I. Introduction.

A colligative property of a solution, freezing-point depression, is used in this experiment to determine the molecular weight of an unknown compound A. The freezing-point of a solution is always lower than the freezing-point of the pure solvent. The magnitude of the freezing-point depression (D Tf) is directly proportional to the molality of the solution. Thus, D Tf = - kf  molality            (1) where kf is a parameter called the freezing-point depression constant. This constant corresponds to the value of D Tf when molality = 1 and is characteristic of the solvent. Some representative values are shown in Table 1.

Table 1.

Solvent	kf (oC/molality)
diphenyl	8.0
water	1.86
benzophenone	9.8
naphthalene	6.90
camphor	37.7
cyclohexane	20.5

The molality of a solution is given by where WA is the mass of compound A, Wsolv is the mass of solvent, and FWA, is the formula weight of A. Hence, by substitution of (2) into (1) and subsequent rearrangement we obtain (3) Equation (3) tells us that we may calculate the molecular weight of A (FWA) provided a solution is prepared containing known masses of solute and solvent (WA and Wsolv), the freezing-point depression of the solution can be measured, and the value of kf for the solvent is known. This is a fairly common method for determining the molecular weights of non-volatile substances and it forms the basis for this experiment. You will dissolve an accurately weighed quantity of an unknown solute in an accurately weight quantity of solvent and measure the freezing-point depression for the resulting solution. Then, using the freezing-point depression constant for solvent (Table 1) and equation (3), you will calculate the molecular weight of the unknown solute.  Sample calculation. A solution prepared by dissolving 1.53 g of an unknown compound in 5.60 g of naphthalene has a freezing point which is 10.1 oC lower than the freezing point of pure naphthalene. The freezing-point depression constant for naphthalene is 6.90 oC/m. What is the molecular weight of the solute? MWsol = 1000× kf×Wsol / WsolDT = - (1000 g/kg)(6.90 oC/m)(1.53 g) / (5.60 g)(- 10.1 oC) = 187 g/mol II. Experimental Procedure. 1. Pipet exactly (± 0.01 mL) 15 mL of cyclohexane into the 10 cm  tube. For cyclohexane, d=0.779 g/mL (View the explanatory video) 2. Prepare a beaker containing ice and water. Stir it until the temperature is close to freezing. Insert the test tube in a beaker containing ice and water.             3. Place a temperature probe in the test tube. Your setup should be similar to that illustrated in the diagram. (View explanatory video) 4. Start data collection  (Click on "Collect" icon) and observe the falling temperature of the molten solvent. Gently stir the liquid with the probe during this period of cooling . The temperature should level off as the solvent begins to crystallize. Record this freezing-point temperature (± 0.1 oC) for your data. Note that you may have some jump of the temperature because of supercooling. 5. In the window data, click at "Store latest run". 6. Repeat steps 3-5. Calculate an average freezing-point temperature (± 0.1 oC) and an average deviation (± 0.1 oC) for your data. Do not forget to click on "Store latest run" before the next run. (View explantory video) 7. Melt the Cyclohexane and carefully remove any that is adhering to the walls of the tube.         Accurately weigh out 0.1 g (± 0.001 g) of the unknown compound on a piece of clean weighing paper. Transfer the sample to the  tube containing the cyclohexane.           You can use an ultrasonic cleaning bath to help the sample dissolve. Here is a of how this works.   (Video) (Video) Video 8. Repeat steps 3 - 6. Be sure that all of the unknown dissolves in the cyclohexane. The finished Cyclohexane/unknown sample ice should look like the demonstration video to the right. 9. Repeat step 7, and make a second addition measurement. As a result, you will have one freezing temperature for a pure solvent and two different freezing temperatures for two different amount of unknown compound.  An example of such experiment is shown below: (Video) 9. Calculate two D Tf  values. 10. Calculate the molecular weight of the solute (3 significant figures). Estimate standard deviation of your result.

To clean glassware, melt the mixture of solvent and unknown. Pour the liquid into the waste container with the appropriate label.  Determination of a Molecular Weight by Freezing-point Depression. Name:________________________________________Date:__________________ III. Data and Results.     Unknown #______ Mass of solvent (g):                                 15 mL x 0.779 g/mL = 11.68 g Freezing-point of solvent (oC): Trial 1: Trial 2: Average value (oC): First addition: Mass of unknown compound (g): Freezing-point of solution (oC): Second addition: Mass of unknown compound (g): Freezing-point of solution (oC): Molecular weight of unknown compound (g/mol): Results received by other students of your group:     II. Questions. How your results will be affected if: 1. The average freezing point of the solution was erroneously recorded as 1 oC higher than the actual value. 2. All thermometer readings were consistently high by 0.2 oC. 3. The unknown contained an impurity which was insoluble in the solvent