Molality: Benzoic Acid: MW g/mol Freezing Point Depression Data: 0.750 g BA/8.00

Question

Molality: Benzoic Acid: MW g/mol Freezing Point Depression Data: 0.750 g BA/8.00 g LA: 1.50 g BA/8.00 g LA: AVERAGE K_f: Save the data in the file fpd2lauric150 acid on your jump drive. Calculate the second freezing point depression constant as in the previous case but be certain to use the mass of the solute in the second solution when calculating molality. Average these results for the value of K_f and report them in your lab report. You can now look at any or all of your data on the same graph. Simply click on the Temperature label on the Y-axis, expand the runs you want to see, and check next to the box next to temperature Wa- Lah! (Do you think it would be possible to obtain the data you collected today using just a thermometer, constantly stirring AND reading the thermometer every second???) Your reports for the next three labs should include the following sections: Date and title of experiment Purpose of the experiment and a synopsis of the procedure Equations and reactions, as necessary Data, calculations, and results, presented in an appropriate and concise manner Discussion of the results, including error analysis Conclusions Your TA will further discuss the lab reports with you. Mp of pure Lauric Acid: Molality Benzoic Acid: MW g/mol Freezing Point Depression Data: 0.750 g BA/8.00 g LA: 1.50 g BA/8.00 g LA: AVERAGE K_f: Now that we know where the normal freezing point of Lauric Acid is, we can begin to determine the freezing point depression constant Before starting the experiment, store and hide your last run again, and then change the title of the graph on Logger Pro to read "Freezing Point Depression of LA and 0.750 g BA". Obtain a test tube that is labeled to contain a solution of 8.00 grams of Lauric acid (remember, this is the solvent) AND 0.750 grams of benzoic acid (this is the solute). Place the test tube in your water bath and melt the mixture. When the solution is fully melted, take it out of the hot water bath, add the temperature probe and click and obtain the freezing curve for this solution. Once the data has started collecting, place the test tube in the room temperature water bath and stir, stir, stir. Determine the freezing point as in the case of the pure Laurie acid. Save these data in a file fpdlauric075 acid on your jump drive. Calculate delta T and the molality of this solution. Use the molecular weight of benzoic acid (MW = 122.12 g/mol) to determine the number of moles of solute. Divide this by the number of kilograms of solvent (Lauric acid) to obtain the molality of the solution. The above procedure should be repeated, but this time obtain a test tube that has a solution with 8.00 grams of solvent and 1.50 grams of solute. Change your graph title and file names to represent this increase in solute. Melt this second mixture and determine the freezing point of this second solution. you stir constantly while the liquid Lauric Acid is cooling. Once you reach the flat area of the cooling curve, you can stop stirring You will probably want to add a few ice chips to your beaker of water to aid in cooling. Record your data up to the point where the temperature starts to trail off and drops below 30 degree Celsius. the experiment and analyze your data (see more info on data analysis below) You can highlight the flat portion of the curve, and then click on "linear fit" The Y-intercept is your freezing point for the Lauric Acid. Move the data box so you can see your data, and then click on Save Data. Save your data on your jump drive with the file named mplauric acid To save the same data file to your partner's jump drive, you can safely remove your drive, and insert your partner's drive. Go to file, select Save as. rename the file m plauric acid2. and select the jump drive to save to ** After each experiment, once you have saved your data, you will need to go to the Experiment menu in Logger Pro and elide on "Store Latest Run to prepare for the next experiment To hide the data from the last run. Click on the "Temperature" label on the Y-axis, select "More", and un-check run 1 to hide it. At this point, return the test tube with the now frozen Lauric acid to the hot water bath, and re-melt the Lauric acid COMPLETELY. Repeat the above experiment once more and record the data and save it 10 your jump drive as before, but with the name mp2lauric acid You will later determine the average of your freezing points for Lauric acid. Return the test tube with the frozen Lauric acid in the hot water bath, and remelt the Lauric acid COMPLETELY once again. When the Lauric acid has melted, you can take the temperature probe out of the test tube. (ALWAYS complete this step after each freezing point determination today-the goal is to leave as much of the liquid in the test tube as possible when removing the temperature probe after each experiment!!) When you pull the temperature probe out of the Lauric acid, you may notice that there is some Lauric acid residue on it. There should not be much if you have fully melted the Lauric acid before removing the temperature probe. If there is some residue, swirl the temperature probe in the warm water to melt the Lauric acid and wipe it off with a paper towel. Carefully blot the temperature probe completely dry. When analyzing your data, the freezing point of the Lauric acid will be the flat portion of the curve. The window below shows some possible data for the freezing point of Lauric acid. On the graph, the "flat portion" of the curve is easy to pick out. Sometimes as we remove heat from a liquid, we can temporarily cool it below its freezing point without forming a solid. This is called super-cooling. Eventually, stirring and agitation will cause the liquid to freeze, and the temperature will return to that of the freezing point The flat portion of the curve occurs immediately after this. Your solution may not supercool, but as you can see, there is a flat portion to the curve. During the freezing process, the flat part of the cooling curve represents the heat of fusion. The temperature does not change during the heat of fusion because the energy "lost" during the freezing process is used in packing the Lauric acid molecules into the dense, solid state. The temperature continues to fall after the flat portion. This is due to further thermal cooling of the solid Lauric acid. Determine the freezing point temperature of the pure Lauric acid by averaging the values you obtained in your two experiments. Today we will use Lab Pro to determine the freezing point depression constant for lauric acid, which is a non-electrolyte [For strong electrolyte solutions, the van't Hoff factor must bee considered, see your text book for more information on the van't Hoff factor. We will ignore the i term in this experiment since lauric acid is a non-electrolyte] The amount that the freezing point is lowered is given by the equation: delta T = K_t m where delta T is the lowering of the freezing point in degree C for the solution (relative to the freezing point of the pure solid) and is commonly reported as a positive number m is the molality of the solution K_f is the freezing point depression constant for the solvent The solution will be a solute (benzoic acid) in the solvent lauric Acid. The solution will have a specific molality Molality = moles solute/mass solvent Molality is defined as the number of moles, of solute, divided by the mass of the solvent. This stands in contrast to molarity, which is the number of moles of volute, divided by the volume of solution. Why is molality used instead of molarity? Volume is temperature dependent, so at different temperatures, the volume of a solution will change, which would also affect the molarity. Molality does not have this problem, since the mass of solvent remains constant. It is a good idea to keep a spare 400 mL beaker of warm (not boiling) water at hand to rinse the temperature probe after each run.** Attach the temperature probe to CH I on the LabPro interface, and connect the interface to the computer. Plug the interface in. Start the LoggerPro software. Click on the "Open File" icon, and select the "Chemistry with Computers" folder. Then find and double click on the file "15 Freezing Pt Depression. cmbl". Double click on the graph title, and correct the title to read "Freezing Point of Laurie Acid" so the title is appropriate for your graph. The temperature probe can now be used to determine the freezing point of pure Lauric acid (the solvent in today's experiment). Laurie acid melts at a temperature slightly above room temperature. You should have a hot plate with a 400 mL beaker of water that is about 60-70 degree Celsius. Obtain a test tube that has pure Lauric acid in it. Put the test tube in the beaker of hot water and melt the Lauric acid. Gently swirling the test tube in the water bath will aid in melting. (Remember, the hot plates do not need to be turned up past three. Our hot plates are research quality, and can obtain temperatures that will melt lead!). When the Lauric acid is fully melted and you are ready to acquire data, remove the test tube from the hot water bath. The TOP of the test tube should be cool enough to bold, but you can also use a loop of folded-over paper towel to hold the test tube by the top (Don't Burn Your Fingers). Put the temperature probe into the test tube, making sure the tip is totally immersed molten Lauric Acid, and begin stirring the melted Lauric acid. Wait 5 seconds and click to begin collecting? data with Lab Pro. Next, put the test tube with the melted Lauric acid into a room temperature water bath. You will need to use the temperature probe to constantly stir the Lauric acid, using an up down and side to side motion. Make sure that the temperature probe stays IN the Lauric Acid and

Explanation / Answer

From the given plot

Freezing point = 41.8 oC

Freezing point of pure lauric acid = 43.2 oC

So,

dTf = 43.2 - 41.8 = 1.4 oC

For, 0.75 g/122.12 x 0.008 = 0.768 m

Kf = dTf/i x m = 1.4/1 x 0.768 = 1.824 oC/m

For, 1.5 g/122.12 x 0.008 = 1.535 m

Kf = dTf/i x m = 1.4/1 x 1.535 = 0.912 oC/m

Average Kf = (1.824 + 0.912)/2 = 1.37 oC/m

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