Osmosis and Cell
Name: ______________________________________ Date: ________________________
Student Exploration: Osmosis
Vocabulary: cell membrane, concentration, diffusion, dynamic equilibrium, osmosis, semipermeable
membrane, solute, solvent
Prior Knowledge Questions (Do these BEFORE using the Gizmo.)
1. Suppose you were trapped on a desert island with no sources of fresh water. Should you drink
water from the ocean? Explain why or why not.
_________________________________________________________________________
_________________________________________________________________________
2. What do you think would happen if you watered your houseplants with salt water? ... Show more
content on Helpwriting.net ...
Observe: Wait until the numbers are not changing very much. What do you notice about the solute
concentrations inside and outside of the cell? _______________________________ This situation
is called dynamic equilibrium.
5. Experiment: Click Reset. Check that the Solute outside is 10 and the Initial cell volume is 40%.
To calculate the solvent concentration, divide the number of solvent particles by the total number of
particles, and then multiply by 100. (Note: The Gizmo only displays the solute concentrations.)
A. What is the solvent concentration inside the cell? ____________________________
B. What is the solvent concentration outside the cell? ___________________________
C. Where is there a higher solvent concentration? _____________________________
D. Click Play. Do most of the solvent particles move into or out of the cell? (Hint: Does the cell
expand or shrink?) ______________________________________________
6. Experiment: Click Reset, and set the Solute outside to 1.

A. What is the solvent concentration inside the cell? ____________________________
B. What is the solvent concentration outside the cell? ___________________________
C. Where is there a higher solvent concentration? _____________________________
D. Do you think the cell will get larger or smaller?
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Diffusion and Glucose
Physioex 9.0
Review Sheet
Exercise 1
Cell Transport Mechanisms and Permeability
Name
Lab Time/Date ___
Activity 1 Simulating Dialysis (Simple Diffusion)
1. Describe two variables that affect the rate of diffusion. Size of material and concentration
2. Why do you think the urea was not able to diffuse through the 20 MWCO membrane? How well
did the results compare with your prediction? The molecules were too large to pass through. This is
what I predicted on my Hypothesis.
3. Describe the results of the attempts to diffuse glucose and albumin through the 200 MWCO
membrane. How well did the results compare with your prediction? The glucose diffused through
the 200 MCWO ... Show more content on Helpwriting.net ...
It increased the filtration rate because the blood pressure increased, but the concentration of solutes
did not because the pore size did not increase.
___
Activity 5 Simulating Active Transport
1. Describe the significance of using 9 mM sodium chloride inside the cell and 6 mM potassium
chloride outside the cell, instead of other concentration ratios. The reason for the use of 9mM of
sodium chloride is because 3 ions are ejected for every 2 K that are added. K hangs outside the cell
unless actively transport inside the cell.
2. Explain why there was no sodium transport even though ATP was present. How well did the
results compare with your prediction? There were no sodium ions because they were ejected by the
cells by active transport.
3. Explain why the addition of glucose carriers had no effect on sodium or potassium transport. How
well did the results compare with your prediction? The introduction of glucose had no effect on the
sodium and potassium transport due to solute pumps.
4. Do you think glucose is being actively transported or transported by facilitated diffusion in this
experiment? Explain your answer. The glucose is transported by facilitated diffusion in the
experiment that is how glucose is transported. The glucose could not be actively transported because
they are lipid insoluble and too large to pass through the membrane pores.
Lab Report 1

Activity 1
The purpose of this lab is to gain a better

Effect Of Lowering The Freezing Point Of Water
Nayla Epps, Mirsha Pierre and Malia Shields Honors Chemistry Period 2 Group:11 Topic:Lowering
the Freezing Point of Water Due Date: 10/6/17 Background Research The scientists topic sentence is
the freezing point point of water will decrease depending on the amount of solute used in the
solvent. In this case it is water. The problem statement is "How much solute is needed in cvde to
decrease the freezing point of water. The independent variable is the amount of solute and the
dependent variable is the temperature at the point when the water will freeze. There is no constant
variable, the controlled variable is no solute (water only). The scientist's hypothesis is if there is
more solute added to the water, then the water's freezing point will decrease based on the number of
grams of solute. To test the hypothesis the scientists will constantly add more salt/sugar to the water
to see how much freezing point decreases per gram. In order to lower the temperature at which
water freezes you would need salt. When you add salt to ice the ice would melt and surprisingly get
colder. Scientist prove salt lowers the freezing point of water. From scientist we learned that salt will
lower the freezing point of water. Adding salt to an ice cube will make the ice cube melt. As the ice
cube melts, energy is being used to separate bonds in the solid state. Breaking down salt in water
causes

Effect of Solute Concentration on Osmosis in Potato Cells...
INTRODUCTION
Osmosis is the movement of water molecules from high concentration to low concentration through
semipermeable membranes, caused by the difference in concentrations on the two sides of a
membrane (Rbowen, L.). It occurs in both animals and plants cells. In human bodies, the process of
osmosis is primarily found in the kidneys, in the glomerulus. In plants, osmosis is carried out
everywhere within the cells of the plant (World Book, 1997). This can be shown by an experiment
with potato and glucose/salt solution. The experiment requires putting a piece (or more) of potatoes
into glucose or salt solution to see the result of osmosis (a hypertonic type of solution is mostly used
as it would give the most prominent visual prove of ... Show more content on Helpwriting.net ...
6. Place each group of potato cylinders into each beakers.
7. Leave for 24 hours, with all of them staying together in the same place (not 3 beakers inside the
building and the other 3 outside, or similar situation) to keep the temperature roughly the same.
8. Use the plastic forceps to remove the potato cylinders from the beakers (keep them together in the
same group), and blot off the excess solution weight on them with the paper towels.
9. Weigh the mass of each group of potato cylinders and record data.
RESULT – (For graph, see Appendix 3)
Data: Effect of Solute Concentration on Osmosis in Potato Cells (for our group)
Solution Conc. Of Glucose (M) Initial Mass (g) Final Mass (g) Change of Mass (%) Class Average
0 4.5 5.3 17.8 17.5
0.2 4.3 4.5 4.7 5.6
0.4 4.2 3.9 –7.1 –6.7
0.6 4.5 3.3 –26.7 –25.8
0.8 4.1 2.7 –34.1 –30.1
1.0 4.4 2.8 –40.9 –35.5
Data: Effect of Solute Concentration on Osmosis in Potato Cells (for the 6 groups of our class) Mass
Changes In Potato–Groups For Each Solution (%)
Solution–Conc. Of Glucose (M) Group 1 Group 2 Group 3 Group 4 Group 5 Group 6 Class Average
0 17.8 11.9 8.4 22.2 14.1 17.5
0.2 4.7 4.8 4.8 5 13.6 5.6
0.4 –7.1 3.2 –2.5 –8 8.1 –6.7
0.6 –26.7 –10.9 –22.8 –28.8 –25.4 –25.8
0.8 –34.1 –16.4 –31.6 –41.7 –20.1 –30.1
1.0 –40.9 –19.4 –41.5 –45.1 –25.7 –35.5
EVALUATION

Discussion
The data shows that the higher the concentration of

How Concentration Of A Solute Affect The Rate Of Osmosis
Biology
How Does the Concentration of a Solute Affect the Rate of Osmosis
BY: Kevin Liu, Prabhat Aluri, Sriram Thothathri, Edward An
Question:
How does the concentration of a solute affect the rate of osmosis?
Hypothesis:
The concentration of a solute affects the rate of osmosis over time, in a way where, the higher the
concentration of a solute, the faster the rate of osmosis. This happens because, in a semi–permeable
membrane the water is the only through that can move through. In this case our solute is sugar, and
sugar cannot fit through the pores in the membrane. When there is more sugar the difference in
concentration is greater. That difference makes all the molecules want to move faster. That results in
osmotic pressure being built. The sugar tries to get through the membrane, but the pores don't
accommodate it. That means that the water can only go through less pores because more sugar is
blocking the pores. The water that goes through the few pores, ends up clumping up near the open
pores and, rushes into the membrane faster. That results in the rate of osmosis being faster.
Procedure:
1. Take a long strand of dialysis tubing, and split the tubing into nine 3 inch pieces.
2. Fill a flask with pure water (0% sugar), another with 0.2M water, another flask with 0.4 M water,
and another flask with 0.8M water.
3. In addition, fill a beaker with pure water.
4. Take a piece of dialysis(as cut earlier), find the mass of the dialysis, and the zero out the scale on

Solute To Dissolve Lab Report
The Effect of Solvent on the Ability of Solute to Dissolve
Problem: How does the solvent effect the ability of a solute to dissolve?
Hypothesis: If the solvent is denser, than the time it takes the sugar to dissolve will decrease.
Materials:
Sugar
Rubbing Alcohol
Vegetable Oil
Water
Scoopula
Beaker(200mL)
Graduated Cylinder(50mL)
Stirring Rod
Balance
Stop Watch
Medicine Cup
Procedures:
1. Measure 25mL of water into a graduated cylinder.
2. Pour the water into a beaker.
3. Measure 2g of sugar into a medicine cup with a scoopula, and pour the cup into the beaker of
water.
4. Take your stirring rod and start stirring as soon as all the sugar is out of the cup.
5. Start your stopwatch as soon as you start stirring.
6. Stop the timer after ... Show more content on Helpwriting.net ...
Three different solvents were used and measured out equally, and then timed to see how fast 2g of
sugar can dissolve into them. It was hypothesized that if the density of the solvent increased, then
the time it takes the solute to dissolve will decrease. The data shows that sugar dissolved only in
water with an average of 42 seconds to dissolve. Vegetable oil and rubbing alcohol both took over 2
minutes and were not completely dissolved yet.
The hypothesis that the denser the solvent is, more solute will dissolve in a quicker time period was
correct. These results occurred because sugar has polar energy, which makes it easy to dissolve in
water which is also polar. Vegetable oil is nonpolar, so these two substances do not mix. Rubbing

alcohol is somewhat polar although it is mostly non–polar, which leads to some sugar dissolving but
not all. A possible source of experimental error could be the mixing of two solvents. When the water
was poured into the beaker, some oil was still in the beaker, causing the two to mix, making less
sugar dissolve. However, the amount of solvent mixed was minimal, so hhit had a small effect on
the results in the data. If this experiment were to be made again, then three different beakers and
graduated cylinders could be used to measure out the solvents separately so two substances will not

Effect Of Solute On The Production Of Magnesium Alloys
Effect of solute addition on grain refinement in cast magnesium alloys.
1. Introduction:
Magnesium is 30% lighter than aluminum and possesses excellent mechanical properties. It has
higher weight to strength ratio, damping capacity, dimensional stability, impact and dent resistance
when compared to aluminum alloys or steel alloys. These properties have increased the usage of
magnesium alloys in the automotive and aerospace industry for weight reduction. But magnesium
alloys are often associated with some limitations such as low ductility, lower strength, poor
workability (due to hexagonal lattice structure), lower creep resistance and lower corrosion
resistance. Alloying with rare earth metals like Gadolinium, neodymium and cerium have ... Show
more content on Helpwriting.net ...
1.2 Melt treatment during casting process:
Inoculation is the successful melt treatment to achieve the grain refinement. Inoculation is referred
to the addition of grain refiners or solute elements to restrict grain growth and promote
heterogeneous nucleation. Further magnesium alloys are categorized into two types in inoculation
treatment. One is aluminum free alloy and other is aluminum bearing alloy. It is observed that
adding 0.15 wt.% zirconium during inoculation of pure magnesium leads to reduction of 80% of the
grain growth [1–21]. However if zirconium is added to the alloys having aluminum, it formed
intermetallic phases with aluminum and loses its grain refining properties. Since most commonly
used magnesium alloy in the casting is Mg–Al alloy, there has been an extensive research on the
solute elements for the Mg–Al alloys for achieving the grain refinement. Some of the solute
element's effect on the grain refinement are discussed in this review paper.
1.2.1 Effect of manganese on the grain refinement in Mg–Al based alloys:
Manganese has been used as iron remover in the magnesium alloys [4–4]. Tiner [4–5] had
investigated the effect of 0.19% or more of Mn in Mg–Al alloys and found that superheating has
given grain

The Solubility As A Measure
Science Fair Essay
–chemist understands solubility as a measure
–a chemist would say that solubility is the max solute that can be dissolved in the solvent until it
reaches equilibrium
–when the solution reaches equilibrium, there can be no more solute added to the solvent: it won 't
dissolve anymore, it will just rest at the bottom as residue
–the chemical formula of salt is NaCI
–1 ATM is standard air pressure for earth
–the unit for solubility of water it for ex: 357.00 g/L
–when the solubility is put in the solvent, the solution reaches equilibrium
–every chemical substance has a fixed solubility
–if something doesn 't dissolve, it 's solubility is 0
–measures of solubility cannot be applied to immiscible substances
–some substances can be mixed and create a homogeneous phase in any proportion: miscible
–the opposite is an immiscible substance
–in dissolving, molecules of the solvent enter the solute and are surrounded many other molecules
–for the solute 's molecules to be surrounded, the molecular bonds between the molecules of the
solute have to be broken and the molecular bonds of the solvent have to be disrupted
–energy given off in the process
–then, when the solute dissolves into the solvent, new molecular bonds are created
–but for the solute and solvent to dissolve, the molecular bonds have to be broken and the energy
output has to be immense
–right when the solute is placed in the solvent, the concentrated solute, breaks into pieces
–then the molecules of the

Solubility Lab Report
Solubility is the amount of solute that can dissolve in a certain amount of solvent to create a
saturated solution at a constant temperature and pressure. The solute can be a gas, a liquid or a solid.
Generally, the solubility of liquids and solids increase as the temperature is increased. However, the
solubility of gases actually decreases as the temperature is decreased. The solubility of solids and
liquids increases as the temperature of increases because the higher temperatures mean higher
kinetic energy which allows for more and stronger collisions between the solute and solvent, which
causes the solute to be surrounded by the solvent molecules and dissolve into the solution. However,
gases become less soluble in higher temperatures because ... Show more content on Helpwriting.net
...
The purpose of this lab was to determine the effect of temperature on the solubility of potassium
nitrate in water as well as to create a solubility curve showing how potassium nitrate's solubility
varies with temperature. Different amounts of solute (potassium nitrate) were brought to
crystallization, meaning that their solutions were formerly supersaturated and brought to saturation.
As a result, the crystallization temperatures could be used to create a solubility curve that indicated
the amount of g KNO3 dissolved per 100g of water in a saturated solution. The experimental results
showed that the solubility of KNO3 increased as the temperature increased, indicating a directly
proportional relationship. The test tubes with the highest solute to water ratio crystallized at the
highest temperatures, showing that the most amount of solute were able to dissolve in the highest
temperatures. The purpose of the lab was fulfilled as the effect of temperature on potassium nitrate's
solubility was determined (higher temperatures increase the solubility) and a solubility curve was
created in Calculations #2. The hypothesis was also supported as raising the temperature did result
in higher solubility of the potassium

Lab Report On Osmosis
In this lab, we performed two experiments to understand the process of osmosis and its relationship
with solute concentration, and how water potential is affected by temperature. In the first part of our
lab (there was no hypothesis for this experiment), where we used dialysis tubing to model osmosis
and its relationship to solute concentrations, we observed how the percent change in mass increased
as the solute concentration of the solution did. The percent change in mass was positive, meaning
that the dialysis tubing had gained mass after being submerged in the distilled water solution. This
means that the dialysis tubing gained water, which was the result of osmosis. The water diffused
from outside the tubing, through the semipermeable membrane, into the tubing in all of the cases.
Because we know that water diffuses from high water concentration to low water concentration, we
know that the sucrose solutions in the dialysis tubing had less water concentration than the distilled
water surrounding the tubing. Furthermore, the molarity of the sucrose solutions inside the tubing
increased from 0.2M to 1.0M, which means that the sucrose (solute) concentration was increasing,
so the water concentration was decreasing. This explains why the percent change in mass increased
as the molarity increased; more water was flowing into the tubing as the water concentration got
lower and solute concentration got higher. In other words, the water was flowing from a hypotonic
solution to a

Molarity Lab Report
The overall purpose of this lab was to determine how the molarity of solution concentration
relations to solutes, solvents, and the total solution volume. The given testable question the
experiment was, "What is the mathematical relationship between the moles of a solute, the volume
of the solution, and the molarity of an aqueous solution?" The simulation gave various aqueous
solutions and dials that giving us the ability to adjust the moles of solute and the solution volume,
which then adjusts the molarity of the solution accordingly. After playing around with the
simulation and getting familiar with the structure of the overall experiment, we started noticing a
relationship between the level of molarity and the settings of the solute amount and solution volume.
Investigating further into these correlations, we determined that the solute amount over the solution
volume represented a mathematical relationship that would produce the molarity of the aqueous
solution. When we started collecting the data we collected the solute amount levels in intervals of
0.25 and the solution volume levels in 0.2 intervals in order to get a wide range of data values, while
keeping a constant (view data table attached). We also used only the CoCl2 aqueous solution in
order to keep the type of ... Show more content on Helpwriting.net ...
We did notice one odd data value that did not fit into the data that our mathematical relationship that
represented our experiment, but after further research we determined that at a certain point our
solution reached saturation point, or the point in which no more solute could be dissolved into the
solute. Our next step was to understand the concept of what each of the terms within the lab

The Effect Of Osmotic Temperature On The Concentration Of...
OSMOTIC DRYING
Introduction
Osmosis is the movement of fluid from a low solute concentration to a high solute concentration
through a permeable membrane. Small solutes such as water and ion molecules can pass through the
membrane easily whereas bigger solutes such as polar molecules are not able to pass through the
membrane (Bitesize, 2014).
Osmosis is the suitable and preferred method to preserve the food quality before the final treatment
such as drying or freezing because it retains the fruit's characteristics such as nutritional
constituents, colour, aroma and flavor (Gekas et al., 1998; Yadav and Singh, 2012).
The aim of this experiment is to show the effectiveness of osmotic drying between two different
solutions set at two different conditions, at room temperature and 40°C. The two most common
solutions used in osmotic drying are salt (sodium chloride) and sucrose (Gekas et al., 1998). Both
are very effective osmotic agent and are very different in terms of their molecular size.
Materials and Methods
Apples were used as a sample for this experiment. The apples were cut to a thickness of 1.5mm.
600g of sucrose and 400g of water to make the 60° Brix solution and 600g of salt and 400g of water
to make the salt solution. Four buckets were used because each solution will be placed at room
temperature and 40°C.
Sample/solution ratio is 1/50. The apple slices were placed inside the bucket containing the solution
but were seen floating so loads such as

Energetically Synthesizing Solutes
All halophiles must maintain their cytoplasm isoosmotic with their surrounding medium. Salt
tolerance requires that compatible solutes accumulate in the cytosol and organelles where these
function in osmotic adjustment and osmoprotection (Rhodes and Hanson, 1993). Some compatible
osmolytes are essential elemental ions, such as K+, but the majority are organic solutesBiological
membranes are permeable to water, and active energy dependent inward transport of water to
compensate for water lost by osmotic processes is energetically not feasible. Moreover, cells that
keep a turgor need even to maintain their intracellular osmotic pressure higher than that of their
environment.
There are two fundamentally different strategies used by halophilic microorganisms to balance their
cytoplasm osmotically with their medium (Galinski and Trüper, 1994; Zahran, 1997). (1)
Accumulation of molar concentrations of potassium and chloride. This strategy requires extensive
adaptation of the intracellular enzymatic machinery to the presence of salt, as the proteins should
maintain their proper conformation and activity at near–saturating salt concentrations. The proteome
of such organisms is highly acidic, and most proteins denature when suspended in low salt. It is
called the'high–salt–in strategy'. They maintain osmotically equivalent internal concentrations by
accumulating high concentrations of potassium chloride. Potassium ions enter the cell passively via
a uniporter. Sodium ions are

Essay about Laboratory #1
Review Sheet 1 Results
1. Describe two variables that affect the rate of diffusion.
Your answer:
Two variables that affect the rate of diffusion are the MWCO membrane and the solute
concentration. Increasing the membrane size and solute concentration will also increase the average
diffusion rate. Decreasing the membrane size and solute concentration will reduce diffusion rates
and can even prevent all diffusion.
2. Why do you think the urea was not able to diffuse through the 20 MWCO membrane? How well
did the results compare with your prediction?
Your answer:
The urea was not able to diffuse through the 20 MWCO membrane because the molecular weight
was too high. For a molecular weight of 60.07, a larger membrane would need to be ... Show more
2. The larger value obtained when more glucose carriers were present corresponds to an increase in
the rate of glucose transport. Explain why the rate increased. How well did the results compare with
your prediction?
Your answer:
The rate increased because the glucose was easier to transport with more carriers. Saturation levels
decrease when carriers are added and glucose concentrations are constant. The results were the same
as my prediction that assumed the glucose transport rate would increase.
3. Explain your prediction for the effect Na+ Cl– might have on glucose transport. In other words,
explain why you picked the choice that you did. How well did the results compare with your
prediction?
Your answer:
My prediction for the effect Na+Cl– might have on glucose transport was that the glucose transport
rate would decrease. I picked this choice because I thought having Na+Cl– in one beaker would
limit the space needed for proper glucose diffusion. All of the other runs involved water, so I
predicted that adding a new solute could slow diffusion. The results opposed my prediction because
the Na+Cl– had no effect on the glucose diffusion rate.
Review Sheet 3 Results
1. Explain the effect that increasing the Na+ Cl– concentration had on osmotic pressure and why it
has this effect. How well did the results compare with your prediction?
Your answer:
The effect that increasing Na+Cl– concentration had on osmotic pressure was that the pressure also

Solute Concentration, Particle Size, And A Membrane 's...
Objective: The overall purpose of this experiment is to determine how solute concentration, particle
size, and a membrane's selective permeability affect the occurrence of diffusion.
Hypothesis: In exercise one, as potassium permanganate's molecular mass (158.03 g/mol) being less
than half that of methylene blue (319.85 g/mol), KMnO4 is expected to further diffuse in the agar
plate due to the negative correlation associated with particle size and rate of diffusion rate with a
semipermeable membrane. With this correlation in mind, it is expected a color change would occur
only within the tube, as starch would remain in a dialysis tube while iodine is able to enter the tube
due to starch's relatively larger molecular size in comparison to iodine. Additionally, a selectively
permeable dialysis tube containing a higher solute concentration in comparison to its surrounding
solution should gain volume and the opposite (decreased volume) with a tube in a hyperosmotic
solution because osmosis of water, rather than diffusion of sucrose, from a high water concentration
(distilled water solution) to that of a lower water concentration (within the sucrose bag).
Background and Introduction Within this experiment's purpose, understanding of the cell
membrane, diffusion, and osmosis must first be known. As a part of all living cells, the membrane is
meant to serve as a semipermeable barrier between what is inside, intracellular, and outside of the
cell, extracellular, as well as

Colloids For Research Paper
Solutions, colloids, and suspensions are all types of mixtures. Suspensions and colloids are
classified as heterogeneous mixtures and solutions are classified as the homogeneous mixture.
Solutions are single atoms that never settle and are transparent which also means clear. Colloids are
aggregates of atoms, molecules, or ions, they can settle and they are transparent with a Tyndall
effect. Suspensions are clumps of particles such as sand and their movement is by gravity with the
translucent effect. An example of a solution is sugar dissolving in water because the solute the sugar
is being dissolved in the solvent which is the water it is basically two substances evenly mixed
together. Some examples of Colloids is plasma proteins, albumins, ... Show more content on
Helpwriting.net ...
An example of a nonelectrolyte is C6H12O6 – glucose because it can not dissociate into ions
whenever it is in an aqueous solution.
Solubility is the quantity of a particular substance solid, liquid, or gas solute that can dissolve in a
particular solvent, yielding a saturated solution(biology online). The solubility of solids increases
with the temperature. The increasing of the temperature allows the solvent molecules to easily
detach the solute molecules that are held by intermolecular attractions. The solubility of gases is a
little more complicated because as the temperature increase the kinetic energy also increases which
causes more motion of the molecules to move more freely. There're two ways for material to be
dissolved the first way is dipole to dipole which is molecular and the second way is dipole to ion
which is ionic.
When an ionic compound like NaCl dissolves in water, we end up with a positive Na and negative
Cl in a

Solute Lab Report
The purpose of this lab was to to learn how temperature and surface area affect how fast a solute
dissolves. This lab relates to what we have recently been learning in class because we have been
going into deeper discussion about solutes, solvents, and the concept of solubility. In this lab, we
first had to have a trial that tests whether or not the surface area of a solute affects the speed at
which it dissolves in the solvent. To do this we first had everyone put on goggles for safety
purposes. Next, a group member opened the packet of Alka Seltzer, keeping one tablet as a whole
and crushing the second tablet into a powder using a spoon. Following that, another group member
measured 5 mL of vinegar in a beaker and then poured that into the ... Show more content on
Helpwriting.net ...
In addition, we recorded that when a whole tablet of Alka–Seltzer was dropped in a canister with
room temperature vinegar, it took around 10.46 seconds to dissolve. However, when another whole
tablet was dropped in the canister with heated vinegar (72 degrees), it only took 4 seconds to
dissolve. The way in which we measure how long the solute took to dissolve with the solvent was
simply by recording the amount of time the cap of the canister took to pop off due to the mixing of
solutes/solvents. This data indicates that the more surface area a solute has, the faster it will dissolve
in the solvent and also that the higher the temperature of the solvent, the faster it takes for the solute
to dissolve. On the other hand, this also indicated that less surface area means the solute takes
longer to mix with the solvent and lower temperature also causes this same effect. These results do
make sense because a greater surface area means that the randomly moving solvent molecules are
more likely to collide with the solutes since they are more spread out, thus easier to get in between.
As these collisions occur, the solvents are breaking up the solute molecules causing them to mix
faster. Furthermore, our data for whether or not temperature increases the rate at which the solute
mixes with the solvent makes sense because as we have learned in class, a higher temperature
causes the particles in a substance to move more rapidly which causes more collisions. As stated
before, these collisions are responsible for the breaking down of the solvent molecules, so the faster
the solvent molecules randomly move and collide with the solute, the faster the dissolving process.
Although we tried to be as careful as possible to avoid error, there were still a few in this lab.

Why Do Solutes Dissolve
Analysis
Before I start the experiment, I predicted that all crystal form solutes would dissolve while powder
form solutes will not, but my prediction was partially supported. A baking soda was a one that
showed an unexpectedly result because even though it is in a powder form, it dissolved in the water.
Except for a baking soda, everything could be observed from what I predicted. Sugar and salt,
which are in a crystal form dissolved in water and had high solubility while flour and cornstarch,
which are in a powder form didn't dissolve and had low solubility.
When we experimented with different solvents, the result showed that water was the best solvent to
dissolve solutes because it dissolved the most of solutes, such as sugar and salt. Alcohol was the
most inefficient solvent because it wasn't able to dissolve any sugar, salt and, margarine. Vegetable
oil was also able to only dissolve margarine partially.
Conclusion
The data shows sugar and salt had high solubility in water. From that result, I can conclude they are
polar. Salt is made up of positively and negatively charged ions and sugar is made with hydrogen
bonds that make them polar. Initially, I predicted that baking soda wouldn't dissolve due to its form
of powder, but it ... Show more content on Helpwriting.net ...
First, we weren't informed about exact amount of solvents. The method said put a spoonful of
solute, but a spoonful doesn't define the exact mass. This might have impacted our data because
when we put a lot of powder or less, solute can takes longer time to dissolve or dissolve right away
due to small amount. Second, we were careless about the way we kept apparatus. For an example,
we didn't properly wash the spatula and stick during the experiment. Therefore, some chemicals
might have left and caused contaminations in solvents and solutes. To improve limitations and
possible source of errors, I can do several trials to improve accuracy in the

Determination Of Solubility Of Polar Solutes
Introduction
In the extraction of certain compounds, solubility plays an important factor in how the procedure is
performed. The identification of which solvent is best for the substance is crucial. The relation of
polarity between the solute and solvent is determined at a molecular level. The relationship between
the solvent and solute needed shared similar characteristics. Polar solutes are dissolved in polar
solvents while nonpolar solutes are dissolved in nonpolar solvents. From a chemical level,
molecules are attracted to other similar molecules closest to their properties. Polar molecules are
only attracted to polar molecules in both the solute and the solvent, and the same goes for nonpolar
molecules. This explanation ... Show more content on Helpwriting.net ...
For example, in some alkaline solvents acidic products will become deprotonated, keeping them
soluble in water. The target compound can then be extracted as a free base. The extract results then
are isolated by the addition of two immiscible solvents. Two solvents that cannot be mixed are used
so that there is a clear separation between the aqueous and organic layer. The component needed for
extraction is contained in the organic layer. This is because the compound is more soluble to the
composition of the organic layer. Multiple extractions with smaller volumes are more efficient than
a single extraction at a one large volume. When an organic solvent is used to extract a compound
from the aqueous solvent, smaller volumes will result in a better extraction. The success upon the
collection of the crude material is depended on how well the water is absorbed by the anhydrous
drying material. The presence of the drying material correlates with the vapor pressure of the other
compound. When the vapor pressure is low, there is a smaller amount of moisture in the gas
produced. Once the anhydrous material is added and clumping was avoided, the evaporation of the
added organic substance can proceed. The final material collected can be physically identified by
the final color of the precipitate. A greenish– white precipitate is most likely identified to be pure,
and a brownish color indicates that the collected material is wet, and not

The Concentration Of Solutes Affects The Rate Of Osmosis
Running head: THE CONCENTRATION OF
The Concentration of Solutes affects the Rate of Osmosis
Chintan Jani
Biology 101–01
Dr. Reynolds
October 6, 2016
Abstract The purpose of this lab was to observe the osmosis rates and mass changes of dialysis
tubes. To this three–dialysis tubes with differing sucrose levels were tested on their rate of osmosis
and weighed at 15–minute intervals. The results found described that as the sucrose level increases
the rate of osmosis increased as well.
Introduction
The purpose of this experiment was to determine if the rate of osmosis has an impact on the
concentration of solutes. During this lab varying levels of sucrose were put into dialysis bags and
then into containers with varying levels of sucrose. This was done to observe how the water moved
through the dialysis tubes. This process is known as osmosis is "...diffusion of water across a
selectively permeable membrane from high to low concentration... (Mader et. al., 2015)". The
movement of water across the membrane can be determined by the concentration of solutes inside
the dialysis bags and the solution surrounding it. If the cell is hypotonic, meaning it has a lower
osmotic pressure, its water concentration is lower than that of the concentration in the cell, which
can cause a rupture. If the cell is hypertonic, meaning it has a higher osmotic pressure; its water
concentration is lower outside the cell then it is inside causing the cell to shrivel up. The hypothesis
tested

Separation Of Solute Lab Report
The purpose of this experiment is to purify a mixture of compounds in order to separate it into its
pure components, using the common methods of extraction, gravity filtration, vacuum filtration, and
drying organic solutions. Extraction is a common laboratory technique used to isolate and purify the
product of a chemical reaction. The principle behind this method is that a solute is transferred from
one solvent into a second solvent that it is more soluble in. Therefore, the two solvents must be
immiscible and form two distinct layers. Generally, one solvent is aqueous and the other is organic.
Also, the extracting solvent should be somewhat easy to separate from the solute. The two layers
can be physically separated in a separatory funnel, where the denser solvent is usually on the
bottom. . ... Show more content on Helpwriting.net ...
The separation of benzoic acid and naphthalene is achieved through dissolution and extraction. By
dissolving the benzoic acid/naphthalene mixture in diethyl ether, a solvent that both are soluble in,
the mixture is completely dissolved and suspended throughout the solvent, forming a solution. By
next adding sodium hydroxide, the benzoic acid is converted into its conjugate base so that it is
soluble in water and insoluble in diethyl ether. Through the process of extraction, an organic layer
with naphthalene as the solute and diethyl ether as a solvent separates from an aqueous layer with
benzoic acid as the solute and water as the

Lab Report Of Freezing Point Of Cyclohexanol
Jacob Winter
Chem 101–103
11/7/16
Freezing Point Depression
Introduction:
The purpose of this lab experiment is to calculate the change in the freezing point of cyclohexanol
and a solution made with cyclohexanol to find the molar mass of the unknown solute. The
colligative properties of solutions are based on the concentration of molecules rather than the
identity of the chemical. Freezing point, a colligative property of a solution, is constant in any
substance and does not rely on the amount given. For any solution composted of two or more
components, there is one solvent and one or more solutes. The major amount is the solvent and the
minor amounts are the solutes. When solutes are dissolved in solvents, the solution's freezing point
is ... Show more content on Helpwriting.net ...
This can be accounted for in the sample of cyclohexanol. The experiment must have had a
contaminated sample of cyclohexanol, rather than a pure sample. This accounts for the difference in
the experimental values and the actual values.
Conclusion:
By demonstrating the application of a colligative property of a solution, the change in freezing point
observed between the cyclohexanol and the solution with the cyclohexanol as the solvent can be
used to calculate the unknown solute's molar mass. The data in the lab was graphed and shown in
the cooling curve. Bases off the data on the graph, the freezing points are shown where the
temperature remains constant. Using the freezing points, the difference of freezing point was
calculated. The molality of solution was calculated in the lab using the molar mass of the unknown
solution from the given previous data. The data in the lab showed the molar mass of the unknown
solute to be off from the given molar mass in the lab. This can be explained through contamination.
Theoretically, it was learned that by calculating the freezing point of an unknown substance, it can
be identified. The lab showed important techniques in using a graph

Potato Osmosis Lab Report
Chloé ten Brink
Biology Block C
Dr Reza
22/11/2017
Bio Lab Report
An Experimental Study on the Estimation of the Osmolarity of Potato Cores
Research Question:
How do different concentrations of sucrose solutions (0M, 0.2M, 0.4M, 0.6M, 0.8M, 1M) affect the
mass of potato cores when in sucrose solution?
Background information:
Osmosis is the diffusion of water across a membrane to create an equilibrium between the levels of
concentration of a solute both inside and outside the cell. In this case the solute will be sugar as the
potato core will be immersed in sucrose solution.
There are 3 potential states that the cell could be in depending on the type of solution:
Either the solution is hypotonic which means that the solution has a lower concentration than the
potato core and this would cause water to flow into the potato and make it larger and therefore
increase its mass. The state the cell is in is called turgid.
It could also be an isotonic solution. This is when the concentration (or molarity) of the cell is
identical with the concentration of the solution. In this case the cell (or potato core) stays the same
in size, which is called flaccid. No water moves either in or out of the cell.
Finally, it could also be a hypertonic solution which is when there is a higher concentration of solute
in the solution than in the cell and therefore the water leaves the cell. This make the cell
plasmolyzed or "shrunken". In our experiment this means that the potato cell would

Solute Concentration of Potatoes Essay
Nov 19/12 Solute Concentration of Potatoes Block 2–2
Purpose: To determine the concentration of solute in the potato's cytoplasm by measuring the
change in mass after the process of osmosis.
Materials and Equipment: Refer to Biology 12 Lab Manual – Investigation 13
Procedure: Refer to Biology 12 Lab Manual – Investigation 13
Data and Observations:
The Potatoes change in mass after the process of Osmosis Test Tube # | Concentration of Sucrose
Solution (mol/L) | Initial mass (g) | Final mass (g) | Change in mass (g) | Percentage change in mass
(%) | 1 | 1.0 | 5.12 | 3.63 | –1.49 | 29.0% | 2 | ... Show more content on Helpwriting.net ...
As the Concentration of the Sucrose Solution decreases, the more the potato's mass increases. This
is due to the solution being hypertonic. So, as the solute concentration gets lower, the potato's water
concentration will get higher, therefore more water particles from the solution will absorbed by the
potato. Some changed very little in mass because the concentrations of the H2O molecules in the
potato and outside the potato were equal. This equality in concentration is called Isotonic.
2) When the concentration was at 0.3M, the potato's cytoplasm and the sucrose solution was
isotonic. The concentration of the potato's cytoplasm was having the same solute concentration as
the surroundings. Therefore, there would be no net movement of materials happening. 3) The mass
of the potato cells would decrease, since the NaCl is a higher concentration than theH20 molecules
inside the potato cells. So, the potato's cells will evacuate and go into the surroundings, making it
shrivel. This is known as a hypotonic solution. 4) The Sucrose molecules cannot pass through the
plasma membrane, but osmosis can still occur. So, the permeability of the potato's cytoplasm only
applies to the H2O molecules. 5) Salt can use as a weed killer because the salt itself can cause the
plant to dehydrate through osmosis. Since the salt has a higher concentration, it will cause the plants
water molecules and nutrients to be absorbed by the salt, leaving the plant all

Resistance Of A Solute From Concentration
Introduction
Cells must move materials throughout membranes and cytoplasm in order to keep on maintaining
homeostasis. This movement is regulated through cellular membranes, which are selectively
permeable. Membranes have phospholipid bilayers, containing embedded proteins. The
phospholipid fatty acids hinder the flow of H2O because of its hydrophobic behavior.
The cellular environment is aqueous, meaning that the solutes dissolve in water (solvent). Water
may cross voluntarily through the membrane by osmosis. Most ions move through protein channels,
while transport proteins import larger molecules like carbohydrates.
The simplest form of movement is diffusion, where solutes move from an area of high concentration
to low concentration. Diffusion is directly related to molecular kinetic energy, taking in no energy
input. The movement of a solute from an area of low concentration to high concentration requires
energy input (ATP).
The process by which water moves through membranes by a special kind of diffusion is called
osmosis. Like solutes, water moves down its concentration gradient. Water moves from areas of
high potential to low potential. The high potential contains low solute concentrations, while the low
potential contains high solute concentrations. In surrounded cells, osmosis is affected not only by
the solute concentration, but also by the defiance to water flow by the cell wall, known as turgor
pressure.
The terms hypertonic, hypotonic, and isotonic are used

Egg Osmosis Lab
YEAR 11 SACE BIOLOGY THE EFFECT OF CONCENTRATION GRADIENT ON THE RATE
OF OSMOSIS Introduction: Water is the most abundant substance in any cell. All the chemical
processes of a cell involve water in some way. Osmosis is diffusion of water between a cell. The
following factors could affect the rate of osmosis: the solute concentration inside and out of the cell.
A raw chicken's egg with the shell removed can be used as a model of a cell to demonstrate the
process of osmosis.
Aim:
To determine the effect of changing the concentration gradient on the rate of osmosis. Hypothesis:
As the solute concentration of the water the egg is placed increases, the egg will shrink as the water
from inside it uses osmosis and moves towards the salt outside. ... Show more content on
Helpwriting.net ...
Pick up an egg with the spoon and carefully rinse it under tap water.
2. Place it on some paper towelling and blot it dry with another piece of paper towelling.
3. Place the egg on a watch glass and weigh it. Record the mass.
4. Place the egg in a beaker containing 10% salt solution ensuring there is enough liquid to cover the
egg.
5. Leave it in the solution for 10 minutes and then remove it with a spoon, pat it dry as before and
weigh it again. Record this mass.
6. Repeat steps 1 to 5 using the 5% salt solution, 2.5% salt solution and then distilled water.
7. Calculate the percentage mass change using the following formula: % mass change = Change in
mass x 100 Initial mass 1
Results:
1. Present the data in a table which includes the % mass change.
The effect of change in solute concentration on the mass of an egg Solute Concentration (%)
Initial Mass (grams)
Final Mass (grams)
Mass Change (grams)
Mass Change (%)
0
78.34
78.34
0
0

10
78.34
75.56
2.78
3.55
5
75.56

The Presence Of A Solute Affects The Boiling Point Of Water
Most people know how to boil water and what the boiling point of water is. Most people don't know
that if you dissolve anything into the water it will increase the boiling point. If you dissolve
anything in anything, it will increase the boiling point. "It's a generic phenomenon dissolve one
substance (the solute) in another (the solvent) and you will raise its boiling point" (Sciencebase, Salt
and the Boiling Point of Water, para. 1). "The presence of a solute affects the boiling point and
freezing point of the solvent. Solutions boil at a temperature that is higher than the normal boiling
point of the pure solvent, and freeze at a temperature that is lower than the freezing point of the pure
solvent," (Liquids, Liquid Mixtures and Solutions, para. 1). The boiling point changes when a
substance is dissolved in there because it takes more energy to move that atom.
Some substances increase the boiling point more than other substances when dissolved in water. In
my experiment, I am going to use salt (NaCl) and sugar (C12H22O11) to see which one will make
the boiling point go higher. Salts and sugars are both things that dissolve in water and ... Show more
I will be boiling water with the substances dissolved in it. There is however, another way to find
what the boiling point of the solution by using a formula. It is ΔTb=iKbm. "The formula used to
calculate the change in boiling point (Δ T b ) relative to the pure solvent is similar to that used for
freezing point depression: Δ T b = i K b m ," (Colligative Properties, para. 5). The Tb is the change
of the boiling point, Kb is the molal boiling point elevation constant, and m is the molal
concentration of the thing that you're dissolving
(http://www.chem.purdue.edu/gchelp/solutions/eboil.html). This equation says that the lower the
molal concentration, the bigger increase in boiling point. Salt does have a lower molal concentration
than

Diffusion And Osmosis Of Potato Cores
Diffusion and Osmosis in Potato Cores By Brittany Pangilinan Abstract Van't Hoff's Law suggests
that the osmotic potential of a cell is proportional to the concentration of solute particles in a
solution. The purpose of this experiment was to determine if there are any differences between the
osmolalities, the no–weight–changes of osmolalities, and the water potentials of potato cores in
different solutions of different solutes. The percent weight change of the potato cores was calculated
through a "change in weight" method. The potato core's weight was measured before and after they
were put into different concentrations of a solute for 1.5 hours. In our experiment, there were no
significant differences from the osmotic potentials of our results and the osmotic potentials of other
scientists work. Ending with chi square values of 2.17 and 2.71, and p values of 0.256 and 0.337,
concluding that there is no difference in water potentials of potato cores in different solutions of
different solutes at varying concentrations. Introduction Water potential is the measure of potential
energy in water, or the difference in potential energy between a water sample and pure water.
According to Kosinski (2015), white potatoes are an experimentally favorite subject for the
determination of water potential in plant tissue. In our lab we use a procedure called "change in
weight" method, which was published by Meyer and Anderson in 1935. Where we use potato cores
to explain and show

What Is The Colligative Properties Of Salt Dissolved In Water
Colligative properties are those properties of solutions that depend on the number of dissolved
particles in solution, but not on the identities of the solutes. For example, the freezing point of salt
water is lower than that of pure water, due to the presence of the salt dissolved in the water. To a
good approximation, it does not matter whether the salt dissolved in water is sodium chloride or
potassium nitrate; if the molar amounts of solute are the same and the number of ions are the same,
the freezing points will be the same. For example, AlCl3 and K3PO4 would exhibit essentially the
same colligative properties, since each compound dissolves to produce four ions per formula unit.
The four commonly studied colligative properties are freezing point depression, boiling point
elevation, vapor pressure lowering, and osmotic pressure. Since these properties yield information
on the number of solute particles in solution, one can use them to obtain the molecular weight of the
solute.
Freezing Point Depression
The presence of a solute lowers the freezing point of a solution relative to that of the pure solvent.
For example, pure water freezes at 0°C (32°F); if one dissolves 10 grams (0.35 ounces) of sodium
chloride (table salt) in 100 grams (3.53 ounces) of water, the freezing point goes down to −5.9°C
(21.4°F). If one uses sucrose (table sugar) instead of sodium chloride, 10 grams (0.35 ounces) in
100 grams (3.53 ounces) of water gives a solution with a freezing point of

Lab Report For Solutes
Water boils at 212 degrees Fahrenheit; it freezes at 32 degrees Fahrenheit. If you add a solute to the
water (solvent), it may change the temperature at which the water will boil or freeze. Depending on
the solute, the solution may boil at an elevated temperature. The solution may also freeze at
depressed temperature. If there is a change in the temperature, it will be because of the solute added
to the solvent.
A solute is added to the solvent, which produces a solution. A solute is a component in a solution
dissolved in the solvent. A solution is the solute and the solvent combined together making the
solution. For example, the water is your solvent,and the solute is your salt. When you combined the
solvent and the solute together you get

Potatis And Its Effect On The Rate Of Osmosis
When osmosis, a specialized process of water moving across a semi–permeable membrane from a
higher concentration to a lower concentration, occurs, scientists can understand the events that occur
when solute is mixed with water (Readel 2000).
Many factors account for the varying ways that osmosis can change during the process. One factor
is temperature, in that the higher the temperature, the quicker rate the rate of osmosis whereas, the
lower the temperature, the slower the rate of osmosis. Osmosis occurs quickly in high temperatures
because the membrane becomes more permeable (as in can pass through more easily) and molecules
move faster as heat increases. On the other hand, Osmosis occurs slowly when it is cold because the
membrane becomes less permeable and the molecules move slower as heat decreases (Traxler
1927). Osmosis is different than diffusion in that while a diffusion can occur in a variety of
mixtures, osmosis can only occur when there is a semi–permeable membrane. Particles are more
limited during osmosis because only water molecules can move, whereas any particle can move
more freely during diffusion (Anne Marie 2017).
With gained knowledge from this experiment, our group performed another experiment and tested
how certain amounts of solute concentration impacted the rate of osmosis. We hypothesized that
high solute concentrations affected the rate of osmosis. Tubes with high solute concentrations were
bound to show higher rates of osmosis than tubes with

How Does Kinetic Energy Change Over The Plasma Membrane
Cells are constantly active, preforming diffusion and osmosis on a daily basis just in order to remain
in equilibrium. Water and other solutes will cross over the plasma membrane in order to get into or
out of the cell. The plasma membrane is made of phospholipids that are in a bilayer. This bilayer
allows the membrane to be selectively permeable, allowing small and nonpolar solutes across. Other
solutes that are larger or are polar either cross at a very slow rate or use transfer proteins to get
across.
Particles in the body that are constantly moving in and out of cells have energy that is known as
kinetic energy, which is the energy of motion. Because these particles have kinetic energy, they are
able to be transferred. Many things can affect kinetic energy, and ultimately the diffusion within the
cell, such as heat, size of the solutes, and turgor pressure. The reason kinetic energy is so important
is because it allows there to be a concentration gradient. This gradient makes it possible for solutes
to diffuse into or out of the cell. Solutes will travel from a high concentration to a low concentration
in order to create equilibrium. However, once at equilibrium the cell will be isotonic to the
extracellular fluids and ... Show more content on Helpwriting.net ...
This is very important in biology. The cells even have tiny convolutions on its membrane in order to
increase its surface area dramatically. This higher surface area allows for the cell to diffuse more
particles in and out of the cell. The higher the rate of diffusion the better it is for cells because, like
stated before, equilibrium can be very bad because no particles are moving. Cells need this constant
movement of particles in and out of the cells in order to remain healthy.
The purpose of this lab is to see how different solutions have different rates of diffusion and how
surface area plays a role in

Trigonometry: Supersaturated Solution
It is necessary to heat the mixture in Step 2 of the procedure because without heat, the mixture is
unsaturated– it can hold more solute. At a higher temperature, however, the amount of solute added
in Step 1 allows for a saturated and then supersaturated solution.
2. The solution is supersaturated because there is more solute than the maximum at the temperature
at the end of Step 2. There were 5 grams of solute in the 10 grams of water. According to the given
solubility chart for sodium sulfate, in order for the solution to be saturated at the given temperature
(approximately 100°C because the water was nearly boiling), the solution could hold about 4.2
grams of solute. Because there was more than 4.2 grams, the solution was supersaturated.

Diffusion Of The Plasma Membrane
Introduction
Diffusion is one of two passive process membrane transports that moves molecules from a region of
their higher concentration to a region of their lower concentration and has a driving force of kinetic
energy. Simple diffusion, which is what this experiment is about, occurs without the assistance of
membrane proteins along the concentration gradient not using any energy from the cell itself–this is
where the kinetic energy come into play. Instead, the energy is coming from the molecules constant
motion. Once the solutes are dispersed evenly through the solution, equilibrium will occur and the
movement will stop (PhysioEx9.1). The plasma membrane in a cell is the physical barrier and is in
charge of determining what goes in and ... Show more content on Helpwriting.net ...
Keep in mind that the larger the MWCO, the larger the pores in the membrane are. The goal is to
understand how solute concentration and different molecular weights affect the rate of diffusion.
Materials:
PhysioEx9.1 Computer Program Exercise 1 Activity 1 (Simple diffusion), 2 beakers, dialysis
membranes with 20, 50, 100, and 200 molecular weight cut offs (MWCO), and deionized water.
Methods:
In this experiment, two beakers were used–the left beaker held the varying solutes and the right
beaker held the deionized water. Placed in the middle of the two beakers was the dialysis membrane
with various molecular weight cutoffs (MWCO), 20, 50, 100, and 200 MWCO. Four different
solutes, sodium chloride, urea, glucose, and albumin, were tested in different trials during this
experiment based on their molecular weight. Sodium chloride has a molecular weight of 58.44
g/mol, the smallest out of the four, so it was tested in the first trial with 20 MWCO. The left beaker
was filled with 9mM of sodium chloride and the right beaker was filled with water. The barrier
between the two beakers was dropped, allowing the solutions to have full access to the dialysis
membrane. At the end of 60 minutes, the amount of solute that passed through the membrane will be
able to be observed on the concentration display. After the amount is recorded, the beakers were
flushed and the next solute was added. The second trial with the 20

The Effect of Solute Concentration on Osmosis in Potatoes.
Martina Oganesyan
Title of investigation: The effect of solute concentration on osmosis in potatoes.
Research question: What is the solute concentration of a potato?
Hypothesis:
Osmosis is the movement of water across the membrane from an area of high concentration to an
area of low concentration. The osmosis continues until the solute concentrations are equal in both
areas. In this experiment, we put pieces of potatoes into test tubes with sugar solutions of different
concentrations and leave them there over night. It was done in order to compare the initial mass of
the pieces and the ones after being placed into the solute concentration. Pieces which were placed
into a hypotonic (lower) concentration of solute should increase in ... Show more content on
Helpwriting.net ...
From this we can conclude that measuring mass gives more logical and accurate results than
measuring length. It is so because a balance measures more precisely than a human eye using a
ruler, which also doesn't even take into account if there is a widening in the potato pieces.
However, there are also some errors in this experiment. Firstly, the potato pieces were not exactly
the same size despite the care taken to cut them. A solution to this would be to use a form cutter,
which would cut exactly equal shapes. Secondly, the temperature is not exactly a constant variable
as it fluctuated throughout the experiment as the test tubes were even put in a refrigerator with a
much lower temperature than in the room. A solution for improving could be to place the test tubes
into an electric cooler in the beginning to make the temperature stable and put potato pieces when
the test tubes are in the cooler. Also, errors could be decreased by using a more precise beaker.
Finally, repeating the experiment several times would allow as to take the average results and have a
more adequate result.
Conclusion:
Thanks to this experiment, we are able to answer our research question. The solute concentration of
a potato is 0.47 mol/l. Also, we can deduce that the method of measuring the mass gives more
accurate results than the method of measuring the length. I say that because the correlation
coefficient obtained was 0.95 which is very close to 1. However, there are

Freezing Point Lab
Observe
To conduct an experiment in order to calculate and compute how a particular solute affects the
overall freezing point of a solvent.
To calculate and determine the molecular weight of a specific solute (unknown) that is utilized
within the experiment.
Procedure
1. Freezing Point of Cyclohexane (Solvent)
Prepare an ice/water bath by obtaining a 400–mL and 600–mL beaker while, following Figure 14.4.
Obtain a ring stand and a thermometer. Secure the thermometer to the ring stand (consider
precautions if a glass thermometer is utilized).
Acquire a 250 mL beaker and a dry 200–mm test tube.
Place the test tube inside the beaker and record the mass of the beaker and the test tube together as
shown in figure 14.5.
Obtain 12 mL of cyclohexane (solvent) and add it to the test tube (use safety precautions as
cyclohexane is an irritant and highly flammable).
... Show more content on Helpwriting.net ...
Compute and record the mass of cyclohexane with the test tube and the beaker.
Obtain a wire stirrer and place it in the test tube with a thermometer. Make sure the thermometer is
immersed into the cyclohexane to achieve an accurate reading.
Stir the solvent within the test tube and record the temperature at specific timed intervals (15
seconds). Continue collecting the temperature until a change occurs within the temperature.
Using a graph, plot the data collected on the freezing point of cyclohexane.
2. Freezing Point of Cyclohexane plus Unknown

Solute Lab Report
After completing my experiment I came to a number of conclusions for each solute, the first solute
that I tested was Potassium chloride and as soon as I finished recording my data I came to a number
of interesting conclusions. First, as mentioned in my observations, in each attempt the numbers were
very similar in conductivity. This means that the data collected was very accurate since the numbers
didn't change to much. Also, the Potassium chloride was the hardest solute to dissolve in water. This
means that the solute is very saturated comparing to the other solutes that I tested. In conclusion the
Potassium chloride was very unique comparing to the other solutes that I tested throughout my
experiment. The next solute that I tested was ... Show more content on Helpwriting.net ...
First, one main reason why is because it is extremely high in conductivity comparing to the other
solutes that I tested earlier on in the experiment. The reason why it is so much higher is because it
has lots of ions, protons, electrons, and neutrons and this causes it to a have a very strong and
powerful electrical current. Also, another unique conclusion that I came to was that the level of
conductivity varied tremendously between each attempt. The reason why this happens is because
since the level of conductivity is much high there is a much wider range that it can go from rather
than the other solutes that I test earlier on in my experiment. In conclusions this certain solute was
the most interesting out of all the other

The Concentration Of A Solute
Colorimeter A colorimeter is a piece of equipment that is used to measure the concentration of dilute
solutions (intensity of light). The colorimeter is mainly used to determine the concentration of a
known solute (chemical that is placed into a solvent and dissolved) in a given solution by using
Beer–Lambert's Law which states that the concentration of a solute is proportional to the
absorbance. According to Beer's law, "when monochromatic light passes through the coloured
solution, the amount of light transmitted decreases exponentially with increase in concentration of
the coloured solution". Equation for Beer's law = It = Ioe–KC According to Lambert's law "the
amount of light transmitted decreases exponentially with increase in thickness of the colored
solution". Equation for Lambert's law = It = Ioe–kt According to Beer–Lambert's Law "the quantity
of light absorbed by a substance dissolved in a fully transmitting solvent is directly proportional to
the concentration of the substance and the path length of the light through the solution". Beer–
Lambert's Law = IE/Io = e–KCT IE = intensity of emerging light K = a constant Io = intensity of
incident light C = concentration e = base of neutral logarithm T = thickness of the solution
Colorimetry can be used for testing water quality. This process involves adding a chemical reagent
to a water sample and then measuring it to identify the chemicals present. The concentration of
certain chemicals can

Molarity Lab Report
In the experiment I performed it was proven that, the moles of a solute, the volume of the solvent,
and the molarity of an aqueous solution share a mathematical relationship. Solutions, solutes and
molarity are all terms that are crucial to chemistry. A solution is defined as a liquid mixture in which
the solute is distributed into the major component, and a solute is defined as the minor component
within a solution. The molarity of a solution is the number of moles of solute per liter of solution
and it is used to express the concentration of a solution. Both the solute amount and the solution
volume can be determined in experiments to aid finding the solution concentration or molarity of a
substance that they are included in. I state confidently ... Show more content on Helpwriting.net ...
The concentration or molarity of a solution can be seen through the ratio of solute amount to
solution volume. This means that when you take the solute amount of a solution in moles and divide
it by its volumes in liters you are able to find the solution's overall concentration or molarity. This is
proven in my data table below labeled, Molarity Simulation Data, and it helps to show that when
you get the moles of a solute and the volume of the solvent you can easily and accurately figure out
the molarity of the solution you are investigating. For example, when there were .75 moles of an
energy drink power, and only .25 liters of water to mix it in, as shown in #3 of my data table, I was
able to identify the molarity as 3 M through the process of division The way I executed this
experiment lets me to see the impressionable relationship between the solute amount in moles and
the solution volume in liters and how they affect molarity of a solution. The data table allows me to
demonstrate this pattern. My argument is, the molarity of an aqueous solution can be found if you
take the moles found in a given solute and divide them by the volume of the solvent. The data
presented in this paper proves that the given formula shows the dependent, mathematical
relationship that all three of these factors

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