1. The problem statement, all variables and given/known data A student finds that it takes 32.17 mL of 0.1048 M NaOH to titrate 5.000 mL of sulfuric acid solution. Determine the molarity and percent mass of the sulfuric acid in the solution (you may assume that the density of the sulfuric acid solution is the same as pure water). ____ M ____% sulfuric acid by mass. 2. Relevant equations Molarity of Acetic Acid in Vinegar First, using the known molarity of the NaOH (aq) and the volume of NaOH (aq) required to reach the equivalence point, calculate the moles of NaOH used in the titration. From this mole value (of NaOH), obtain the moles of HC2H3O2 in the vinegar sample, using the mole-to-mole ratio in the balanced equation. Finally, calculate the molarity of acetic acid in vinegar from the moles of HC2H3O2 and the volume of the vinegar sample used. Mass Percent of Acetic Acid in Vinegar First, convert the moles of HC2H3O2 in the vinegar sample (previously calculated) to a mass of HC2H3O2, via its molar mass. Then determine the total mass of the vinegar sample from the vinegar volume and the vinegar density. Assume that the vinegar density is 1.000 g/mL (= to the density of water). Finally, calculate the mass percent of acetic acid in vinegar from the mass of HC2H3O2 and the mass of vinega ============================================================== above two are with given examples of our lab that we did in class. but since the problem that i could not solve was similar to what we did in lab, i thought i could use the formula. however, i could not understand what the whole calculation was about, since there is no implication of whether multiply or divide, etc. could anybody help me out please? 3. The attempt at a solution what i did was given: volume base (Vb) = 32.17 mL molarity (M) = 0.1048 M volume acid (Va) = 5.000 mL volume of acid solution = 5.000 mL moles of acid present verified by titration: volume of base x molarity of base used in acid titration = 32.17 mL x 0.1048 M = 3.371 mol acid solution the molarity of the acid solution is 3.371 mol / 5.000 mL = 0.674 M and then following % would be 67.4%, i thought.