M moles ba kg aa moles ba m kg aa the mass of acetic

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m = moles B.A/ ? kg A.A moles B.A = m *? kg A.A ←← the mass of acetic acid ?kg A.A = V A.A *D A.A = 10 ml * 1.049g*cm 3 A .A = 10.49*10 -3 kg A.A moles B.A = 0.77mole*kg -1 A.A * 10.49*10 -3 kg A.A = 8.08*10 -3 mole 3. Experimental Molecular mass of B.A Exp. M.M B.A = mass B.A/moles B.A = 0.5003 g /8.08*10 -3 mole = 61.42 g/ mole 4. Actual Molar mass of B.A (C 6 H 5 COOH) Act. M.M B.A = 7*12.01 + 6*1.01 + 2*16 = 122.13 g/mole 5. Percent Error % error = {(Act. M.M B.A - Exp. M.M B.A ) / Act. M.M B.A }*100 = % error = (122.13 – 61.42)*100 /122.13 = 49.71 % Conclusion Overall, the results of the lab didn’t quite support the expected outcome. The concept of freezing depression was shown as expected. The addition of Benzoic acid to the pure acetic acid dropped it freezing temperature from 14 ° C to11 ° C but in that process we first cooled the pure acetic acid by stirring until it froze. In the second trial, we didn’t stir hence we observed supercooling and the temperature later stailized at 14 ° C. That made sence because we were still using the pure acetic acid. The third trial was with the benzoic acid mixed with the pure acetic acid. Without stirring, supercooling came about and then the solution froze and stabilized at 11 ° C. This last part is where we must have done some errors which affected the molecular mass of
the benzoic acid. We had to use hot water after each trial to melt the acid back to liquid. We must have overheated it in such a way that almost half of the benzoic acid evaporated hence giving us a 50 % error in the calculation of molar mass.

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