Chemistry of Ice-Cream Making: Lowering the Freezing Point of Water

Difficulty	Advanced
Time Required	1 day :)
Prerequisites	Understanding the concepts of molecular weight and moles. Access to a gram balance accurate to 0.1 gram.
Material Availability	Specialty items are needed. See the Materials tab for details
cost	Average($40 - $80)
Safety	No issues

Contents[]

02:36

41:20

Abstract[]

Have you ever made your own ice cream? If you have, you probably surrounded the ice cream container with ice and rock salt to get the mixture cold enough to freeze. But why does that work? How does adding salt (or other substances) affect the freezing point of water? Find out with this ice-cold science project.

Objective[]

In this science project, you will investigate how dissolving chemicals in water changes the freezing point of a solution.

Materials and Equipment [1][]

These specialty items can be purchased from the Science Buddies Store:

Chemistry of Ice Cream Making kit (1). Includes:
- Test tubes, about 18 millimeters (mm) by 150 mm (7)
- Test tube rack
- 250 mL beakers (6); disposable cups can be substituted if not purchasing the kit
- 100 mL graduated cylinder
- Gram balance. If you are not purchasing the kit make sure you use a balance accurate to 0.1 grams, such as the Fast Weigh MS-500-BLK Digital Pocket Scale, 500 by 0.1 G, available from Amazon.com
- Thermometer capable of reading at least as low as −10°C
- Stir rods (2); spoons can also be used if not purchasing the kit
- Table salt (sodium chloride, NaCl) (500 g or 18 oz.)
- Granulated sugar (sucrose) (360 g or 13 oz.)
- Large Styrofoam® cup (12 ounces [oz.] or more)
- Small spoon or scoop, for scooping up salt and sugar
- Lab notebook

You will also need to gather these items:

Water
Ice

Order Product Supplies[]

Buy KitProject Kit: $64.95==Introduction==

If you have ever made ice cream with an old-fashioned hand-crank machine, you probably packed a mixture of ice and rock salt around the container holding the cream. The salt allows the ice/salt mixture to get colder than pure water ice. This extra-cold ice/salt mixture freezes the cream in the ice cream machine.

If you live in a cold climate, you may also have seen trucks spreading salt and sand on the streets to prevent roads from getting icy after a snowfall. The salt lowers the temperature at which water freezes, so ice can melt even when the temperature is below the normal freezing point of water.

In both cases, salt is used to lower the temperature at which water freezes, a temperature called the freezing point. The decrease in freezing point that happens when salt is added to water is called freezing point depression. Pure water freezes at 0° Celsius (C), but water mixed with salt will freeze at less than 0° C.

Freezing point depression is not unique to water and salt; it happens with all solutions. To make a solution, we dissolve a solute in a solvent. The solvent does the dissolving, and the solute is the thing that is dissolved. In this science project, you will use water as the solvent and sodium chloride (table salt) and sucrose (granulated sugar) as the solutes.

Adding a solute, like salt, to a solvent, like water, lowers the freezing point of the solvent. But, by how much? One degree? Ten degrees? The answer to that question depends on three things: the molality of the solution, the van't Hoff factor of the solute, and the molal freezing-point-depression constant of the solvent. We will take a look at each of these factors.

Molality, m, is defined as moles (mol) solute per kilograms (kg) solvent, as shown in Equation 1, below:

Equation 1:

Molality (moles/kg) = Moles (mol) of Solute/Kilograms (kg) of Solvent

We care about molality because freezing point depression is a colligative property, a property that depend on how many solute particles are in the solvent, not the kind of solute particles. Molality, m, is one piece of this "how many solute particles are present?" question. The van't Hoff factor is the second part of the "how many solute particles are present?" question.

The van't Hoff factor, i, deals with how a molecule of solute dissociates, or breaks apart, in the solvent. Covalent compounds, like sucrose (C12H22O11), do not dissociate in solution. These compounds have van't Hoff factors i = 1. Ionic compounds, like table salt (NaCl), dissociate when in solution. Table salt (NaCl) has a van't Hoff factor i = 2 because it dissociates into two ions in solution, Na+ and Cl-.

The third factor, the molal freezing-point-depression constant, Kf, is different for every solvent. It has units of (° C/m), and it tells us how much 1 mol of solute added to 1 kg of solvent will lower the solvent's freezing point. For pure water, Kf = 1.86° C/m.

We can combine these three factors—molality, m, van't Hoff factor, i, and molal freezing-point-depression constant, Kf—into an equation that predicts how much the freezing point of a solvent will decrease, ΔT, when a certain amount of solute is added. Equation 2, below, is the freezing point depression equation:

Equation 2: Degrees Freezing Point is Depressed (° C) = Molal Freezing-Point-Depression Constant (° C/m) × molality of solution (mol solute/kg solvent) × van't Hoff Factor (unitless)ΔT = Kf m i{| class="formulatext" style="margin:0pxauto;border-style:none;" |

ΔT is the freezing point depression in degrees Celsius (° C)
Kf is the molal feezing-point-depression constant in degrees Celsius per molal (° C/m)
m is the molality of the solution in moles per kilogram (mol/kg)
i is the van't Hoff factor of the solute, which does not have units

|}

If you want to know the new freezing point of a solution, Tn, you subtract the change in temperature, ΔT, from the original freezing point, Tf, as shown in Equation 3, the solution freezing point equation:

Equation 3: Solution Freezing Point (° C) = Solvent Freezing Point (° C) - Degrees Freezing Point is Depressed (° C) Tn = Tf - ΔT{| class="formulatext" style="margin:0pxauto;border-style:none;" |

Tn is the freezing point of the solution in degrees Celsius (° C)
Tf is the freezing point of the solvent in degrees Celsius (° C)
ΔT is the freezing point depression in degrees Celsius (° C)

|}

In this science project, you will investigate how the freezing point of a solution changes with the concentration and kind of solute in the solution. You will also compare the results from your experiments with the results predicted by Equation 3, the new freezing point equation.

Terms and Concepts[]

To do this project, you should do research that enables you to understand the following terms and concepts:

Freezing point
Freezing point depression
Solution
Solvent
Solute
Molality
Mole (chemistry)
Colligative properties
van't Hoff factor
Dissociate
Molal Freezing-Point-Depression Constant
Molecular weight

Questions[]

When sodium chloride dissolves in water, how many solute molecules result from each molecule of solid dissolved?
When sucrose dissolves in water, how many solute molecules result from each molecule of solid dissolved?

Experimental Procedure[]

Preparation of Ice Bath[]

Fill the Styrofoam cup (or 400 mL beaker) 3/4 full with ice.
Cover the ice with 1/4 to 1/2 inches of table salt.
Stir this ice-salt mixture with a spoon or stirring rod.
Use the thermometer to check the temperature of the ice-salt mixture. It should be at least −10° C.
You will use this ice bath to freeze many samples of test liquids. During the course of your experiments, you may need to pour melted water out of the ice bath and replenish the ice and salt. When you do this, wait until the temperature of the ice bath drops to at least −10°C before continuing your experiment.
1. Tip: Check the temperature of the ice bath before freezing each of your test liquids.
Always carefully rinse and dry the thermometer before using it to measure the freezing point of your test liquids. You do not want the thermometer to carry salt water into your test liquids!

Determination of Freezing Points of Test Liquids[]

You will measure the freezing point of seven different test liquids: three salt solutions, three sucrose solutions, and a pure water control. The pure water control is important for determining the freezing point of the water in your experiment, which may or may not be exactly 0° C.
Use a permanent marker to label one of the test tubes and one of the 250 mL beakers (or 8 oz. disposable cups, if you are using those instead) as "#1".
Prepare your first test liquid in the beaker labeled #1.
1. Put beaker "#1" on the gram scale.
  1. Tip: Some scales, like the one in the Science Buddies kit, have a cover over the weighing surface. This should be removed before using the scale.
2. Zero the gram scale. This is usually done by pressing a button that says "Zero/Tare".
3. Put 2.9 g of table salt (NaCl) into the beaker.
4. Take the beaker off the scale and place the beaker on a stable surface.
5. Measure 100 mL of water in a graduated cylinder, and pour it into the beaker with the salt. Stir with a spoon or stirring rod until all of the crystals are dissolved.
6. Rinse the spoon or stirring rod with water. You do not want to contaminate the other test liquids.
Fill test tube #1 half-way full with test liquid #1. Place the test tube in the Styrofoam cup with the ice and salt.
1. The liquid in the test tube should be below the level of the ice and salt in the cup.
2. Do not allow any ice or salt from the cup to get into the test tube.
Stir the test liquid in the test tube gently with a thermometer while keeping track of the temperature.
When the first ice crystals appear on the inside wall of the test tube, record the temperature in your lab notebook. This is the freezing point of the test liquid. Figure 1, below, shows what ice crystals look like.

Figure 1. The opaque, white regions of this liquid are frozen. You know that you have reached the freezing point when the transparent liquid develops opaque, crystalline regions.

Empty the test tube (you can pour the liquid down a drain in your house), and then refill the same test tube with fresh test liquid #1. Repeat steps 4 through 6 with a fresh sample. Then empty and refill the test tube again, repeating steps 6 through 8 for a third time. You should have a total of three replicates for test solution #1, all carried out in the same test tube.
1. If you need to make more of the test liquid, simply repeat step 3.
Repeat the steps above for each of the remaining test liquids, making sure to use a different beaker and test tube for each of the test liquids. You do not want the residue from one test liquid to contaminate the other test liquids. Make sure to label the beakers so that you know which test liquid is in which beaker.
1. Test liquid #2 = 5.8 g NaCl in 100 mL water
2. Test liquid #3 = 11.7 g NaCl in 100 mL water
3. Test liquid #4 = 17.1 g sucrose in 100 mL water
4. Test liquid #5 = 34.2 g sucrose in 100 mL water
5. Test liquid #6 = 68.5 g sucrose in 100 mL water
6. Test liquid #7 = 100 mL water (simply measure 100 mL of water in the graduated cylinder and store the water in the graduated cylinder)
Calculate the freezing point depression, ΔT, for all three replicates of each of the NaCl and sucrose solutions. Record these values for ΔT in your lab notebook.
1. First, average the freezing point temperatures you observed for plain water, test liquid #7. This averaged temperature will be the solvent freezing temperature.
2. Re-arrange the terms in Equation 3 of the Introduction to figure out how to calculate ΔT.
Average the freezing point depression, ΔT, for each of the NaCl and sucrose solutions. Record these averages in your lab notebook.

Calculation of Expected Freezing Point Depression[]

Calculate the molalities of the NaCl and sucrose solutions.
1. Molality is defined as the number of moles of a substance, divided by the weight (in kg) of the solvent, as shown in Equation 1 in the Introduction.
2. The number of moles of a substance is defined as the weight of the substance (in g) divided by the gram molecular weight of the substance.
3. The gram molecular weight of NaCl is 58.443 g.
4. The gram molecular weight of sucrose is 342.3 g.
5. 100 mL of water weighs 0.1 kg.
Make a table of your results, like the example shown in Table 1, below.

Solution	g Substance	Molecular Weight Substance (g)	Amount Water (kg)	Molality (mol/kg)
Test liquid #1	2.9 g NaCl	58.443	0.1	0.50
Test liquid #2	5.8 g NaCl	58.443	0.1	0.99

Table 1. Table for calculating the molality of NaCl and sucrose solutions.#How do the molalities of the NaCl and sucrose solutions compare?

For each of your NaCl and sucrose solutions, graph the molality of the solution (x-axis) versus the average amount of freezing point depression, ΔT (y-axis). What trend do you notice?
Use Equation 2 in the Introduction to calculate the expected freezing point depression for each of the NaCl and sucrose solutions. Make a second table to help you calculate with these calculations. Table 2, below, shows what that table might look like.

Solution	Substance	van't Hoff Factor	Molality	Kf for water (° C/m)	Expected Freezing Point Depression (° C)
Test liquid #1	NaCl	2	0.50	1.86
Test liquid #1	NaCl	2	0.99	1.86
Test liquid #1	NaCl	2		1.86
Test liquid #2	Sucrose	1		1.86
Test liquid #2	Sucrose	1		1.86
Test liquid #2	Sucrose	1		1.86

Table 2. Table for calculating expected freezing point depression using Equation 2.#Compare the experimentally-measured freezing point depression to the calculated freezing point depression for each of the NaCl and sucrose solutions. How close are the experimental and calculated values? What do you think causes any differences between the measured and calculated values?

Hint: The real van't Hoff factors for solutes in solution are often less than their ideal values due to interactions between the ions in solution.