Molar Mass by freezing point depression? Help
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Molar Mass by freezing point depression? Help

[From: ] [author: ] [Date: 12-01-23] [Hit: ]
8927m solutionThis decreased the freezing point by 7.3°C Kf for BHT = 7.3/0.8927 = 8.177°C/m.kgThe freezing point depression for solution #2 was 3.......
Molality of solution = 216.419/242.44 = 0.8927m solution

This decreased the freezing point by 7.3°C
Kf for BHT = 7.3/0.8927 = 8.177°C/m.kg

The freezing point depression for solution #2 was 3.1°C
Therefore the solution was 3.1/8.177 = 0.379 molal

Mass unknown in 1.0kg BHT = 0.9960/0.5019*1000 = 1,984.45g in 1.0kg

Molar mass of unknown = 1,984.45/0.379 = 5,236g/mol

We both get the same answer.

What is the problem: I am concerned about the very high amount of unknown you used in solution #2. 0.996g in 0.5055g solvent. I do not know much about the solubility of stearic acid in BHT, but I wonder if you have produced a true solution or some form of saturated solution or an emulsion. Under these conditions you cannot use Kf to determine the molar mass of the unknown. Was there a special reason that you used such a large amount? I would prefer you to have made a test solution using a similar quantity as you used for the cetyl alcohol. You will see that the molar mass of cetyl alcohol and stearic acid are very similar - so solutions of similar concentration would have produced a better result , I think. You have used about 9 times the mass concentration of the unknown compared to the cetyl alcohol. I think that this is where your problem is.
Recheck you figures for possible mistake in recording the masses or mistake in recording freezing points.
Sorry , cannot help you beyond this.
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