TEACHER GUIDE

TOPICS:  Chemistry, freezing point depression, vapor pressure, solutions, concentrations

LEVEL: High school

TIME: One period

ADVANCE PREPARATION:  Obtain class set(s) of test tubes that may be reserved for subsequent use only in this experiment. Prepare a cream mixture for the class.  The following is a simple recipe that makes enough for 25 students:

1 cup heavy whipping cream

1 cup milk

1/3 cup sugar

1 teaspoon vanilla

Makes 2 1/2 cups (about 25 ml/student).

Limit the use of additional flavoring extracts; the alcohol in these reduces the freezing point of the ice cream mixture and the students may not finish within one period .

MATERIALS: Each student makes ice cream, sharing containers and thermometers with a partner.  Coffee cans or other containers may be used for the large beakers.  Some groups should use sodium chloride, others should use calcium chloride.  One or two students may be sent to bring in pails of snow to be stored in the sink for class use .

INTRODUCTION: Be sure students understand the meaning of the equation. For example, 1 molal salt solution lowers the freezing point of water 1.86^oC from 0^oC to -1.86^oC .

SAFETY NOTE: Stress to students that these test tubes are reserved for use in this experiment only. Under normal circumstances, nothing in the laboratory should be eaten .

STEP D: Have students flex and bend the end of the straw; the straw may pull out as the ice cream is removed unless the crimped end of the straw is frozen into the ice cream .

STEP E-G: The ice cream mixture will freeze much faster if the test tube is completely surrounded by snow.

QUESTIONS 1-9 require higher order thinking skills.  You may wish to help students with the calculations in questions 1-5, depending on the ability of your students.  We recommend that you discuss questions 6-9 in class before having students write out their answers .

QUESTIONS 1-5: A typical calculation is provided, based on the following data:

NaCl = 50.0 g

Melted snow = 750 g

Freezing point depression = 3.4^oC

1. 50g / 58.5 (g/mole) = 0.85 moles NaCl .

2. 0.85 moles / 0.750 kg H₂O = 1.14 molal NaCl .

3. 1.14 m NaCl 2 moles ions / 1 mole NaCl = 2.28 molal ions .

4. D T = m ^. K_f, so m = D T / K_f = 3.4 /1.86 = 1.83 molal particles .

5. Let x equal the concentration of salt molecules that dissociate.  This gives x molal Na⁺ and x molal Cl^-, leaving (1.14 - x) molal undissociated NaCl.  The total concentration of particles is [x + x + (1.14 - x)] molal, which simplifies to (1.14 + x) molal.  Since 1.83 molal particles was obtained from the freezing point depression, the equation is:

1.83 = 1.14 + x ,           x = 1.83 - 1.14 = 0.69

The fraction of NaCl molecules dissociated is proportional to the number of NaCl molecules dissociated divided by the number of NaCl molecules initially available:

Percent of NaCl dissociated = (0.69 /2.28) X 100 = 30.3% .

6. Yes, the solubility of the solute limits the amount the freezing point can be lowered.  The lowest temperature will be reached at saturation .

7. Equal molal concentrations of CaCl₂ will cause it to lower the freezing point more because it releases 3 moles of ions per 1 mole of solute.  NaCl produces 2 moles of ions per mole of solute.  With equal weights, 50 g of NaCl yields 1.7 moles of ions; 50 g of CaCl₂ yields only 1.36 moles of ions .

8. The water and anti-freeze form a solution which lowers the freezing point of both .

9. The snow and salt formed a solution with a freezing/melting point well below the temperature of the environment in which the solution formed.  The snow, therefore, melted; absorbing 334 joules of energy per gram of snow.  This heat energy was absorbed from the snow, ice cream mixture, and environment around the beaker, resulting in a decrease in temperature.