Question

Calculate the amount of oxalic acid used to prepare 100 ml of 0.1N and 0.1M solution.

Answer 1

Normality

• Normality is defined as the number of gram equivalents of solute per liters of solution.
  Normality = $\frac{\mathrm{Number}\mathrm{of}\mathrm{gram}\mathrm{equivalet}\mathrm{of}\mathrm{solute}}{\mathrm{Volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{liters}}$
• Molarity is defined as number of moles of solute per volume of solution in liters.
  $\mathrm{Molarity}=\frac{\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{Volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{liters}}$

Step 1: Calculating the equivalent weight of solute
Molar mass of oxalic acid ( ${\mathrm{C}}_2{\mathrm{H}}_2{\mathrm{O}}_4.2{\mathrm{H}}_2\mathrm{O}$) = 126 g/mol
Basicity of oxalic acid = 2
Number of gram equivalents = $\frac{126}{2}$
Number of gram equivalents = 63 g/mol

Step 2: Calculation of the amount of solute needed for 0.1 N solution
Normality = $\frac{\mathrm{weight}\mathrm{of}\mathrm{solute}\times 1000}{\mathrm{Equivalent}\mathrm{weight}\times \mathrm{volume}}$
$0.1=\frac{\mathrm{weight}\mathrm{of}\mathrm{solute}\times 1000}{63\times 100}$
Weight of solute = 0.63 g

Step 3: Calculating the amount of oxalic acid in 0.1 M solution
Molar mass of oxalic acid ( ${\mathrm{C}}_2{\mathrm{H}}_2{\mathrm{O}}_4.2{\mathrm{H}}_2\mathrm{O}$) = 126 g/mol

Calculaton of the amount of solute needed for 0.1 M solution
$\mathrm{Molarity}=\frac{\mathrm{mass}\mathrm{of}\mathrm{solute}\times 1000}{\mathrm{molar}\mathrm{mass}\times \mathrm{volume}}$
$$\begin{gathered} 0.1=\frac{\mathrm{mass}\mathrm{of}\mathrm{solute}\times 1000}{126\times 100} \\ \mathrm{Mass}\mathrm{of}\mathrm{solute}=1.26\mathrm{g} \end{gathered}$$