}$is the n-factor or the valency which is equal to 2.
${\text{E}} = \dfrac{{126{\text{g}}}}{2} = 63{\text{g}}$
Number of gram equivalence${\text{G}}.{\text{eq}} = \dfrac{{\text{w}}}{{\text{E}}}$, ${\text{w}}$is the given weight of solute which we have to calculate here.
${\text{G}}.{\text{eq}} = \dfrac{{\text{w}}}{{63{\text{g}}}}$
Now normality can be calculated using these values.
${\text{N}} = \dfrac{{{\text{G}}.{\text{eq}}}}{{\text{V}}}$
Substitute the value of number of gram equivalence in the above equation, it becomes
${\text{N}} = \dfrac{{\dfrac{{\text{w}}}{{63{\text{g}}}}}}{{\text{V}}} = \dfrac{{\text{w}}}{{63{\text{g}}}} \times \dfrac{1}{{\text{V}}} = \dfrac{{\text{w}}}{{63{\text{g}}}} \times \dfrac{1}{{250{\text{mL}}}}$
Given that ${\text{N}} = 0.1{\text{N}}$in one litre.
i.e., ${\text{N}} = \dfrac{{0.1}}{{1000{\text{mL}}}}$
Substituting the value of normality, we get
$\dfrac{{0.1}}{{1000{\text{mL}}}} = \dfrac{{\text{w}}}{{63{\text{g}}}} \times \dfrac{1}{{250{\text{mL}}}}$
${\text{w}} = \dfrac{{0.1 \times 63{\text{g}} \times 250{\text{mL}}}}{{1000{\text{mL}}}}$
Now we get the amount of solute to be taken.
${\text{w}} = \dfrac{{1575{\text{gmL}}}}{{1000{\text{mL}}}} = 1.575{\text{g}}$

Hence option A is correct.
$1.575{\text{g}}$of oxalic acid has to be taken to prepare $250{\text{mL}}$of $\dfrac{{\text{N}}}{{10}}$solution.

Additional information:
When oxalic acid is dissolved in water it dissociates into two protons. Therefore its n-factor tends to be two. Normality measures the concentration of solution. It is also known as the equivalent concentration of solution.

Note : Normality and molarity have an important relation. Normality is the product of molarity and acidity or basicity. Acidity is the number of hydroxyl ions the molecule can give. Basicity is the number of protons the molecule can give.