freezing point depression and molecular weight A. Pure benzophenone freezes at 4
ID: 986230 • Letter: F
Question
freezing point depression and molecular weight
A. Pure benzophenone freezes at 48.1°C. A solution of 1.25g of biphenyl (C12H10) molar mass = 154.2 g/mol) plus 25.38 g of benzophenone froze at 16.8°C. Calculate the Kf of benzophenone. (I belive this is the correct answer)
B. A mixture of 1.03 g of unknown plus 23.94 g of benzophenone had the following cooling behavior:
B1. Prepare a graph showing temperature on the ordinate and time on the abscissa. Identify the freezing temperature of the solution.
B2. Calculate the molality in mol/Kg using the formula ?T= Kfm. Kf was determined in problem A.
B3. Calculate the moles of the unknown using the molality and the mass, in Kg, of benzophenone solvent.
B4. Calculate the molecular weight of the unknown in g/mol.
K ((48.1°C - 168°C) * (1542g/mol) * (25.38g) – 97.99°C/m (1000 * (1.25g)Explanation / Answer
A. Freezing point depression = molality x Kf
1.25 g / 154.2 g/mole = 0.008106 mole
0.008106 mole / 0.02538 kg = 0.3194 m
(48.1 - 16.8) = m Kf = 0.3194 Kf
Kf = 97.99°C/m = 98°C/m.
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