Question

6.31 M 7. Molarity of acetic acid, HC2H30in vinegar sample (show your calculations) Volume of vinegar sample = 0.003L 0.003L Vinegar contain 0.01893 moles IL 0.01893 = 6.3 IM 0.003 Vinegar contain 8. Grams of Acetic Acid, HC,H,O2 (show your calculations) 1.1368 g HC2H5O2 Grams acid = 12 *MW = 0.01893 mot x 60.052 g/ mot = 1-1368 g 31.90 % L 3 100 9. Percent (m/v) of HC,H,O, in vinegar sample (show your calculations) 3.0 mL vinegar contain 0.01893 x60.052 gram CHs cool I vinegar contain 0.0189 3x 60.052 L vinegar contain 0.01893 860.052 x100 3 Question 1: Is your experimentally calculated value of percent of HC H2O2 is high or low compare to the percent of acetic acid value listed on the label of your vinegar sample bottle? Explain, why do you have a different calculated percent concentration? Labelled value = B) and experimentally obtained Value à 37-90% Question 2: You have added distilled water to the sample of vinegar solution before you started the titration. Does this addition of water to the vinegar sample affect the calculated molarity of acetic acid? Support your answer with proper reasoning. 2