Number of millimoles of NaOH consumed in present titration =
33.23 X 0.09892
= 3.287
Thus, number of millimoles of vinegar present in sample = 3.287
Now, Molarity of vinegar solution =
![\frac{\text{number of millimoles of vinegar}}{\text{volume of solution (ml)}}](https://img.qammunity.org/2019/formulas/chemistry/college/e9fvg5hypzc3tcp0uc0t556v009cetv1aj.png)
=
![(3.287)/(26.89)](https://img.qammunity.org/2019/formulas/chemistry/college/t37p4r94692gj8q6xfc7evlk18fua2o44s.png)
= 0.1222 mol/dm3
Molar concentration of vinegar present in sample solution is 0.1222 M