Respuesta :
Answer: The volume of NaOH required is 22.39 mL
Explanation:
We are given:
Mass of sample = 0.6543 g
Mass percent of oxalic acid = 53.66 %
This means that 53.66 grams of oxalic acid is present in 100 grams of sample
Mass of oxalic acid in the given amount of sample = [tex]\frac{53.66}{100}\times 0.6543=0.351g[/tex]
- To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of oxalic acid = 0.351 g
Molar mass of oxalic acid = 90 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of oxalic acid}=\frac{0.351g}{90g/mol}=0.0039mol[/tex]
The chemical equation for the reaction of oxalic acid and NaOH follows:
[tex]C_2H_2O_4+2NaOH\rightarrow Na_2C_2O_4+2H_2O[/tex]
By Stoichiometry of the reaction:
1 mole of oxalic acid reacts with 2 moles of NaOH
So, 0.0039 moles of oxalic acid will react with = [tex]\frac{2}{1}\times 0.0039=0.0078mol[/tex] of NaOH
- To calculate the volume of solution, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Moles of NaOH = 0.0078 moles
Molarity of solution = 0.3483 M
Putting values in above equation, we get:
[tex]0.3483M=\frac{0.0078\times 1000}{V}\\\\V=\frac{0.0078\times 1000}{0.3483}=22.39mL[/tex]
Hence, the volume of NaOH required is 22.39 mL
