Chemistry Report 3 Freezing Point Depression
Chemistry Report 3 Freezing Point Depression CHM 113
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This 3 page Class Notes was uploaded by Pooja Dave on Thursday September 29, 2016. The Class Notes belongs to CHM 113 at University of Miami taught by Dr. Tegan Eve in Fall 2015. Since its upload, it has received 5 views. For similar materials see General Chemistry 1 lab in Chemistry at University of Miami.
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Date Created: 09/29/16
Pooja Dave CHM 113, Section FY Freezing Point Depression with Lauric Acid Introduction: The freezing point depression is used in determining the temperature at which the system freezes. The molecules within a substance get closer together as it solidifies, the exception being water as its molecules are farther away when solid. The addition of another substance creates an impurity and can alter these molecular interactions. Freezing point lowers as solutes are added. This freezing point is found within the flat portion of the graph Time vs. Temperature in Celcius. The flat portion represents the heat of fusion during the cooling process. The temperature does not change and the molecules are being packed tightly into a solid. The freezing point of Lauric acid changes with the addition of benzoic acid, as the molecules cannot be packed as tightly into the solid state. However, it can be estimated by finding the average freezing point constant using molality and the change in temperature. Molality is used because it does not change with volume which is temperature dependent, unlike molarity. The freezing point constant for both solution is averaged together to get an approximate value for pure Lauric acid. The differences in freezing point constant for the two solutions also show how different amounts of solute can lead to different freezing points. Procedure: 1. Set up LoggerPro, labeling the x and y axis Time in second and Temperature in Celcius and connecting the temperature probe to the data collector 2. Fill a 400mL beaker with water and heat up using a hot plate until approximately 60 to 80⁰C 3. Obtain a test tube containing a solution of 0.75 g of Benzoic Acid with 8.00 g of Lauric Acid 4. Place the test tube in water bath and melt the mixture 5. Once fully melted take it out of hot water bath and add the temperature probe, hitting collect 6. Place test tube in the room temperature water bath and stir as data has started collecting 7. Determine freezing point from the graph, looking for the point when the graph straightens out 8. Repeat steps 3 through 8 but with a test tube containing the solution with 8.00 g Lauric Acid with 1.50 g Benzoic Acid Equations: ΔT = K f molessolute m = masssolvent(kg) Observations: During our experiment, the greatest error was in the data collection. The beginning of the graph was not smooth, making freezing point harder to find. There was also a gap in time from when the test tube left the water bath to the time data started collecting. Data: Molecular Weight of Benzoic Acid = 122.12 g/mol Melting point of pure Lauric Acid = 43.2⁰C Moles of 0.75 g BA = 0.00614 mol Molality of 0.75g BA/8.00g LA = 0.768 mol/kg Change in Temperature = 3.2 ⁰C Kfof 0.75g BA/8.00g LA = 4.17 Moles of 1.5g BA = 0.0122 mol Molality of 1.5g BA/8.00g LA = 1.54 mol/kg Change in Temperature = 6.1 ⁰C Kfof 1.5g BA/8.00g LA = 3.84 Calculations: 0.75g BA/8.00g LA: 1mol × Moles of BA = 0.75g 122.12g = 0.00614 mol 0.00614mol Molality (m) = 0.008kg = 0.768 mol/kg ΔT = 40 – 43.2 = 3.2 ⁰C 3.2 ⁰C = (0.768) K f K = 4.17 f 1.50g BA/8.00g LA: × 1mol Moles of BA = 1.50g 122.12g = 0.0122 mol 0.0122mol Molality (m) = = 1.54 m 0.008kg ΔT = 37.1 – 43.2 = 6.1 ⁰C 6.1 ⁰C = (1.54) K f K f 3.84 Average K = f.003 Discussion and Conclusion: The average K value obtained from this experiment shows the value when f Lauric acid freezes. It represents the average freezing point constant for Lauric acid. This freezing point constant is the moment when the molecules of Lauric acid form close bonds, making it a solid. With the known melting point of Lauric acid and the molecular weight of Benzoic acid, the moles and change in temperature can be calculated. These two pieces of data can be applied to the equation to solve for freezing point constant. The differences between the solution with 0.75g Benzoic acid and the one with 1.50g Benzoic acid show how the K value chafges with more solute. The solution with 1.50g of BA had a lower K value fhan the one with 0.75g of BA. Because the K valfe is lower, the freezing point is lower meaning that it takes more energy for the solution to freeze. Our data then supports the idea that the one with the most solute interferes the most with the molecular arrangement during solid formation. Errors: The temperature probe was first placed in the heat bath to test its temperature. This may have interfered with the collecting of data when testing the temperature as Lauric acid freezes. The initial temperature then would have been higher than in reality, and the warm temperature probe may have delayed the freezing of Lauric acid. The time gap between taking the test tube out of the heat bath and collecting data may have cause data to be missed. This would then have altered the freezing point, making the initial temperature lower than it actually was. Some of the Lauric acid may not have completely melted when we took the test tube out of the water bath. This would cause the freezing point to be higher than what it actually was. Stirring the warm test tube in room temperature water bath may have disrupted the cooling of Lauric acid, as there were inconsistencies in the stirring. This would have delayed the freezing, making the freezing point lower than it actually is. Graph:
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