***I have the values to complete the following problems***

Final Results

1. From your titration curve, your plot of Conductivity versus Volume of NaOH added, you’ll determine the milliliters of NaOH added at the equivalence point, Step #4 under Analysis. Using the value for these mL, you’ll be required to report the molarity of acetic acid in your unknown vinegar solution.
2. The calculation will look something like, for example:$\left(\frac{1.17molNaOH}{L}\right)\left(\frac{{10}^{?3}L}{mL}\right)\left(6.82mL\right)\left(\frac{1molC{H}_{3}COOH}{1molNaOH}\right)\left(\frac{1}{7.05mL}\right)\left(\frac{mL}{{10}^{?3}L}\right)=1.13\text{M}C{H}_{3}COOH$where the 1st factor is your calculated molarity of NaOH, the 2nd is a SI prefix from L to mL, the 3rd is your mL of NaOH from the equivalence point from your plot as in the previous step, the 4th is the mole-to-mole conversion from the balanced reaction, the 5th is forcing the calibrated mL of unknown vinegar measured for your titration, and the last, 6th, is the SI prefix from mL to L.
3. And finally, converting your molarity of vinegar, latter step, to percent by mass (%(w/w) or “% by mass”) would follow as,$\left(\frac{1.13molC{H}_{3}COOH}{L}\right)\left(\frac{{10}^{?3}L}{mL}\right)\left(\frac{1.00g}{mL}\right)\left(\frac{60.053g}{mol}\right)x100=6.79\text{\%}\left(w/w\right)C{H}_{3}COOH$where the 1st factor is the molarity of acetic acid in vinegar from the previous step, the 2nd is the SI prefix from L to mL, the 3rd is the density of vinegar, the 4th is the molar mass of acetic acid, and the 5th times 100 to convert to a percentage.
4. Report both of your final values below.