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 = [tex] \frac{\text{number of millimoles of vinegar}}{\text{volume of solution (ml)}} [/tex]
= [tex] \frac{3.287}{26.89} [/tex]
= 0.1222 mol/dm3
Molar concentration of vinegar present in sample solution is 0.1222 M