Question

What will be the volume of $\dfrac{N}{{10}}$ solution of oxalic acid obtained by dissolving 63g of hydrated oxalic acid?

Answer 1

Normality of oxalic acid dihydrate solution is the number of gram equivalents of oxalic acid divided with volume of solution.
${\text{Normality }} = {\text{ }}\dfrac{{{\text{Number of gram equivalents of oxalic acid dihydrate}}}}{{{\text{Volume of solution}}}}$

Complete step by step answer:
Hydrated oxalic acid is oxalic acid dihydrate. One molecule of oxalic acid is associated with two molecules of water. The chemical formula of the oxalic acid is ${{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}{\text{ }}.{\text{ 2}}{{\text{H}}_{\text{2}}}{\text{O}}$
The atomic masses of hydrogen, carbon and oxygen are \;{\text{1 g}}/{\text{mol}},{\text{ 12 g}}/{\text{mol & 16 g}}/{\text{mol}} respectively. Calculate the molecular weight of oxalic acid dihydrate
${\text{2}}\left( {\text{1}} \right) + {\text{2}}\left( {{\text{12}}} \right) + {\text{4}}\left( {{\text{16}}} \right) + {\text{2}}\left( {{\text{18}}} \right) = {\text{126 g}}/{\text{mol}}$
The mass of hydrate oxalic acid is $63{\text{ }}g$ .
Divide the mass of hydrated oxalic acid with its molecular weight to obtain the number of the moles of hydrated oxalic acid.

$\dfrac{{{\text{63 g}}}}{{{\text{126 g}}/{\text{mol}}}}{\text{ }} = {\text{ }}0.{\text{5 mol}}$

Oxalic acid has a basicity of two. This is because one molecule of oxalic acid has two replaceable protons. One molecule of oxalic acid reacts with two hydroxide ions.
Divide the number of moles of oxalic acid dihydrate with its basicity to obtain the number of gram equivalents of the oxalic acid dihydrate.

$\dfrac{{0.{\text{5 mol }}}}{2}{\text{ }} = {\text{ }}0.{\text{25 g eq}}$

Normality of oxalic acid dihydrate solution is the number of gram equivalents of oxalic acid divided with volume of solution.

${\text{Normality }} = {\text{ }}\dfrac{{{\text{Number of gram equivalents of oxalic acid dihydrate}}}}{{{\text{Volume of solution}}}}$

Rearrange above equation to obtain the volume of solution
${\text{Volume of solution }} = {\text{ }}\dfrac{{{\text{Number of gram equivalents of oxalic acid dihydrate}}}}{{{\text{Normality}}}}$

${\text{Volume of solution }} = {\text{ }}\dfrac{{0.{\text{25 g eq }}}}{{\dfrac{1}{{10}}}}$

${\text{Volume of solution }} = {\text{ }}2.5{\text{ L}}$
Hence, the volume of $\dfrac{N}{{10}}$ solution of oxalic acid dihydrate is $2.5{\text{ }}L$ .

Note: Another approach to solve the problem is to determine the number of gram equivalents, and the number of grams present in one litre of $\dfrac{N}{{10}}$ solution of oxalic acid. Then correlate this mass with $63{\text{ }}g$ and calculate the required volume.

Normality is different from molarity. Normality is the number of gram equivalents of solute divided with volume (in liter) of solution. Molarity is the number of moles of solute divided with volume (in liter) of solution.