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Lab 4 Diffusion and Osmosis
BIO201L
Student Name: Access Code (located on the lid of your lab kit): Click here to enter text.
Pre-Lab Questions:
”
1. Compare and contrast diffusion and osmosis.
”
Similarities: Passive transport process Differences: Osmosis occurs in liquid medium, diffusion occurs in any medium.
”
2. What is the water potential of an open beaker containing pure water? ”
None or zero
”
3. Why don’t red blood cells swell or shrink in blood? ”
Plasma solution is isotonic Experiment 1: Diffusion through a Liquid
Table 1: Rate of Diffusion in Corn Syrup
Time (sec)
Blue Dye
Red Dye
10
1.6
2.7
20
1.8
3.2
30
2.0
3.5
40
2.2
3.7
50
2.3
4.0
60
2.6
4.1
70
2.7
4.2
80
3.1
4.3
90
3.3
4.3
100
3.1 4.9
110
3.3
5.2
120
3.6
5.3
Table 2: Speed of Diffusion of Different Molecular Weight Dyes
Lab 4 Diffusion and Osmosis
BIO201L
Structure
Molecular Weight
Total Distance
Traveled (mm)
Speed of Diffusion
(mm/hr)*
Blue Dye
793 g/mole
2cm = 20mm 600 mm/hr
Red Dye
496 g/mol
2.6cm 26mm
780 mm/hr
*To get the hourly diffusion rate, multiply the total distance diffused by 30.
Post-Lab Questions
”
1. Examine the plot below. How well does it match the data you took in Table 1? ”
My data and the graph look very similar. Based on my data, red dye has a faster diffusion rate than the blue dye and this is what the graph is showing.
”
2. Which dye diffused the fastest? ”
Red dye
”
3. How does the rate of diffusion correspond with the molecular weight of the dye? ”
Per the data, rate of diffusion is faster if molecular weight is lighter
”
4. Does the rate of diffusion change over time? Why or why not? ”
Yes it can due to surface area and temperature Experiment 2: Diffusion - Concentration Gradients and Membrane Permeability
Table 3: Indicator Reagent Data
Lab 4 Diffusion and Osmosis
BIO201L
Indicator
Starch Positive
Control (Color)
Starch Negative
Control (Color)
Glucose Positive
Control (Color)
Glucose
Negative
Control (Color)
Glucose Test
Strip
n/a
n/a
Brown
blue
IKI
brown
yellow n/a
n/a
Table 4: Diffusion of Starch and Glucose Over Time
Indicator
Dialysis Bag After 1 Hour
Beaker Water After 1 Hour
Glucose Test Strip
Light brown
Dark blue
IKI
Dark brown yellow
Post-Lab Questions
”
1. Why is it necessary to have positive and negative controls in this experiment? ”
to reduce variables and to do the expirement accurately ”
2. Which substance(s) crossed the dialysis membrane? Support your response with data-based
evidence. ”
Glucose and Iodine as they are permeable ”
3. Which molecules remained inside of the dialysis bag? ”
Starch as they are too large to pass through the tube
”
4. Did all of the molecules diffuse out of the bag into the beaker? Why or why not? ”
No, as starch molecules is too large to diffuge out of the bag into the beaker Experiment 3: Osmosis - Direction and Concentration Gradients
”
Hypothesis: ” Displacement is dependent on the amount of sucrose. More sucrose higher displacement.
Table 6: Sucrose Concentration vs. Tubing Permeability
Band
Color
Sucrose %
Initial Volume
(mL)
Final Volume
(mL)
Net Displacement
(mL)
Yellow
30%
10
14
4
Red
15%
10
13
3
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Lab 4 Diffusion and Osmosis
BIO201L
Band
Color
Sucrose %
Initial Volume
(mL)
Final Volume
(mL)
Net Displacement
(mL)
Blue
3%
10
11
1
Green
3%
10
9
-1
Post-Lab Questions
”
1. For each of the tubing pieces, identify whether the solution inside was hypotonic, hypertonic, or isotonic in comparison to the beaker solution it was placed in. ”
Yellow: Beaker solution/tube = hypotonic/hypertonic
Red: Beaker solution/tube = hypotonic/hypotonic
Blue: Beaker solution/tube = hypotonic/hypotonic
Green: Beaker solution/tube = hypertonic/hypotonic
”
2. Which tubing increased the most in volume? Why? ”
Yellow because sucrose solution is more compared to all
”
3. What does this tell you about the relative tonicity between the contents of the tubing and the solution in the beaker? ”
Most of the solution is hypotonic in beaker and tube
”
4. What would happen if the tubing with the yellow band was placed in a beaker of distilled water? ”
Volume increase as the solution would be hypotonoic in beaker
”
5. Osmosis is how excess salts that accumulate in cells are transferred to the blood stream so they can be removed from the body. Explain how this process works in terms of tonicity. ”
During osmosis molecules move from low to high concentration thrugh a semi permeable membrance ”
6. How is this experiment similar to the way a cell membrane works in the body? How is it different? Be specific with your response. ”
Both cell memberance and cell wall prevent toxic and unwanted materials from entering the body. It is different because with this experiment cell wall prevent from bursting but cell memberance do not
”
7. If you wanted water to flow out of a tubing piece filled with a 50% solution, what would the minimum concentration of the beaker solution need to be? Explain your answer using scientific evidence. ”
Above 50% because water move from high to low concentration
Experiment 4: Osmosis - Tonicity and the Plant Cell
Table 7: Water Displacement per Potato Sample
Lab 4 Diffusion and Osmosis
BIO201L
Potato
Potato Type and
Observations
Sample
Initial
Displacemen
t (mL)
Final
Displacement
(mL)
Net
Displacement
(mL)
1
Sweet potato with brown color and hard exterior 1A
7
9
2
1
Sweet potato with brown color and hard exterior 1B
6
5
1
2
Russet potato with brown color and soft exterior
2A
5
4
1
2
Russet potato with brown color and soft exterior 2B
4
3
1
Post-Lab Questions
”
1. How did the physical characteristics of the potato vary before and after the experiment? Did
it vary by potato type? ”
Sweet potato was hard and it became soft whereas Russet was soft and became hard
”
2. What does the net change in the potato sample indicate? ”
Amount of solute ”
3. Different types of potatoes have varying natural sugar concentrations. Explain how this may influence the water potential of each type of potato. ”
More sugar level will obsorb or take more water
”
4. Based on the data from this experiment, hypothesize which potato has the highest natural sugar concentration. Explain your reasoning. ”
Sweet potato because more water will enter if potatoes have more sugar
”
5. Did water flow in or out of the plant cells (potato cells) in each of the samples examined? How do you know this? ”
Based on the expirement water flowed both in and out of potatoes ”
6. Would this experiment work with other plant cells? What about with animal cells? Why or why not? ”
Yes it will work because plant cell have cell wall to prevent from bursting
”
7. From what you know of tonicity, what can you say about the plant cells and the solutions in the test tubes? ”
Water move from high to low concentration area
Lab 4 Diffusion and Osmosis
BIO201L
”
8. What do your results show about the concentration of the cytoplasm in the potato cells at the start of the experiment? ”
The concentration of the cytoplasm in the potato is low
”
9. If the potato is allowed to dehydrate by sitting in open air, would the potato cells be more likely to absorb more or less water? Explain.
Sweet potato would obsorb more water because of sugar content
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Question 26 of 40
What volume (in mL) of 0.3350 M HBr is required to neutralize 50.00
mL of 0.7000M KOH?
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Use the information from the table provided (IMAGE ATTACHED) to answer Question 1 (Part 1 AND 2).
The data in blanks a and c show the effects of what variable?
What are the 2 constants for the datas (in blanks a and c)?
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Each row of the table below describes an aqueous solution at 25 °C.
The second column of the table shows the initial components of the solution.
• Use the checkboxes in the third column to explain the type of the initial solution.
The fourth column describes a change in the solution.
• Use the fifth column to predict how the change in the solution will change its pH.
solution initial components
initial type
(check all that apply)
change
O acidic
A
H₂O
basic
add H Cl
O neutral
acidic
B
H₂O, HI
basic
add NaOH
O neutral
acidic
C
H₂O, HI
basic
add NaI
neutral
acidic
effect of change on pH
(check one)
O pH higher
O pH lower
O pH the same
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O pH lower
O pH the same
O pH higher
O pH lower
O pH the same
O pH higher
D
H₂O
basic
add K Cl
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O neutral
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Please solve!!!!!!!!!!!!!!
arrow_forward
Each row of the table below describes an aqueous solution at 25 °C.
The second column of the table shows the initial components of the solution.
• Use the checkboxes in the third column to explain the type of the initial solution.
The fourth column describes a change in the solution.
• Use the fifth column to predict how the change in the solution will change its pH.
initial type
solution initial components (check all that apply)
acidic
basic
neutral
acidic
basic
neutral
acidic
basic
neutral
acidic
basic
neutral
A
B
09
C
D
H₂O
H₂O
H₂O, HCIO
H₂O, HCIO4
0000000
change
add HI
add KI
add NaClO
add NaOH
effect of change on pH
(check one)
O pH higher
O pH lower
OpH the same
O pH higher
O pH lower
OpH the same
O pH higher
O pH lower
O pH the same
O pH higher
O pH lower
O pH the same
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Super fast answer super fast!!!!!!
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This the last question u guys gotta help me on so please hurry up!!
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5
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[Review Topics]
(References)
Use the References to access important values if needed for this question.
What volume of a 0.101 M HCl solution is required to neutralize 28.5 mL of a 0.150 M KOH solution?
HCI+ KOH →KCI + H2O
mL HCI
Submit Answer
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Question 4 of 12
Submit
A neutral solution of water at a
particular temperature has a
concentration of OH¯ of 9.4 × 10-7
M. What is Kw at this temperature?
1
4
6.
7
8
+/-
x 10 0
Tap here or pull up for additional resources
3.
N LO
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Diagrams of solution conductivity versus the total volume of three solutions are shown (VINITIAL = 1.0 L before additions).
Which diagram best represents the solution conductivity when 1.0 M NH3 is added to 1.0 L of 1.0 M HF?
Question 11 options:
2
1
3
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ENB Assume the densities of all solutions are 1.0 g/ml and their specific heat capacities 4.184 J/g.
Experime
The enthalpy change of neutralization
Part I
3
NH4OH(aq) +
HCl(aq)-
NH4CI(aq) +
H2O(E)
2.1
Equation of reaction:
2.0 M HCI
2.0 M NH.OH
Volume
Number of moles 0.05
25 ml
25 ml
0.05
Temperature
15 °C
15 °C
eaction
Initial Temperature,T
Final Temperature, T
Change in temperature, AT
15 °C
16.5 °C
1.5 °C
Calculation of q- in J (q- = - qsa)
2.2
What is the enthalpy change (AH)
in kJ.mol, when 1 mol of acid is
2.3
neutralized?
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number of significant figures. Record these values on the spreadsheet template before the lab.
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