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AP Biology Lab: Osmosis and Potatoes
Transcript of AP Biology Lab: Osmosis and Potatoes
What was the initial task?
1. To determine the isotonic point of a sucrose solution and a potato
2. To determine an unknown concentration of a sucrose solution using a potato.
To determine the isotonic point and various unknown concentrations of a sucrose solution with the aid of potatoes.
If the concentration of the sucrose solution is 0.5M, then the isotonic point has been reached.
If the data collected from the various sucrose concentrations tested are used, then unknown concentrations of sucrose can be determined accordingly.
What's in an experiment?
: molarity of the various concentrations of sucrose in the different beakers
: percent change in mass of the slice of potato
: size and weight of the potato, the sizes of the beakers, the time, the temperature, the amount of solution
Sources of Error
Ants infested the solutions
Error with the weighing scale (trial 1)
Different sizes (thickness) of potatoes
Make a 1L solution of 1M
Create 200mL solutions in with the molarities of 0,
0.1, 0.2, 0.3 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, and 1 and pour each into a labeled 200mL beaker
Pour 50mL of each 200mL solutions into
individual 50mL beakers labeled accordingly to the molarity
Repeat previous step for a second set
Cover the beakers and put each of the solutions in the refrigerator for 7 and a half hours
Cut potatoes relatively the same size (length 2, height 2, width 1cm) that relatively weigh 3 grams
Weigh each of the potato slices and record
Take out 50mL solutions
Put 1 potato in each solutions leave it in overnight (16 hours and 2 minutes)
Sandra, Natalia, and BoRa
Take the potatoes out with a spoon and gently dry to potato without squeezing any water out
Weigh each one and record in table 1 (for trial 1) and table 2 (for trial 2)
, do the exact same as trial 1 and trial 2 (pour 50mL of solution for each molarity, cut and weigh potato, put in solutions). After 7.41 hours, dry, weigh, and record (table 3)
Trial 4 (dried potato trial)
For trial 4:
a. Pour last remaining 50mL of sucrose
b. Slice and weigh potatoes
c. Leave potatoes out for 4 hours
d. Put dried potatoes into the solutions for
e. Take out, dry, and weigh potatoes
f. Record in Table 4
How can it be fixed?
What does it mean to be isotonic?
"having equal water potential since the two solutions have an equal concentration of water molecules." (Isotonic, 2011)
Where will water move?
Water will move from a higher water potential to a lower water potential (more will be explained later) (Water Potential, Slide 7)
What is water potential?
Potential for water molecules to do work (Isotonic, 2011)
How do we make 1M sucrose solution?
Find the molecular weight, and that is 342.3g/M (Molecular, 2005)
If the concentration of the sucrose solution is 0.5M, then the isotonic point has been reached. If the data collected from the various sucrose concentrations tested are used, then unknown concentrations of sucrose can be determined accordingly.
First part of hypothesis above is incorrect
The isotonic point of the potato is not 0.5M, it is 0.2M (and 0.4 for dried potato)
Second part is correct because through the data collected, we can use our equation to find concentrations of unknown solutions
0.28 is the isotonic point
0.212M is the isotonic point
0.204M is the isotonic point
Closer Look: What happened and why?
Water potential = Solute Potential + Pressure Potential
Water moves from higher water potential to lower water potential
What happened to Trial 4? Let's compare!
In trial 4, dried potatoes were used, and the isotonic point was 0.4M while all the rest's were 0.2M
Why? How did the dryness affect the movement of water?
"Isotonic." - Definition from Biology-Online.org. N.p., 8 Dec. 2011. Web. 2 Sept. 2013. <http://www.biology-online.org/dictionary/Isotonic%20>.
"Molecular Weight and the Mole." Molecular Weight. N.p., 20 Oct. 2005. Web. 02 Sept. 2013. <http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/M/Mole.html>.
"Water Potential"; Sept. 2, 2013 < http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=6&cad=rja&ved=0CFwQFjAF&url=http%3A%2F%2Ffaculty.muhs.edu%2Fklestinski%2FWaterPotential.pps&ei=GLUvUu7CGoKFrAeK9oHwDQ&usg=AFQjCNF5q8MxCzSaAJtBEkf8nbRW6PuQCw&sig2=D8Q30DmGlC9MRwg59GgmnQ&bvm=bv.51773540,d.bmk>
0.473M is the isotonic point