Lab #5: Osmosis, Tonicity, And Concentration.
Lab #5: Osmosis, Tonicity, and Concentration.Background.The internal environment of the human bodyconsists largely of water-based solutions. Alarge number of different solutes may bedissolved in these solutions. Since movement ofmaterials across cell membranes is heavilyinfluenced by both differences in theconcentration of these various materials acrossthe cell membrane and by the permeability ofthe lipid bilayer to these materials, it is criticalthat we understand how the concentration of aparticular solute is quantified, as well as howdifferences in concentration influence passivemembrane transport.Diffusion, Osmosis, and TonicitySimple diffusion.Particles in solution are generally free to moverandomly throughout the volume of the solution.As these particles move about, they randomlycollide with one another, changing the directioneach particle is traveling.If there is a difference in the concentrationof a particular solute between one region of asolution and another, then there is a tendency forthe substance to diffuse from where it is moreconcentrated to where it is less concentrated.This is because of the random collisions amongparticles which eventually will evenly distributethe solute throughout the volume of the solution.Thus net diffusion occurs “down” aconcentration gradient, from an area of highconcentration to an area of low concentration,until a state of equilibrium is reached throughoutthe volume of the solution. At this equilibrium(typically when the concentration is uniform),there will be no more net diffusion of solutefrom one area to another, although randommovement of particles will continue.If such diffusion is unimpeded by thepresence of some barrier (e.g., a membrane thatis impermeable to the solute), it is referred to assimple diffusion. In the case of cells, solutes thatcan readily pass through the lipid bilayer of cellFig 5.1. An example of simple diffusion. Molecules ofred dye gradually diffuse from areas of higherconcentration to areas of lower concentration until theconcentration of dye is uniform throughout the volumeof the solution.membranes (i.e., small uncharged molecules ormoderate-sized nonpolar molecules) aretransported across cell membranes via simplediffusion. For example, the exchange of gasessuch as O2 and CO2 across the plasma membraneoccurs through simple diffusion.OsmosisLike some other small, uncharged molecules,water (H2O) can pass quite readily through a cellmembrane, and thus will diffuse across themembrane along its own concentration gradientindependent of other particles that may bepresent in the solution. However, often solutesthat are dissolved in water (e.g., glucose, Na ,Cl- , etc.) cannot pass through the lipid bilayer ofthe cell membrane, and thus cannot move acrossthe cell membrane even if there is a difference inconcentration.The “concentration” of water in a solution isinversely related to solute concentration – thegreater the concentration of total solutes in thesolution (regardless of what those specificsolutes are), the lower the number of watermolecules per unit volume of the solution. Thuswater tends to diffuse from more dilute areas(i.e., with lower solute concentration) to moreconcentrated areas (i.e., with higher soluteconcentrations).Cells are surrounded by a semi-permeablemembrane which will allow water to pass
semipermeable membranesemipermeable membraneFig 5.2. During osmosis, water diffuses across a semipermeable membrane from the less concentratedsolution to the more concentrated solution (blockarrow) until the concentrations across the membraneare equal. Note that the volume of the solution to theright increases and the volume of the solution on theleft decreases (black arrows) as water diffuses from leftto right.through but prevent many hydrophilic solutesfrom passing through. Thus, the diffusion ofwater into or out of the cell is driven bydifferences in the total concentration of solutesthat cannot pass through the plasma membrane.The diffusion of water across a semi-permeablemembrane such as a cell membrane is termedosmosis. Osmosis is thus a special case ofdiffusion involving only the movement of water(the solvent of the solution) across a barrierpermeable to water but impermeable to certainsolutes.It is the difference in the soluteconcentration across a semi-permeablemembrane that provides the driving force forosmosis (Fig 5.2). Therefore, osmosis can onlyoccur under specific conditions where a) twoaqueous solutions are separated by a membranethat is permeable to water but impermeable to atleast one of the solutes in the solution1 and b)there is a difference in the total concentration ofimpermeable solutes between the two solutions.1If water flows from one solution intoanother, then the volume of the second solutionwill tend to increase while that of the firstsolution decreases (Fig 5.2). This change involume can only occur if the semi-permeablemembrane (or one of the other walls of thecontainer holding each solution) is compliantenough to accommodate this change in volume.If water is redistributed across the membrane butthe membrane will not stretch or reposition itselfto accommodate an increase in volume, thenpressure will build up inside of the solutiongaining water, and likewise decrease in thesolution that is losing water. As more water isdrawn into the more concentrated solution,pressure will build up, pushing outward on thewalls of the container. Eventually, this pressureexerting outward will become high enough toequal the force that is driving water from the lessconcentrated solution to the more concentratedsolution. At this point, no further osmosis willoccur, since these two equal and opposite forcesacting on the movement of water are cancelingeach other out.The amount of force that would need to beexerted on a solution to prevent osmotic uptakeof water by that solution is called the osmoticpressure. In effect, it is a measurement of howstrongly a solution will draw water into itselffrom an adjacent solution across a semipermeable membrane. Osmotic pressure isdirectly related to the total solute concentrationof a solution. As solute concentration increases,more and more pressure would need to beexerted on a solution to prevent osmotic uptakeof water from an adjacent solution.To illustrate this, consider a situation wherewe have a solution containing a particularconcentration of solutes that is separated from apure water by a semi-permeable membrane (Fig5.3). The membrane and walls of the containerare rigid, but the top of the container is open, sothe solutions can change in volume by changingvertical height. Notice that as water moves fromthe pure water into the solution, the height of thepure water decreases and the height of thesolution increases. Will all of the pure water be– if the membrane is not impermeable to any of the solutes within the solutions, then the solutes will simply diffuse across themembrane to equalized their concentrations. This would be a case of simple diffusion, not osmosis.
[solute] 8 Osm[solute] 20 OsmIsotonicHypertonicHydrostatic pressure( Osmotic pressure)Hydrostatic pressure( Osmotic pressure)HypotonicFig 5.3 A comparison of the osmotic pressures of twosolutions. Notice that the hydrostatic pressure neededto balance the osmotic uptake or water by the solutionfrom pure water is higher in the more concentratedsolution to the right than for the less concentratedsolution to the left.drawn into the solution? No. As the height ofthe solution increases, so does hydrostaticpressure as a result of the pull of gravity on thewater. This pressure tends to push water backinto the pure water solution. Equilibrium (no netosmosis) will occur when this hydrostaticpressure is equal to the osmotic pressure.Now compare this solution with one with aconsiderably higher concentration of solute(Figure 5.3). Notice that equilibrium is achievedonly after much more water is transferred fromthe pure water to the solution, and thus a highercolumn of solution is supported. Thehydrostatic pressure needed to balance theosmotic pressure is much higher, indicating thatthe osmotic pressure itself is much higher.TonicityLiving cells have the potential of gaining orlosing water from the surrounding extracellularfluid through osmosis. The net movement ofwater into or out of the cell is driven bydifferences in osmotic pressures between theextracellular fluid and the intracellular fluid.Thus, the effect that an extracellular solution hason the osmotic movement of water into or out ofthe cell is described by the tonicity of theextracellular fluid. For example, if the osmoticconcentrations (i.e., total solute concentrations)Fig 5.4. Examples of erythrocytes suspended inisotonic, hypertonic, and hypotonic saline solutions.Images are from www.visualsunlimited.com.of the intracellular fluid and the extracellularfluid are the same, and none of these solutes canpass through the cell membrane, then theosmotic pressures of the intracellular fluid andthe extracellular fluid will be the same, and nonet osmosis will occur. The extracellularsolution is said to be isotonic (“equal tension”,referring to the equal osmotic pressures). Cellsremoved from the body and placed in isotonicsolutions will retain the “normal” shape theyhave in the body (Fig 5.4).If a cell is placed into a solution with ahigher osmotic concentration than theintracellular fluid (for example, human bloodcells in sea water), then the osmotic pressure ofthe extracellular fluid will exceed that of theintracellular fluid. As a result, water will flowout of the cell and into the extracellular fluid,causing the cell to shrink and crenate (Fig 5.4).In this case, the extracellular fluid is said to behypertonic (“greater tension”).Conversely, if a cell is placed in a solutionwith a lower osmotic concentration (forexample, distilled water), then the osmoticpressure of the extracellular fluid is less thanthat of the intracellular fluid. As a result, waterflows into the cell, causing it to swell (Fig 5.4)perhaps so much that the cell may undergo lysis(burst). In this situation, the extracellular fluidis said to be hypotonic.
Measurements of ConcentrationMolarityThere are many different ways of quantifyingthe concentration of a solution. In the clinicalsciences, concentrations are often expressed as amass of solute dissolved in a given volume ofsolution (e.g., g/dL, mg/ml, etc.). A ratio ofsolute mass to solution volume is also the basisof “percentage” solutions (for example, a 2%saline solution). A percent solution is thenumber of grams of solute in 100 ml of solution(or 1 dL of solution).To understand the effects of soluteconcentration on diffusion and osmosis,however, it is important to know the actualnumber of solute particles present in a givenvolume of solution. Using solute mass as asurrogate presents a problem, since differentsolutes have different particle sizes the samemass of solute could have very differentnumbers of particles for two different solutes.Therefore, scientists often express concentrationbased on moles of particles in a given volume ofsolution. One mole is equal to 6.02 1023particles, and the mass of one mole of particlesof a substance is equal to that substance’s atomicor molecular weight (molecular weight being thesum of the atomic weights for each atom in themolecule). To determine the number of molesof a given mass of some substance, divide themass of the substance by the molecular weightof that substance:Molarity can be defined as the number of molesof a solute per liter of solution volume. Insolutions with more than one solute, each solutewill have its own specific molarity. The unit ofthis measurement of concentration is molar (M).1 M is equal to one mole solute per liter ofsolution.Moles Mass (g)Molecular weight (g/mole)For example, to calculate the number of molespresent in 117g of NaCl, first determine themolecular weight of NaCl. The atomic weightof Na is 23.0 g/mole, and that of Cl is 35.5g/mole, so the molecular weight would be thesum of the two (23.0 35.5 58.5 g/mole). Thenumber of moles of NaCl is then calculated bydividing the mass by the molecular weight, so117 g / 58.5 g/mole 2 moles.There are a number of differentexpressions of concentration than use moles toquantify the amount of solute in the solution.Herein we will describe three of them: molarity,molality, and osmolality.Molarity (M) Moles SoluteSolution Volume (L)Determining the molarity of any solution iseasy given the mass of solute, the molecularweight of that solute, and the total volume of thesolution. First, determine the number of molesof solute used to make the solution by dividingthe mass of the solute by its molecular weight.1) Moles Solute Mass (g)Molecular weight (g/mole)Then divide the number of moles solute by thevolume of the solution in liters.2) Molarity (M) Moles SoluteSolution Volume (L)For example, if we were to mix 90 g ofglucose (MW 180 g/mole) in 2 L of water, wewould calculate the molarity of this solution firstby determining the number of moles of glucose(90 g / 180 g/mole 0.5 moles), then dividingthe number of moles of glucose by the volume inliters (0.5 moles / 2 L 0.25 M).If the molarity of a solution is less than 1M(and particularly if it is less than 0.1 M), it maybe appropriate to express the concentration inmillimolar (mM) where 1 mM is equal to1/1000th of a mole per liter solution. Conversionfrom M to mM is performed the same as anymetric conversion, so a concentration of 0.25 Mis equal to a concentration of 250 mM.MolalityWhereas molarity is often the preferredmeasurement of concentration for chemists,physiologists often prefer to use a somewhatdifferent measure of concentration called
MOLARITY1 mole solute AMOLALITY1LAdd water to 1L solution1 mole solute B1 mole solute A1LPre-measure 1L (1kg) water1LAdd water to 1L solutionSolute/Solvent ratio variesVolume constant1 mole solute B1LPre-measure 1L (1kg) waterSolute/Solvent ratio constantVolume variesFig. 5.5 A comparison of molarity and molality using two solutes of different molecular weights. When mixing 1 L of a 1M solution of each, notice that a different amount of water is used to bring the total volume of the solution to 1L. Thusalthough the final volume of the two solutions is the same (1L), the solute to solvent ratio is different. When mixing a 1 msolution with a specified solvent mass of 1 kg, notice that in both cases the total solution volume exceeds 1L, and that thesolution with the larger solute particles has a greater total volume. However, since both the amount of solute and theamount of solvent are specified, the two solutions have equal solute/solvent ratios.molality. Molality is a ratio of moles of aparticular solute per kg solvent used to make thesolution. The unit used to express molality ismolal (m), where 1 m one mole solute per kgsolventMolality (m) Moles SoluteSolvent Mass (kg)Calculation of molality is very similar tocalculation of molarity. First, one mustdetermine the number of moles of solute (solutemass/molecular weight).1) Moles Solute Mass (g)Molecular weight (g/mole)Then divide the number of moles of solute bythe mass of solvent in kg.2) Molality (m) Moles SoluteSolvent Mass (kg)In physiological studies, water is usually the2solvent used, and since the density of liquidwater is 1.0 kg/L, a 1 L volume of water willhave a mass of 1 kg.As an example, if 2.3 g of ethanol (MW 46 g/mole) were added to 50 ml of water, wecould determine the molality as follows: first,calculate the number of moles of ethanol (2.3 g /46 g/mole 0.05 moles) then divide by the massof water in kg (0.05 moles / 0.05 kg water2 1m).The difference in calculating molarity andmolality seems subtle, but this difference doeshave very important implications (Fig 5.5). Inpreparing a solution based on molarity, thenumber of solute particle and the total volume ofthe solution is specified, but the amount ofsolvent is not. Total volume of the solution isthe volume of the solvent PLUS the volume ofthe solute. Therefore, since different soluteshave different volumes per mole, the amount ofsolvent used to make a solution of a particularmolarity will vary depending on what solute isused. Therefore, the solute to solvent ratio willvary depending on the solute used. In contrast,– at this point you should be comfortable enough with conversions to see how I derived this mass from 50 ml of water.
BOTH the number of solute particles and theamount of solvent are specified. Therefore thevolume of solution occupied by the solvent isconstant, so total volume will vary based on thevolume of the specific solute used in thesolution. However, since both the amount ofsolvent and the number of solute particles arespecified, the solute to solvent ratio will beconstant in any solution of a particular molality,regardless of what specific solute is used.The osmotic movement of water is driven bydifferences in solute to solvent ratio between onearea and another, and is relatively independentof what specific solutes are present in solution.Thus molality is often the preferredconcentration measurement for physiologistsbecause not only does it quantify theconcentration of specific solutes in a solution,but also indicates the osmotic properties of thatsolution.OsmolalityOsmosis is driven by differences insolute/solvent ratios that exist across a semipermeable membrane. Importantly, it isdifference in the total solute concentration ofthese solutions (i.e., the osmotic concentration)that influences osmosis, not the specific type ofsolute particles found in that solution (Fig 5.6).The total solute concentration of a solution (andthus the osmotic concentration) can bequantified through the osmolality of the solution.Osmolality is the ratio of moles of total soluteparticles per kg water. The unit for thismeasurement of concentration is osmolal (Osm),where 1 Osm is equal to one mole of total soluteper kg water.Osmolality (Osm) Moles Total SoluteSolvent Mass (kg)Calculation of osmolality can be somewhatmore challenging than the calculation ofmolality. If the solution contains a single,nodissociating solute, then the osmolality of thesolution is equal to the molality of the solution.However, if the solution contains multiplesolutes, the number of moles of each solute mustbe calculated independently, then the totalnumber of solutes particle determined by addingvFig 5.6. Differences in osmolality drive the osmoticmovement of water. The solutions above differ in themolality of each of the two solutes. However, sincetotal solute concentration (Osm) is equal in bothsolutions, no net osmosis occurs.together the number of moles for each individualsolute.For example, let us calculate the osmolalityof a solution containing 18 g of glucose (MW 180 g/mole) and 23 g of glycerol (MW 92g/mole) mixed with 500 ml of water. Todetermine the osmolality, we must first calculatethe number of moles of each individual solute.We find there are 0.1 moles of glucose (18 g /180 g/mole) and 0.25 moles of glycerol (23 g /92 g/mole). We then add up the total number ofsolute particles (0.1 mole 0.25 mole 0.35moles). By dividing the moles of total solute(0.35 moles) by the mass of water used to makethe solution (0.5 kg) we determine theosmolality of the solution to be 0.7 Osm.An additional consideration must be madefor ionic compounds. Recall that ionic bondscan be broken when an ionic compound isplaced in water. This is because watermolecules have an electrostatic attraction to eachof the ions in the ionic compound, and thecombined draw of many water molecules oneach ion can exceed the draw the ions have forone another. Thus, ionic compounds have atendency to dissociate (break apart) whendissolved in water.This can have a considerable influence onthe osmotic concentration of a solution.Consider the ionic compound NaCl, whichreadily dissociates in water into Na and Cl-.Thus for each mole of NaCl mixed into thesolution, TWO moles of solute particles (1 mole
vvcontaining an ionic compound, first calculate thenumber of moles of the ionic compound used tomake the solution. Then multiply the number ofmoles solute by the number of particles formedwhen the compound dissociates in solution.Finally, divide the total number of moles ofsolute particles in solution by the mass of waterused to make the solution.For example, 22.2 g of CaCl2 (MW 111g/mole) are dissolved in 3 L of water. Note thatCaCl2 dissociates into three separate particles(Ca2 2Cl-). To calculate the osmolality of thissolution, first determine the number of moles ofCaCl2 (22.2 g / 111 g/mole 0.2 moles).Second, determine the number of solute particlesin solution that would be formed as a result ofdissociation (0.2 moles of CaCl2 3 particlesformed per CaCl2 molecule 0.6 moles totalsolute). Finally divide the total number of soluteparticles by the mass of water used to make thesolution (0.6 moles / 3 kg 0.2 Osm).vFig 5.7. Effect of ionic compound dissociation onosmotic concentration. The two solutions separated bya semipermeable membrane above have the samemolality. However, the solution on the right containsan ionic compound as its solute. The ionic compounddissociates, which increases the osmotic concentrationof the solution on the right. As a result, water flowsfrom the solution on the left to the solution on the right.of Na and one mole of Cl-) are present insolution. So if one was to mix together a 1molal solution of NaCl, the osmoticconcentration would be 2 Osm3. This, in turn,would give this solution a markedly higherosmotic pressure than that of a solution with thesame molality but containing non-dissociatingsolutes (Fig. 5.7).To determine the osmolality of a solution3– this assumes 100% dissociation—that every single NaCl particle breaks into Na and Cl-. In reality, no ionic compound willcompletely dissociate in solution, and the degree to which ionic compounds do dissociate depends on the specific ion compound.For sake of simplicity, though, we will assume 100% dissociation for calculation in this course.
Fig 5.8. Thistle tube osmometer we will use in this exercise. On the left is the assembled thistle tube with dialysis membranecovering the base and a transfer pipette. On the right is an illustration of the fully assembled osmometer. Water is drawnthrough the dialysis membrane into the sucrose solution. This results in an increase in the volume of the sucrose solution,which expands up the stem of the thistle tube.Experiment I: Assembling an Osmometer.In this activity you will construct an apparatus that will enable you to observe osmotic movement of waterfrom one solution to another over time.Procedures1. Obtain a thistle tube assembly (consisting of a solution bell covered on the bottom with a piece ofdialysis membrane and a tube-like stem—see Fig 5.8), a specimen cup with lid, and a smallrubber “o-ring”.2. Pull the stem apart from the bell. Using a transfer pipette fill the bell of the osmometer as full aspossible with 40% sucrose solution (this sucrose solution has been dyed green to make it morevisible).Hint: As you are filling the bell, hold it up off the countertop so that the dialysis membraneon the bottom sags downward under the weight of the sucrose solution. This will allow youto fill even more sucrose into the bell. Make sure you fill it all the way to the top, so that thesucrose is about to spill over our of the opening at the top of the bell.3. Reassemble the thistle tube by FIRMLY inserting the stem into the opening at the top of the bellHint: It is VERY important that you have a good seal between the bell and stem. If there isany leakage at that joint then either air will flow in or sucrose solution will flow out, and theexperiment will not work. The stem should be inserted as far down into the bell as it will go,and should not be easily rotated once in place.
4. Insert the stem of the thistle tube through the hole in the lid so that the top of the bell is flushagainst the bottom of the lid. Fasten into place by sliding the o-ring down the stem until it isflush with the top of the lid.5. Fill the specimen cup most of the way full with water. Place the lid onto the specimen cup so thatthe bell of the thistle tube is submerged in the water. Seal the seams of the lid around the lip ofthe cup.Hint: sucrose solution should be pushed up into the stem of the thistle tube. If you cannotsee sucrose solution in the stem, then either there is a leak in the tube or not enough sucrosesolution was placed in the bell.6. Mark the position of the meniscus of the sucrose solution with a piece of masking tape. Set aside.7. At a time specified by your instructor, examine your osmometer, and record any changes thatoccur in the amount of sucrose solution present in the stem.Experiment II: Effects of Varying Tonicity on Erythrocyte MorphologyIn this exercise we will expose erythrocytes to saline solutions of different concentrations andmicroscopically examine the effect of varying extracellular tonicity on the shape of the cells.READ THIS BEFORE YOU START!!!You will be using high quality microscopes for your examination of cell morphology. These scopesare very expensive and could easily be damaged by careless usage. Therefore, it is critical that youfollow the instructions carefully in order to observe the cells. Moreover, you MUST follow theproper procedures for putting away the microscopes at the end of this exercise. Before you leave,you must have your lab instructor inspect your scope to be sure it has been properly put away.Failure to do so will result in the loss of 10 points from your homework assignment (yeah, we’repretty serious about this).ProceduresA) Preparing the slidesNote: Make one slide and perform your microscopical observations on it before making the next slide1. Obtain a clean microscope slide and a cover slip2. Apply a small drop of blood to the center of the microscope slide (see Fig 5.9)3. Place a drop of one of the five saline solutions on the slide over the blood.o Available saline solutions (NaCl) include: 0.2%, 0.45%, 0.85%, 3.5%, and 10%4. Immediately mix the blood and saline using the edge of your cover slip.
Fig. 5.9. Illustration of a microscope slide showing the proper amount of blood that should be placed on the slide.5. Gently place the cover slip over the blood/saline mixture, taking care to avoid trapping largebubbles undeneath.6. Observe cell morphology under the microscope.7. Repeat with the other four saline solutions.B) Using the microscope.1. Familiarize yourself with the different parts of the scope labeled in Fig 5.10.2. Before inserting a slide into your microscope make sure that ooThe lowest power objective lens has been rotated into the locked position over the slide stage.The stage has been lowered all the way down using the coarse focus knob.3. Lock the slide into the slide holder on the stage.4. Using the slide position controls, position the slide so that the sample is immediately above theFig 5.10. Illustration of the Leica CM E microscope used in this lab, with essential components labeled.
opening in the bottom of the stage that leads to the light source.5. Turn on the light source for the scope and look through the eyepiece. If the 4x objective lens isproperly locked in place, you should see light (although probably not in focus).6. Using the coarse focus knob, focus the image you see in the eyepiece. You will be observing thecells at 40x magnification (the product of the 10x eyepiece lens and the 4x objective lens).7. Once the image is in focus with the 4x objective lens, rotate the objecive lenses to bring the 10xobjective lens into place. Refocus using the FINE FOCUS KNOB ONLY. Once focused, youwill be observing the cells at 100x magnification.8. Rotate the objective lenses once again to bring the 40x objective lens into place, which will giveyou a 400x magnification of the cells. Again, refocus using the FINE FOCUS KNOB ONLY—this is critical now—since the stage has now lifted the slide so close to the objective lens it wouldbe very easy to smash the objective lens through the slide!9. Observe the shape of the red blood cells under the scope and describe their shape on your datasheet. Based on the shape of the cells, is the saline solution isotonic, hypotonic, or hypertonic?Hint – in one saline solution it may be very difficult to find any cells at all, and you may justsee little bits of “flotsum” on the slide. Recall that under very hypotonic conditions the cellsmay swell up so much that they undergo lysis.10. When you are done with the slide, rotate the 4x objective lens back into place, lower the stage allof the way down with the coarse focus knob, and remove the slide. UNDER NOCIRCUMSTANCES ARE YOU TO REMOVE A SLIDE WITHOUT SETTING THEOBJECTIVE LENS BACKTO 4X AND LOWERING THE STAGE. Wipe off any fluidfrom the stage with a kimwipe.11. Repeat with the other four slides.12. When you are finished, be sure that oooooThe lowest power objective lens has been rotated into the locked position over the slide stage.The stage has been lowered all the way down using the coarse focus knob.There is no blood on the stage of the scope (or anywhere else)That the power has been turned offThe slides have been properly disposed of in a biohazard box13. Summon your lab instructor to inspect your scope to be sure that it has been properly shut down.
the cell membrane and by the permeability of the lipid bilayer to these materials, it is critical that we understand how the concentration of a particular solute is quantified, as well as how differences in concentration influence passive membrane transport. Diffusion, Osmosis, and Tonicity Simple diffusion.
Lab 4: Diffusion and Osmosis (Revised Fall 2009) Lab 4 - Biol 211 - Page 1 of 23 Lab 4. Diffusion and Osmosis in Selectively Permeable Membranes Prelab Assignment Before coming to lab, read carefully the introduction and the procedures for each part of the experiment, and then answer the prelab q
Biology Lab Notebook Table of Contents: 1. General Lab Template 2. Lab Report Grading Rubric 3. Sample Lab Report 4. Graphing Lab 5. Personal Experiment 6. Enzymes Lab 7. The Importance of Water 8. Cell Membranes - How Do Small Materials Enter Cells? 9. Osmosis - Elodea Lab 10. Respiration - Yeast Lab 11. Cell Division - Egg Lab 12.
1.1.3 Recent Advances in RO Membrane Technology 9 1.1.4 Future Advancements 12 References 12 . 2 . Reverse Osmosis Principles 2.1 Osmosis 2.2 Reverse Osmosis 2.3 Dead-End Filtration 2.4 Cross-Flow Filtration 3 Basic Terms and Definitions 3.1 3.2 Recovery 3.3 Rejection 3.4 Flux 3.5 Concentration Polar
(EFW) in Parenteral Fluids IV fluid Na Osm Tonicity % EFW (mOsm/L) D5W 0 252 0 100 D5W ¼ NS 34 321 78 D5W ½ NS 77 406 150 50 LR 130 273 16 D5 LR 130 525 16 D5 0.9% NS 154 560 308 0. Osmolarity vs Tonicity Osmolarity is the number of osmoles of solute per liter of solution
AP BIOLOGY NAME_ CELL UNIT ACTIVITY #6 DATE_HOUR_ DIFFUSION AND OSMOSIS LAB INTRODUCTION: In this laboratory you will investigate the processes of diffusion and osmosis in a model membrane system. You will also investigate the effect of solute concentration on w
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energy –referring to active transport, exocytosis and endocytosis. The correct processes that the molecules Louise was may utiliseinclude A. osmosis, simple diffusion and active transport. B. osmosis and simple diffusion. C. osmosis and exocytosis. D. simple diffusion, exocytosis and endocytosis
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