Lab 3.docx - Experiment 3 Determination of a Molar Mass by...

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Experiment # 3 Determination of a Molar Mass by Freezing Point Depression Student Name: Gabriella Selestiyanta Student Number: 1004627914 Lab Partner: Konrad W. Section Number: 132 Date the experiment was performed: February 16 Date the lab report was submitted: March 2 Submitted to George Li Pre-lab Questions Pre-lab Quiz Introduction Experimental Method Results and Calculations Accuracy Discussion Learning Objectives Summary References Experimental Technique Report Presentation
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Introduction The purpose of this experiment is to determine the molar mass value of an organic compound. It was done by freezing test tubes filled with either the pure solvent or the three solutions with different concentrations inside a beaker filled with ice, in order to determine their freezing point. The temperature decrement of the pure solvent and the solutions were noted down for every 15 seconds. Next, the freezing point of both the solvent were achieved when all of them have reached a specific temperature and stayed constant in that temperature. Further decrement was observed to make sure that the solid solvent or solutions underwent more cooling. This experiment is very useful in the field of forensic science, since it aided scientists to identify a large number of unknown samples that can be used as evidences just by determining its molar mass 6 . Experimental Method The complete method of this experiment can be found in the “CHM120H5S Chemical Principles 2 2018 Course Manual” on page 41-43 4 . There were no modifications made during the experiment. Results Graphs 0 200 400 600 800 1000 1200 1400 0 5 10 15 20 25 Temperature vs. Time of Pure Solvent Temperature Time (s) Temperature ( ) 1
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Figure 1 0 100 200 300 400 500 600 0 5 10 15 20 25 Temperature vs. Time of Solution 1 Temperature Time (s) Temperature ( ) Figure 2 2
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0 50 100 150 200 250 300 350 400 450 500 0 5 10 15 20 25 Temperature vs. Time Graph of Solution 2 Temperature Time (s) Temperature ( ) Figure 3 3
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0 200 400 600 800 1000 1200 -5 0 5 10 15 20 25 Temperature vs. Time of Solution 3 Temperature Time (s) Temperature ( ) Figure 4 Calculations A. Pure Solvent a. T f (average of constant temperature): 7.38167 C B. Solution 1 a. T f (average of constant temperature): 1.68 C b. T f : 7.38167 C - 1.68 C = 5.70167 C c. Value of Van’t Hoff factor (i): 1 d. m = T f /(i x K f ) = 5.70167 C/(1 x 20 C.kg/mole) = 0.285m e. m = moles solute/kg of solvent moles of solute = 0.285m x 9.2880 x 10 -3 kg = 0.002647 moles f. Molar Mass = mass solute/moles of solute = 0.1483g/0.002647 moles = 56.024 g/mol C. Solution 2 a. T f (average of constant temperature): 1.273842 C b. T f : 7.38167 C - 1.2736842 C = 6.1079858 C c. Value of Van’t Hoff factor (i): 1 d. m = T f /(i x K f ) = 6.1079858 C/(1 x 20 C.kg/mole) = 0.3054m e. m = moles solute/kg of solvent moles of solute = 0.3054m x 9.2880 x 10 -3 kg = 0.0028365 moles f. Molar Mass = mass solute/moles of solute = 0.2731g/0.0028365 moles = 96.28 g/mol D. Solution 3 4
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a. T f (average of constant temperature): -0.6525 C b. T f : 7.38167 C – (-0.6525 C) = 8.03417 C c. Value of Van’t Hoff factor (i): 1 d. m = T f /(i x K f ) = 8.03417 C/(1 x 20 C.kg/mole) = 0.4017085m e. m = moles solute/kg of solvent moles of solute = 0.4017085m x 9.2880 x 10 -3 kg = 0.00373 moles f. Molar Mass = mass solute/moles of solute = 0.4212g/0.00373 moles = 112.9 g/mol E. Average molar mass from the 3 solutions:88.40 g/mol
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