Posted by **Adam** on Monday, January 31, 2011 at 12:24am.

Thanks for your help; I am still have a few problems, could you check my work:

Given: Volume of vinegar analyzed-5mL

Con'c of NaOH-0.09890M

Avg. Volume of NaOH from titration-43.75mL

Density CH3COOH: 1.049g/mL

Calculations:

1.Moles of NaOH required to reach equivalence point= M X L

(0.09890M)(0.04375L)= 0.004327 moles.

2.Moles of acetic acid 5.00mL of vinegar= M X L

(what do I do if the con'c of acetic isn't given)(0.0500L)=?

3. Mass of acetic acid in the 5.00mL sample of vinegar= Mol X MM (0.004327 mol)(60.01g/mol)=0.2529g

4. Volume of the acetic acid in the 5.00mL sample of vinegar

(1.049g/mL)(0.2597g)=0.2724mL

5. % by volume of acetic acid in the vinegar (0.2724mL)/(5.00mL)X 100%= 5.45%

Please let me know if this looks correct.