All Chemistry Classes
Ice Cream Lab – Freezing Point Depression
Name __________________
ICE CREAM LAB
Item
Large Ziploc
Milk
Vanilla Extract
Sugar
Amount per Group
2
2 Cup
2 teaspoon
4 tablespoons
Salt
¼-½ cup
Conversions
1 Gallon of Milk about 20 cups
6 tsp = 1 bottle of 29mL
2 tbs is about 24 grams (1 bag of sugar =
2270g)
15 tbs = 180 grams (1 container of salt =737g)
Procedure:
1. In one of the large Ziploc bags add the milk, vanilla extract, sugar and any toppings
2. Seal the bag trying to remove as much air as possible.
3. Place this bag in another large Ziploc bag. Seal the 2nd small bag trying to remove
as much air as possible.
4. In one of the coffee cans fill it half full of ice.
5. Measure out about ¼ c of salt and pour directly onto the ice.
6. Place the Ziploc bags into the coffee can containing the salty ice.
7. Roll the can back and forth on the ground. The temperature of the ice should drop
well below zero. Do this for about fifteen minutes.
8. At the end of the fifteen minutes, squeeze the ice cream. If it is hard, the ice cream
is ready to be eaten. If not, continue to move, press, and turn the bags for another
five minutes. Repeat the checking process until the ice cream is done.
9. When the ice cream is ready, obtain one cup and one spoon per person. Eat the ice
cream.
10. Check the temperature of the ice (NOT the ice cream) and record. ______________
This lab is based on the concept of freezing point depression, which is the difference
between the freezing points of a pure solvent and a nonelectrolyte solution in it. The
formula for this is ∆tf = kf (m)(x), where x = # of ions produced when the solute
dissolves and kf water = -1.86 oC/m.
In class we discussed Boiling point elevation – when solutes dissolve in liquids, they
raise the boiling points. It is the same concept as freezing point depression except
boiling point increases. The formula is similar ∆tb = kb(m)(x), but kb water = 0.512 oC/m.
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All Chemistry Classes
Ice Cream Lab – Freezing Point Depression
Name __________________
Perform the calculations below for the freezing point depression and boiling point
elevation of the following Non-electrolytes and electrolytes. In all problems, use x
= 1. Calculations: Show all work!
1. NON-ELECTROLYTES in solution:
a. 10.0 g urea (CH4N2O) in 125. g water
b. 27.6 g glucose (C6H1206) in 200. g water
2. What are the new freezing points and boiling points of the solutions in a-b above?
3. ELECTROLYTES in solution:
a. 18.0g MgCl2 250. g water
b 80.0 gNH4 100. g water
4. What are the new freezing points and boiling points of the solutions in in a-b above?
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