Question

Freezing Point Depression: Determination of the Molar Mass of an Unknown Substance Laboratory Record Unknown Identifier #4 (All unknowns are nonelectrolytes, soi-1) Determination of the freezing point of pure l-butyl alcohol: (Record time-temperature data on the next page) 29.62 1. Mass of empty test tube 2. Mass of test tube and t-butyl alcohol after determination of freezing point of pure t-butyl alcohol 50.25 20.71 3. Mass of t-butyl alcohol Determination of the molar mass of an unknown: (Record time-temperature data on the next page) 4. Mass of test tube, t-butyl alcohol, unknown 52.25 2.258 5. Mass of the unknown 6. Print one graph with the statistics for your group of the t-butyl freezing point determination that shows both runs. Also print one graph for your group of the unknown freezing point that shows both runs. 7. Average freezing point of solvent (worked out on next page): 8. Average freezing point of solution (worked out on next page): 9. AT, between t-butyl alcohol and mixture My unknown # has a molar mass of approximately Work for molar mass calculation:

Answer 1

Average freezing point of solvent : 25.3 C

Average freezing point of solution :

             1. T = (-0.348 + 2.139 ) /0.1263    = 14.18 C

            2. T = (-0.351 + 1.313 ) /0.0621    = 15.49 C

Average = 14.84 C

Freezing point depression, $\Delta T$ f : 10.46 C

#4 unknown ,   i = 1

we have , $\Delta T$ f= i*m*Kf

for given solvent : Kf = 8.37 C/m

m = $\Delta T$ f / Kf = 10.46 C / 8.37 C/m

thus , molality = 1.25 m

we have, m = # mol solute / # Kg solvent = # mol solute / 0.02071 kg = 1.25

mol solute = 0.0259 moles

# mole = mass of unknown / molar mass = 2.258 g /molar mass = 0.0259 moles

thus, molar mass of unknown = 87.25 g/mol