Respuesta :
Answer: The mass of acetic acid present in the vinegar sample is 0.370 grams
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Molarity of NaOH solution = 0.140 M
Volume of solution = 44.0 mL
Putting values in above equation, we get:
[tex]0.140M=\frac{\text{Moles of NaOH}\times 1000}{44.0}\\\\\text{Moles of NaOH}=\frac{0.140\times 44}{1000}=0.00616mol[/tex]
The chemical equation for the reaction of acetic acid and NaOH follows:
[tex]CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O[/tex]
By Stoichiometry of the reaction:
1 mole of NaOH reacts with 1 mole of acetic acid
So, 0.00616 moles of NaOH will react with = [tex]\frac{1}{1}\times 0.00616=0.00616mol[/tex] of acetic acid
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of acetic acid = 60 g/mol
Moles of acetic acid = 0.00616 moles
Putting values in above equation, we get:
[tex]0.00616mol=\frac{\text{Mass of acetic acid}}{60g/mol}\\\\\text{Mass of acetic acid}=(0.00616mol\times 60g/mol)=0.370g[/tex]
Hence, the mass of acetic acid present in the vinegar sample is 0.370 grams
