Developers:
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Joseph J. Leo
Log College Middle School
Centennial School District
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Allen P. Marks
Research Scientist
Rohm and Haas Company
Spring House, PA
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Grade
Level:
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6 to 8
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Discipline:
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Physical Science, General
Science
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Objectives:
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Students will have the opportunity to
observe, measure and analyze the density of aerosol shaving
creams and foams generated from shampoo and dishwashing
liquids or detergents.
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Background:
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The experiment is designed to measure
the density (mass per unit volume) generated by the
foam(ing) content of each product.
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Materials:
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1/2 - or 1-gallon plastic jugs
250- or 300-mL beakers (2 for each product tested)
25 mL and 50 mL graduated cylinders
safety goggles or glasses
shampoo (Head and Shoulders � , Prell � )
shaving cream (Barbasol � , Gillette � )
liquid soap - (Dawn � , Ivory � , Palmolive �
)
triple beam balance
spatula or wooden tongue depressors
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Procedure:
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- Weigh and record the tare weight
of each beaker to be used. (Mass) (Grams)
- Fill each beaker with water and
find the volume of the beaker by subtracting the tare
weight from the beaker and water weight. Since the
density of water is 1.0 g/mL, the weight of the water in
grams is equal to the volume of the beaker in
milliliters. The volume of water within the beaker will
be the actual volume of foam that the beaker will hold.
(Note that a 250-mL to 300-mL beaker has a volume greater
than noted). (Volume) (Milliliters)
- Prepare a 5% by volume solution of
the soap products or shampoo to be used:
A 5% solution is made up in a plastic jug by adding 5
milliliters of soap to 95 milliliters of water. Gently
swirl the contents to disperse the soap in the water.
Then, shake the jug to develop the foam.
Aerosol shaving cream may be used in place of preparing
foam, but it doesn�t work as well as soap foam because it
does not pack well into the beaker without trapping air
pockets.
- Fill two empty beakers with foam,
by inverting the jug and squeezing.
- Weigh the beakers and subtract the
weight of the beaker from the beaker packed with foam.
The remainder is the mass of the foam.
- Remember that 1 gram of water is
equal to one milliliter.
- Calculate the density of the foam
in the following manner:
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Example:
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Density =
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Mass of foam in
beaker
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Volume of water held by
beaker
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Density =
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Wt of the beaker and
foam minus Wt of beaker
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Wt of beaker and water
minus Wt of beaker
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Density =
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Weight of Foam
(Grams)
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Volume of beaker
(mL)
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Density =
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1379 -
1109
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=
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27g Foam
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270 mL
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270 mL water
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Problems:
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- What is the density of the foam
produced by 2% solution by volume if the container holds
270 mL. and its tare weight is 110 grams? The full
container is 130 grams. Calculate the density.
- Calculate the density of a 5%
Dawn� detergent solution using the same container
with the full container weighing 148 grams.
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Extensions:
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- The density of foam generated may
be varied by preparing several different solutions of 2%,
10% and 20% by volume. Also, the solution�s density will
vary by the type of product you use. Soap detergents and
shampoo were found to be the best test products.
- An economical analysis can be made
by testing several different brands of dishwashing
liquids or shampoos at a constant volume (2%, 5%, etc.).
Calculate the densities of each product and then
calculate the cost/volume (milliliters or ounces) as to
price of product.
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Economic
Analysis
(Data made up but you can supply real
data.)
Product A:
Cost:
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$.50 for 16 fl. oz. ( a fl.
oz. is about 29.6 mL, I think, so 16 fl. oz. - 47.4
mL)
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5 % solution makes foam of
density 0.10 g/mL (The 5% solution will have a
density of about 1.0 before foaming since it's
mainly water.)
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So 100 g of 5% solution will
generate
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100
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= 100 mL of foam
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.07
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The whole container
contains
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47.4
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= 95 aliquots of 5 mL
samples
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.50
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So the bottle will make
95,000 mL or 95 liters of foam for $.50, or
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.50
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= 0.0053 cents/liter
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95
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Product B:
Cost:
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$ .60 for 16 fl. oz.
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5 vol % solution makes foam
of density 0.07 g/cc
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So 100 g of 5% solution will
generate
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100
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= 1429 mL of foam
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.07
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The whole container makes
1429 x 95 = 136,000 m = 136 liters 136
liters
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So the cost is
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.60
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= 0.0044 cents/liter
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136
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Please note that this experiment works
well in cooperative learning groups, with every member
having a task to perform.
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