If 11.5 ml of vinegar sample (d=1g/ml) is titrated with 18.5 ml of standardized Sodium hydroxide
solution. What is the molarity of acetic acid in vinegar sample ?

10.35 M

Explanation:

Molarity = moles solute/ L solution

18.5 mL = 0.0185 L solution

11.5 mL (1g/1mL)= 11.5 grams vinegar (1 mol / 60.052 grams) = 0.1915 mol vinegar

M = 0.1915 mol vinegar/ 0.0185 L solution = 10.35M

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