Question

Part 1

Answer 1

The moles of NaOH used in the titration would be 0.00177 moles while the molarity of the vinegar will be 0.885 M

Titration calculation

Recall that: mole = molarity x volume

In this case, the molarity of the NaOH= 0.1 M

Volume of NaOH = 17.7 - 0.0 = 17.7 mL

Mole of NaOH used = 0.1 x 17.7/1000 = 0.00177 moles.

Since NaOH and vinegar have 1:1 mole ratio, the mole of vinegar will also be 0.00177 moles.

Molarity of vinegar = mole/volume = 0.00177/0.002 = 0.885 M

More on titration calculations can be found here: brainly.com/question/9226000