Person 1 ________________________________
Person 2 ________________________________
Date ______ Period ____________
Diffusion is defined as the movement of molecules from an area of high concentration to an area of low concentration. The diffusion of water molecules through a selectively permeable membrane is known as osmosis. Movement through membranes is called transport. Diffusion and osmosis are passive forms of transport; this means that they do not need energy to move from areas of high concentration to areas of low concentration. Active transport requires energy to transport molecules from low concentration to high concentration.
Question : How will soaking Gummy Bear candies in distilled water affect their size?
Prediction (explain your prediction based on the background information) :
Procedure :
1. Use a pen to label your cup with your name & class period.
2. Use a ruler to find the volume of your gummy bear. (Height x width x depth)
3. Use an electronic balance to find the mass of your gummy bear. a. Use a petri dish to protect the pan of the balance. b. Remember to subtract the mass of the dish.
4. Record descriptive observations about the candy bear.
5. Fill your cup ½ way full with distilled water.
6. Put your gummy bear in the water.
7. Set the beaker aside for 24 hours.
8. Tomorrow you will be repeating steps 2 – 4.
Data:
Volume (mm
3
)
Before soaking in water After soaking in water
Mass (grams)
Observations
Part 2: The next day
1. After the gummy bear has been in the distilled water overnight,
take it out of the water and pat it dry. Be very careful because the candy is now extremely breakable.
2. Follow steps 2 through 4 from yesterday and record your results on the data table above.
Remember, your gummy bear will be very soft and break easily.
3. After data is recorded you need to complete the calculations at the top of page 2.
1

Calculate the percent change in the size of the candy:
% Change in volume = (After soaking vol. – Before soaking vol.) / Before soaking vol. x 100
( _________ - _________) / ________ x 100 = _____%
% Change in mass = (After soaking mass – Before soaking mass) / Before soaking mass x 100
( _________ - _________) / ________ x 100 = _____%
4. If you know that water weighs 1 gram for every cubic centimeter (1 cm
3
= 1000 mm
3
) then calculate the following:
The volume of water gained based on mass :__________________________
(Today’s mass – yesterday’s mass, then convert to mm
3
)
The volume of water gained based on volume :__________________________
(Today’s volume – yesterday’s volume, this should already be mm
3
)
5. Question: Is the gain all due to water?____________________________
6. Before soaking the Gummy Bear in distilled water, the distilled water tested negative for glucose.
Now perform a glucose test using Benedicts solution on the water that the bear was soaking in. a. What were the results? ___________________________ b.
SHOW ME THE R ESULTS FOR CREDIT!!!!! ______________(my initials) c. Explain why glucose (is/isn’t) found in the soaking water: _________________________
___________________________________________________________________________
___________________________________________________________________________
7. Before soaking the Gummy Bear in distilled water, the distilled water tested negative for starch.
Now perform a starch test using Lugol’s solution on the water that the bear was soaking in. a. What were the results? ___________________________ b.
SHOW ME THE R ESULTS FOR CREDIT!!!!! ______________(my initials) c. Explain why starch (is/isn’t) found in the soaking water: __________________________
__________________________________________________________________________
__________________________________________________________________________
8. Now, look at the question in step 4 again. In the space below, explain how the difference in mass may or may not be an accurate measure of the amount of water gained by the gummy bear. Explain!
I am not impressed by one or two word answe rs. (Hint: Look at the results from step 6.)
2