<u>Answer:</u> The mass of acetic acid present in the vinegar sample is 0.370 grams
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
![\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%5Ctimes%201000%7D%7B%5Ctext%7BVolume%20of%20solution%20%28in%20mL%29%7D%7D)
Molarity of NaOH solution = 0.140 M
Volume of solution = 44.0 mL
Putting values in above equation, we get:
![0.140M=\frac{\text{Moles of NaOH}\times 1000}{44.0}\\\\\text{Moles of NaOH}=\frac{0.140\times 44}{1000}=0.00616mol](https://tex.z-dn.net/?f=0.140M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20NaOH%7D%5Ctimes%201000%7D%7B44.0%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20NaOH%7D%3D%5Cfrac%7B0.140%5Ctimes%2044%7D%7B1000%7D%3D0.00616mol)
The chemical equation for the reaction of acetic acid and NaOH follows:
![CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O](https://tex.z-dn.net/?f=CH_3COOH%2BNaOH%5Crightarrow%20CH_3COONa%2BH_2O)
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 =
of acetic acid
To calculate the number of moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Molar mass of acetic acid = 60 g/mol
Moles of acetic acid = 0.00616 moles
Putting values in above equation, we get:
![0.00616mol=\frac{\text{Mass of acetic acid}}{60g/mol}\\\\\text{Mass of acetic acid}=(0.00616mol\times 60g/mol)=0.370g](https://tex.z-dn.net/?f=0.00616mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20acetic%20acid%7D%7D%7B60g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20acetic%20acid%7D%3D%280.00616mol%5Ctimes%2060g%2Fmol%29%3D0.370g)
Hence, the mass of acetic acid present in the vinegar sample is 0.370 grams