Question

CALCULATIONS Show all work. 1. Mass of acetic acid, g 19.66g 2. Mass of unknown sample in acetic acid solution.g 2.00 g 3. Generate a cooling curve by plotting temperature vs time data on a single graph Plot on graphing paper or use a computer program and print plot. See Below 4. Freezing point of pure acetic acid solvent (from graph). "C 16°C 5. Freezing point of acetic acid solution (from graph). °C 13°c 6. Determine the freezing point depression of the acetic acid solution. °C 7. Determine the mass of acetic acid, kg 1. 19.00g = 0.01966kg Kg Tooog 8. Determine the molar mass of the unknown.g/mol K (acetic acid) = 3.90 °C.kg/mol on calculations page 9. Identity of unknown sample (from Unknown Table on page 18) 10. Calculate the percent error in molar mass for the experiment.

i cant figure out #8 please help. thank you.

also if you could help with #6 it would be greatly appreciated

Answer 1

given that mass of acetic acid (9) - 19.66 g mass of unknown sample in acetic acid solution Freezing Point of pure acetic acid solvent (Te (solvent)) = 16°C freezing point of pure acetic acid solution (TF (solution]) = 13°C mass of acetic acid solution (kg) = 0.01966 kg we know that Freezing point depression (ATF) = i krom where. ATF = TE Cpure solvent) - Te (solation) - Van't Hoff Factor = number of per molecule of lons solute KF cryoscopic constant m m olality - solute Moles of mass of solvent in kg 6 freezing point depression of solution. LATE) the acetic acid - Telpure solvent) - Tf (Bolution) ATF = 160 - 13°c = 3.0) Hence, freezing point depression of acetic acid solution (ATF)=[3.00°c

Molar mass of the unknown = ? the unknown Kf(acetic acid) - acetic acid) = 3.90 °c.kg/mol dow, ATF = i.krom so i of unknown Since, solute is unknown sointe can't determined. dow, above equation become ATE = Krom so, m= OTG - 3.02 3.90 ot. kglmoe KE So, molality (m) = 0.77 mol Hence, molality of cm) = 0.7z mol acetic acid solution How, cholality of acetic acid solution (m) = moles of solute or unknown - Sample Mass of solvent (acetic acid) in kg moles off unknown sample 0.77 mol kg 0.01966 Kg Moles off unknown Sample - -0.77X0.01966 = 0.01514 mol mol choles of unknown Sample -lo.oisuu 10.01514 Moll

we know that moles of compound - given mass of compound Molar mass of compound ، که لى given mass of unknown moles off unknown Sample = molar mass of F unknown Since, mass of acetic acid given in unknown solution = sample 2.00g Son o-01514 mol - 2.oog Molar mass of unknown molar mass of unknown - 2.oog (molar mass of o.o1514 orol 132.10 glmal unknown = Hence, the molar mass of the unknown =1132.10 glmol]