Today in Chemistry we did the Colligative Properties Lab. The goal of this lab was to use boiling point elevation data to identify an unknown salt. We began by labeling four 100 mL beakers A,B,C, and D and recording their masses in our super neat data tables. Next we filled each of the beakers about half full with distilled water. We recorded the mass of the beaker and water in our data table. We then placed the beakers full of water on the hot plate and heated them to about 85°C at which point we removed them from the hot plate. We determined the boiling point of the water in beaker A by noting the plateau on our lab pro. We recorded this number and proceeded to add about 5.0 grams of the unknown ionic solid to beaker B. We placed it on the hot plate and recorded its boiling point in our data table. We then added about 10.0 grams of the unknown solid to beaker C. We placed it on the hot plate and recorded its boiling point once again. Finally we added 15.0 grams of the unknown solid to beaker D, placed it on the hot plate, and recorded its boiling point in our data tables.
As a part of the data/calculations we were asked to calculate the molality of each solution and were given the molal boiling point elevation constant: 0.51°C kg/mol. To calculate the molality of each solution, we used the equation ΔTb = Kb · m · i. To calculate change in temperature, we subtracted the boiling point of beaker A from beaker B, beaker B from beaker C and so on. We were told the ionic sold has 2 ions, so i=2. We plugged these values into the equation and solved for m to get a value for molality.
To calculate the moles of solute in beakers B, C, and D, we used the molality equation which says that molality = mol solute / kg solvent. We have the value for molality and to find the kg of solvent we simply subtracted the mass of the beaker from the mass of the beaker and water and divided the answer by 1000 to obtain the value in kilograms. To find the moles of solute, we multiplied the molality by kg solvent.
To find the molar mass of the solute in beakers B,C, and D we divided the number of grams of solute in each mixture by the moles of solute obtained in the previous step. We added 5.0 grams to beaker B, 10.0 grams to beaker C, and 15.0 grams to beaker D. These values were divided by the moles solute to obtain the molar mass of the solutes. To calculate the average of the molar masses, we just added them together and divided by three.
The conclusion asks you to decide the formula for the unknown solute, so choose the formula with the molar mass closest to your average. The choices are NaCl, KI, NaNO3 or NaBr. Support your claim with evidence and then calculate the percent error:
| actual value – theoretical value | x 100 %
theoretical value
The lab is due Friday and so are all of the worksheets and Webassigns. Study for the Test Friday!
The next scribe will be Emilio I!