When a cell and a solution have the same concentration of solutes that are considered?

Truly understanding osmolarity and tonicity is one of the more challenging endeavors undertaken by students of the natural sciences.1 In 2012, Cheng and Durairajanayagam (3) reported that 41% of students participating in the International Intermedical School Physiology Quiz were unable to predict correctly what would happen to the volume of an erythrocyte when placed in an isosmotic solution of urea. So how can we teach osmolarity and tonicity more effectively? One strategy for improving student understanding is to let them experience the effects of different solutions on erythrocyte hemolysis, a classic experiment in physiology [see, for example, the recent Sourcebook paper by Goodhead and MacMillan (5)]. But teaching osmolarity and tonicity in a classroom setting can be tough. Nonetheless, mastering these concepts is essential for those interested in pursuing careers in medicine, physiology research, and other related fields. For example, health professionals need to understand that not all isosmotic intravenous (IV) fluids are isotonic (15).

The approach we present here has been simplified for teaching osmolarity and tonicity to undergraduate health professions students. It is specifically designed to assist them in the proper selection of IV fluid therapy in clinical settings. As with any simplification of a complex physiological topic, the nuances of the topic are glossed over to facilitate student comprehension. The selection of appropriate IV solutions is not an exact science, and it does not require understanding or application of the van’t Hoff equation or reflection coefficients (1, 4, 18). The clinician needs to choose IV solutions based on their ability to increase effective circulating volume, to expand or contract cell volume, and to affect osmolarity. Having the skills to calculate approximations of these changes using broad assumptions helps prevent iatrogenic errors in clinical settings.

The difficulties students and teachers alike encounter while learning osmolarity and tonicity are at least partially accounted for by the differing approaches taken by many physiology texts. A lack of consistent terminology and detail often leaves room for confusion and misconceptions (Table 1).

Because osmolarity and tonicity are fundamental and often confusing, we thought it important to review the topic for instructors, as well as outline some teaching approaches we have found effective in the classroom. We start with our definitions of the two terms, then give you a chance to check your understanding of the basics of osmolarity and tonicity through a few short questions and a brief review. We then present several aspects of this topic that our students find most confusing. Finally, we outline our approach to teaching osmolarity and tonicity and give examples of questions that we have used to develop our students’ understanding of the topic.

Clarifying definitions.

The varying presentations of tonicity in Table 1 are tricky to understand at first glance. Let us start by clarifying the definitions of osmolarity and tonicity. Our definitions apply to biologically relevant situations only, so we are ignoring solutions such as concentrated acids and bases. One note on units: in this paper, milliosmole is expressed in units of mosmol, whereas milliosmolar (mosmol/l) is expressed as mosM. (Note the capitalizations.)

Osmolarity is a measure of the concentration of osmotically active particles in a solution. It is sometimes called a “colligative” property of the solution by chemists because it depends on the number of particles in a volume of solution rather than the identity of the particles. Osmolarity is closely related to molarity, a concept that most students learn in introductory chemistry. One mole of any substance has Avogadro’s number (6.02 × 1023) of particles. Molarity (M) of a solution is an expression of concentration, with one mole of solute per liter of solution.

However, the molarity of a solution is not always the same as the solution’s osmolarity. This is because some solutes, such as ionic compounds like NaCl, dissociate into separate particles (e.g., Na+ and Cl−) when dissolved in water. For an ideal solution, the solution’s osmolarity equals its molarity times a dissociation factor, the number of ions formed from one particle of the solute when placed in water:

osmolarity(osM)=molarity(M) ×dissociation factor

Every solute has a dissociation factor, or osmotic coefficient (5). For any substance that does not dissociate into ions when dissolved in water, such as glucose or urea, the dissociation factor is 1 and the solution’s osmolarity is the same as its molarity (e.g., 1 M glucose = 1 osM glucose).

The dissociation factor for an ideal solution of NaCl is 2, given that the compound is made up of 2 ions. However, many ionizable substances do not completely dissociate in water. For example, the particles formed when NaCl dissolves in water include Na+, Cl−, and some undissociated NaCl. The calculated dissociation factor for NaCl at 25°C is 1.8 particles per NaCl (13). This means that a 1 mM solution of NaCl would be a 1.8 mosM solution. We prefer to use the term “dissociation factor” when teaching, because students understand the relationship between dissociation of ionic compounds and osmolarity.

You can measure the osmolarity of a solution using a machine called an osmometer. The most common commercial osmometers measure either freezing point depression or vapor pressure of a sample of solution (12, 16). An educational activity for student laboratories with access to an osmometer is to have the students calculate how to make a 150 mM solution of NaCl, make it using good laboratory technique, then measure the osmolarity of the resultant solution. (Answer: Use 8.775 g NaCl/liter of solution. Predicted osmolarity using the dissociation factor of 1.8 would be 270 mosM.)

Osmolarity is an important property of any biological solution. But just as knowing the molarity of a solution does not tell us its osmolarity, knowing the osmolarity of a solution does not tell us its tonicity. This is because tonicity is a term that requires two compartments: the solution being described and a cell. In addition to knowing the concentration (osmolarity) of the solution, you must know the composition of the solution: what the solutes are and whether or not they can enter the cell. Solutes that enter a cell by any means (simple diffusion, protein-mediated transport, and so on) are said to be “penetrating” solutes. Solutes that do not enter the cell are said to be “nonpenetrating” solutes.

The tonicity of a solution predicts the effect of the solution on cell volume at equilibrium and depends on the relative concentrations of nonpenetrating solutes in the cell and the solution. At equilibrium, water movement into the compartment with the higher starting concentration of nonpenetrating solutes will increase that compartment’s volume. Net water movement stops when the concentrations of nonpenetrating solutes in the cell and solution are equal.

The cell’s volume change in response to the solution tells us the tonicity of the solution:

If cell volume at equilibrium has increased, the solution is said to be hypotonic to the cell.

If cell volume at equilibrium has decreased, the solution is said to be hypertonic to the cell.

If cell volume at equilibrium has not changed, the solution is said to be isotonic to the cell.

In other words,

If the cell has a higher initial concentration of nonpenetrating solutes than the solution, at equilibrium water will have moved into the cell. The solution is hypotonic to the cell.

If the solution has a higher initial concentration of nonpenetrating solutes than the cell, at equilibrium water will have moved out of the cell. The solution is hypertonic to the cell.

If the cell and solution have equal concentrations of nonpenetrating solutes, at equilibrium there will be no net movement of water. The solution is isotonic to the cell.

Because clinicians are usually selecting IV fluids based on their effect on intracellular fluid (ICF) and extracellular fluid (ECF) volumes, describing tonicity in terms of cell volume change is the most useful approach for teaching health professions students.

Freely penetrating solutes in a solution can be ignored for the purpose of determining tonicity. They will contribute to the osmolarity of the solution but distribute throughout the cell-solution system as if the cell membrane were not present. They, therefore, do not ultimately contribute to water movement between compartments. This is discussed more below.

There are several important points to keep in mind about tonicity:

Tonicity has no units like mM or mosM. It is a comparative term that predicts changes in cell volume at equilibrium after exposure of the cell to a solution.

By convention, we speak of the tonicity of a solution relative to a given cell. We never speak of the cell being hypo-/hyper-/isotonic to the solution.

With one exception, knowing only the osmolarity of the solution tells you nothing about the tonicity of the solution. The exception is that all hyposmotic solutions are hypotonic.

Tonicity describes what has happened to cell volume at equilibrium but does not tell you what happens to cell volume on the way to equilibrium. We will return to this concept later.

It is important to be clear about the cell that is the frame of reference when discussing tonicity. Solutes that can enter one type of cell may not be able to enter a different type of cell. One example where this is true involves the disaccharide sucrose (table sugar). Mammalian cells do not have transporters for sucrose, so sucrose is a nonpenetrating solute for mammals. But plant cells do have sucrose transporters, making sucrose a penetrating solute for plants.

When teaching health professions students, we assume that the idealized cell is a typical human cell, and we use urea and NaCl as our prototypical solutes. Urea is the classic example of a penetrating solute. It freely crosses most cell membranes, especially the erythrocyte membrane, via diffusion through urea transporters and (to a small degree) through the phospholipid bilayer (10). NaCl is the primary nonpenetrating solute of the ECF. Its ions carry a charge, making it difficult for them to freely pass through the phospholipid bilayer of the cell membrane. As noted later, while some Na+ does leak into cells, Na+-K+-ATPases pump Na+ out at roughly the same rate, making this solute functionally impermeable. We also make the assumption that all solutes in the cell are nonpenetrating and will not leave the cell as long as the cell membrane is intact.

Test your understanding.

Below is a set of basic questions meant to check your understanding of osmolarity and tonicity. For each of the following questions, there are five assumptions we will make to simplify the matter:

1.

Our idealized cell has an internal concentration of 300 mosM. (We will continue this assumption through the rest of the text, unless otherwise noted.)

2.

The solutes present in the cell are assumed to be nonpenetrating and cannot leave the cell.

3.

NaCl is a functionally nonpenetrating solute that behaves as if it cannot cross the cell membrane. (As noted above, in reality, Na+ that leaks into the cell is removed by the Na+-K+-ATPase at a rate that closely matches the rate of leakage in.)

4.

Urea is a freely penetrating solute that easily crosses the cell membrane. Because of this, urea will distribute throughout body compartments until its concentration is equal in all compartments.

5.

Water freely crosses all cell membranes, dividing the intracellular and extracellular compartments.

Try to answer these questions based on your understanding of osmolarity and tonicity.

1.

You place a cell with an internal concentration of 300 mosM in a 400 mosM solution. What happens to the volume and osmolarity of the cell at equilibrium?

Answer: The first part of this question asks about cell volume and, therefore, addresses tonicity of the solution. Tonicity depends on the relative concentrations of nonpenetrating solutes in the solution and cell. There is no way to tell what will happen to the cell’s volume from this question as written, because all you are told is that the solution is 400 mosM. To predict tonicity, you must know if the solution solutes are penetrating or nonpenetrating, and the question did not give you that information.

The second part of the question asks about the osmolarity of the cell at equilibrium. This is a question about concentration, not cell volume. The 400 mosM solution is hyperosmotic to our idealized cell. Thus you can predict that the cell will become more concentrated, either because water moves out into the solution, or because penetrating solute moves into the cell. You do not know the solutes in the solution, and you do not know the relative volumes of the cell and the solution. Without this information, you cannot say what the actual osmolarity will be at equilibrium. You can only say that osmolarity has increased.

2.

Which would have a greater effect on the volume of a cell at equilibrium: solution A composed of 300 mosM NaCl and 200 mosM urea, or solution B composed of 200 mosM NaCl and 300 mosM urea?

Answer: This question asks about cell volume and is, therefore, addressing tonicity, which depends on the concentrations of nonpenetrating solutes. The nonpenetrating solute in these solutions is NaCl, so we compare the NaCl concentrations in the solution to the cell’s osmolarity, assumed to be due only to nonpenetrating solutes. At equilibrium, water will have moved to equalize the nonpenetrating solute concentrations. We can ignore the urea, which is penetrating and will not cause water to shift between compartments once equilibrium is reached.

Solution A has 300 mosM NaCl compared with the cell’s nonpenetrating solutes at 300 mosM. The nonpenetrating solute concentrations are equal in cell and solution, so there is no gradient to cause net water movement. Cell volume at equilibrium does not change, making solution A isotonic to the cell.

Solution B has 200 mosM NaCl compared with the cell’s nonpenetrating solutes at 300 mosM. The nonpenetrating solute concentration is higher in the cell, so water moves into the cell until the nonpenetrating solute concentrations are equal. Water movement into the cell increases the cell volume at equilibrium, making solution B hypotonic to the cell.

So although both solutions have an osmolarity of 500 mosM and are hyperosmotic to the cell, hypotonic solution B exerts a greater effect on the volume of the cell at equilibrium than isotonic solution A.

This example shows how a hyperosmotic solution can be isotonic or hypotonic, depending on the concentration of nonpenetrating solutes in the solution. Saying “a cell will ultimately shrink after being placed in a hyperosmotic solution” is not always correct, although this is frequently heard in the classroom. A cell will shrink when placed in a hypertonic solution. This example is useful for showing students that, while osmolarity and tonicity are related, they are not the same thing.

3.

Given the same solutions as in the previous question (300 mosM NaCl/200 mosM urea or 200 mosM NaCl/300 mosM urea), which solution would have a greater effect on the osmolarity of the cell at equilibrium?

Answer: This question, like the second part of question 1, asks about the osmolarity of the cell at equilibrium. If you place a 300 mosM cell in a 500 mosM solution, you know that the final concentration of the cell will increase. Because solutions A and B have equal osmolarities, you know that they will have the same effect on the final concentration of the cell.

One important take-home message from these three questions is that a student should be able to look at the concentrations of penetrating and nonpenetrating solutes in a solution, figure out its relative osmolarity and tonicity compared with a cell, and recognize how the solution will affect the volume and osmolarity of a cell at equilibrium. For example, given our hypothetical 300 mosM cell, describe the osmolarity and tonicity of solution C with 200 mosM NaCl and 500 mosM urea. The answer is that solution C at 700 mosM is hyperosmotic but hypotonic, because the solution has 200 mosM nonpenetrating solutes vs. 300 mosM in the cell.

How can a hypotonic solution like C increase the osmolarity of the cell? If the cell takes up water and swells, you might expect that the cell’s internal concentration would decrease (concentration = solute/volume). However, this explanation fails to account for the penetrating solute urea, which is also able to get into the cell. Urea, spreading by diffusion throughout the solution and cell, will ultimately add to the overall solute in the cell. Thus the cell in solution C will swell, but also become more concentrated due to movement of urea into the cell down its concentration gradient.

An important concept to emphasize with students is that the cell’s final concentration will follow that of the osmolarity of the solution, regardless of the solution’s effect on cell volume. If you add a cell to a hyperosmotic solution, its osmolarity will increase, regardless of whether the cell swells, shrinks, or stays the same volume. The cell’s concentration will always follow that of the solution. Placing a cell in a hyposmotic solution will cause the cell’s concentration to decrease. Placing a cell in an isosmotic solution will not change the cell’s concentration.

Student points of confusion.

Considering that understanding osmolarity and tonicity seem to require more effort on the part of the student, we thought it important to uncover what aspects students perceive as most complicated or confusing. We asked the students who had completed a course in Animal Physiology at the University of Belgrade (Serbia) during the last 4 academic years (2014–2017) to submit a short essay explaining what they found most challenging about osmolarity and tonicity and what in-class activities proved most useful to them. After analyzing essays submitted by over 80 students, we found that confusion stems from one or more of the three main issues:

1.

Distinguishing tonicity from osmolarity

2.

Determining tonicity based on the solution’s composition

3.

Distinguishing units expressing amount vs. concentration of solutes (e.g., osmol vs. osmol/l or osM)

Given these very common points of confusion, we designed a set of teaching activities that enable students to fully understand and apply the concepts of osmolarity and tonicity.

A starting point in the classroom.

Beginning with clear and concise definitions of the terms establishes a distinction between osmolarity and tonicity. Whereas most students are quick to understand that osmolarity is a concentration, most of them struggle with differentiating osmolarity from tonicity. Osmolarity and tonicity are certainly related, but they are not the same thing as noted above, and this should be stressed in the classroom. According to our students’ reports, they found it helpful to remember that tonicity is defined by the effect a solution has on cell volume at equilibrium, and that tonicity is determined by comparing the concentrations of nonpenetrating solutes in the solution and the cell. In these essays, students also mentioned they found it especially difficult to remember that tonicity describes how the solution affects cell volume at equilibrium. Furthermore, the fact that a hyperosmotic solution can be hypo-, iso-, or hypertonic proved perplexing to many of them.

One of the tasks that seems very effective at clarifying these two muddy points is asking students to graph the time-dependent effects of three hyperosmotic solutions, all with the same concentration of 400 mosM, on the volume of an idealized cell:

Solution A contains 200 mosM NaCl and 200 mosM urea (400 mosM solute)

Solution B contains 300 mosM NaCl and 100 mosM urea (400 mosM solute)

Solution C contains 400 mosM NaCl (400 mosM solute)

We ask students to draw the cell’s change in volume from the moment of placing the cell in the solution until osmotic equilibrium is reached. The complexity of this task is reflected in the fact that, although the three solutions have the same osmolarity (400 mosM), they exhibit different tonicities due to varying concentrations of nonpenetrating and penetrating solutes (Fig. 1).

When a cell and a solution have the same concentration of solutes that are considered?

Fig. 1.The time-dependent effects of three hyperosmotic solutions of different tonicities. Solution A contains 200 mosM NaCl and 200 mosM urea (400 mosM solute); solution B has 300 mosM NaCl and 100 mosM urea (again, 400 mosM solute); and solution C is 400 mosM NaCl (once again, 400 mosM solute). Considering that water molecules move faster across the membrane than solutes, all of the solutions will initially cause a cell to shrink. However, once the osmotic equilibrium is reached, the cell’s volume will increase in hypotonic solution A, stay the same in isotonic solution B, and decrease in hypertonic solution C.

To simplify the matter, students are instructed to assume that the cell’s osmolarity is 300 mosM and that the cytosol only contains nonpenetrating solutes. To successfully complete this task, students have to demonstrate that they understand the following points:

1.

The solution’s tonicity depends on the solution’s concentration of nonpenetrating solutes (e.g., NaCl) relative to that in the cell, not on the total osmolarity of the solution.

2.

Tonicity provides the information on how the solution affects the cell’s volume once osmotic equilibrium is reached; that is, when the osmolarities of the solution and the cytosol become equal.

3.

Water molecules move faster through the cell membrane than particles of solute do. Therefore, a cell placed in a hyperosmotic solution will always shrink initially, regardless of the solution’s tonicity.

This type of problem-solving can be equally as effective when presented in reverse. For example, students can be asked to predict a solution’s osmolarity and tonicity based on the effect it has on the volume of a cell, as depicted in a graph (Fig. 2). To successfully complete these assignments, students must simultaneously consider both the osmolarity and tonicity of a solution.

When a cell and a solution have the same concentration of solutes that are considered?

Fig. 2.The task that requires students to predict a solution’s osmolarity and tonicity based on how the solution affects the cell’s volume before reaching osmotic equilibrium. Considering that the cell’s volume did not change after being placed in solution A, this solution is isosmotic and isotonic. In the second scenario, the cell’s volume increased once the osmotic equilibrium was reached, indicating the solution is hypotonic. Since the cell’s volume was continuously increasing, solution B could be either hypo- or isosmotic.


Osmolarity and tonicity in context.

To ensure that students completely understand the significance of osmolarity and tonicity in the light of systemic homeostasis, a lecturer might design tasks that put these processes in different physiological contexts while giving them relevance. For example, it is essential that pre-health professions students understand that the appropriate choice of IV or oral fluid therapy depends on the particular clinical scenario they will encounter.

An in-class activity that lends itself well to showing how different solutions influence osmotic concentration and compartment volumes is to have students calculate numerical values after adding a solution to the body. To successfully complete such an assignment, we ask our students to do the following:

1.

Compare the solution’s osmolarity to the osmolarity of the body or cell.

2.

Determine the solution’s tonicity.

3.

Predict how adding this solution will affect the volume and osmolarity of body fluids.

a.

Body concentration: increase, decrease, or no change?

b.

Total body volume: increase, decrease, or no change?

c.

ECF volume and osmolarity: increase, decrease, or no change?

d.

ICF volume and osmolarity: increase, decrease, or no change?

4.

Perform calculations that mathematically demonstrate how the solution affects the body volumes and osmolarity (described more below).

5.

Check whether their theoretical predictions (step 3) are in accordance with their calculations (step 4).

Let us take a solution and add it to an idealized body with simple volumes and concentration to work through these steps. For our example, students are asked to describe the effects of adding 1 liter of a 100 mosM NaCl and 300 mosM urea solution to a person whose normal total body volume is 30 liters, with two-thirds of body fluid volume (20 liters) inside the cells. We work through each of the above steps in sequence.

1.

The solution is hyperosmotic to the body because the solution has a concentration of 400 mosM, and our idealized body has a concentration of 300 mosM.

2.

The solution is hypotonic to the body because the solution has a concentration of nonpenetrating solutes of 100 mosM, and the concentration of nonpenetrating solutes in the body is 300 mosM.

3.

The concentration of all body compartments will increase, because the solution is hyperosmotic. This is important for the student to understand. The body compartments are in osmotic equilibrium. Thus increasing the concentration of the body with a hyperosmotic solution will increase the concentration of all compartments.

4.

Total body volume will increase, because we are adding volume to the body.

5.

ECF volume will increase because we are adding volume to the body and because the NaCl will remain in the ECF, keeping some water with it. ECF osmolarity goes up because we added a hyperosmotic solution to the body.

6.

ICF volume will increase because the solution is hypotonic. ICF osmolarity also goes up because we added a hyperosmotic solution to the body and it contained a penetrating solute.

It might initially be difficult for students to think their way through steps 1–3 above, as some of the predictions seem at first to make no sense. How can the ICF concentration increase if the volume of the ICF is increasing? Mathematically describing how the body compartments are affected (step 4) can demonstrate these processes in a concrete manner. We call the series of tables used for the calculations in step 4 “box problems,” and they are shown in Tables 2–7.

The first step is filling out a chart that contains the initial values of an idealized and simplified body (Table 2). Columns include total body fluid (TBF) as well as the ECF and ICF compartments. Each column contains rows for solute amount, compartment volume, and compartment osmolarity. Unknown values can be calculated by using solute amount/volume = osmolarity.

Table 2. Starting fluid volumes table

Starting TableTotal Body Fluid (TBF)Extracellular Fluid (ECF)Intracellular Fluid (ICF)
Amount of solutes (mosmol) 30 liters × 300 mosmol/l = 9,000 mosmol 10 liters × 300 mosmol/l = 3,000 mosmol 20 liters × 300 mosmol/l = 6,000 mosmol
Volume (liters) 30 liters 1/3 × 30 liters = 10 liters 2/3 × 30 liters = 20 liters
Osmolarity (mosM = mosmol/l) 300 mosM 300 mosM 300 mosM

The starting table uses volume and osmolarity values as they are before the solution is added. Our example assumes that normal total body volume is 30 liters, and that two-thirds of the volume (20 liters) are inside the cells. The initial body osmolarity is 300 mosM. For students, using numbers like 300 mosM and volumes of 10, 20, and 30 liters simplifies the work.

As students fill in the chart, they must differentiate between the amount of solute and the concentration of solute as well as the units for amount (mosmol) and concentration (mosM). Students are instructed to always fill in the TBF column first.

The problem asks the effect of administering 1 liter of a solution composed of 100 mosM NaCl and 300 mosM urea. In the second step (Table 3), students add the volume of the solution and the amount (mosmoles) of nonpenetrating solute in the solution to the starting conditions. (The milliosmoles of urea are held in reserve and added later, once water has redistributed.) This step requires students to calculate the amount of solute to add, based on the solution’s concentration and volume. In this example, 1 liter of a solution with 100 mosM NaCl was added. Students calculate that 1 liter of 100 mosmol/l NaCl = 100 mosmol NaCl.

Table 3. Adding volume and nonpenetrating solutes to the total body

Add Volume and Nonpenetrating SolutesTotal Body Fluid (TBF)Extracellular Fluid (ECF)Intracellular Fluid (ICF)
Solutes (mosmol) 9,000 + 100 = 9,100
Volume (liters) 30 + 1 = 31
Osmolarity (mosM)

Students use the changed volume and solute values to calculate the new TBF osmolarity (Table 4). They apply their understanding of osmotic equilibrium in the body by recognizing that the calculated total body osmolarity will be the same as the osmolarities of both the ECF and ICF compartments.

Table 4. Calculating new total body osmolarity after adding volume and nonpenetrating solutes

Calculate New OsmolarityTotal Body Fluid (TBF)Extracellular Fluid (ECF)Intracellular Fluid (ICF)
Solutes (mosmol) 9,100
Volume (liters) 31
Osmolarity (mosM) 9,100/31 = 293.5 293.5 293.5

Students then need to figure out that the added 100 mosmol of NaCl stay entirely in the ECF because NaCl is nonpenetrating and cannot enter cells (Table 5). The amount of solute in the ICF remains unchanged at this point in the calculations.

Table 5. Adding nonpenetrating solute

Add NaCl to ECFTotal Body Fluid (TBF)Extracellular Fluid (ECF)Intracellular Fluid (ICF)
Solutes (mosmol) 9,100 3,000 + 100 = 3,100 6,000
Volume (liters) 31
Osmolarity (mosM) 293.5 293.5 293.5

The revised values for the ECF and ICF volumes can now be calculated by dividing the amount of solute each compartment contains by its osmolarity (Table 6).

Table 6. Calculating new ECF and ICF volumes

Calculate New ECF and ICF VolumesTotal Body Fluid (TBF)Extracellular Fluid (ECF)Intracellular Fluid (ICF)
Solutes (mosmol) 9,100 3,100 6,000
Volume (liters) 31 3,100/293.5 = 10.6 liters 6,000/293.5 = 20.4 liters
Osmolarity (mosM) 293.5 293.5 293.5

The last part of the problem is to add the amount of the penetrating solute contained in the solution to the body. In this case, the 1 liter of the solution had 300 mosM urea, in addition to the 100 mosM NaCl. This means that 300 mosmol of urea (300 mosmol/l × 1 liter) were added to the body. Because urea is a freely penetrating solute, it will not cause water to shift between the ECF and ICF compartments. The urea contributes to the osmolarity of the solution but not its tonicity.

Students first add the urea to the total body solute cell and calculate the new total body osmolarity (Table 7). The total body osmolarity can once again be copied to the ECF and ICF columns because the two compartments are always in a state of osmotic equilibrium. Considering that the entire volume of the solution was already added in the preceding step, the values for the TBF, ECF, and ICF volume have not changed and may be copied from the previous chart.

Table 7. Adding penetrating solutes

Add Urea to Get Final OsmolarityTotal Body Fluid (TBF)Extracellular Fluid (ECF)Intracellular Fluid (ICF)
Solutes (mosmol) 9,100 + 300 = 9,400
Volume (liters) 31 10.6 20.4
Osmolarity (mosM) 9,400/31 = 303.2 303.2 303.2

Finally, based on the ECF and ICF volumes and new osmolarity, the amount of solutes contained in each of the two subcompartments can be calculated. If concentration = amount/volume, then amount = volume × concentration. Depending on the student population, this final calculation can be skipped because clinically we are usually only interested in the compartment volumes and their osmolarity. Once the tables are complete, students check whether their theoretical predictions (step 3) are in line with the calculations (step 4). The predictions were that osmolarity of all compartments would increase and that volume of all compartments would increase. By comparing Table 2 with Table 7, a student should observe the following:

1.

Body osmolarity increased while the ICF volume increased. This is in accordance with the initial predictions: the solution added to the body was hyperosmotic and hypotonic. Because the solution was hypotonic, the ICF volume increased. (In our tables, we see ICF volume increase from 20 to 20.4 liters). It might seem that this would dilute the ICF, causing the ICF concentration to decrease; however, the solution contained enough urea to make it hyperosmotic. Urea, a penetrating solute, passes through cell membranes and adds to the solute in the cells, increasing the overall concentration of the ICF from 300 to 303.2 mosM.

2.

The mosmoles of penetrating solute can be distributed between ECF and ICF in proportion to their respective compartment volumes after the volumes changed from the addition of nonpenetrating solute. This puts an equal concentration of urea into each compartment. Tell students to avoid thinking “distributes equally,” because that makes them inclined to divide the urea amount into two equal batches rather than distributing the urea proportionately according to compartment volume.

3.

A change in total body osmolarity is reflected in a similar change in ECF and ICF osmolarity. ECF and ICF will always be in the state of osmotic equilibrium (in other words, the osmolarity of all compartments are equal).

Insisting that students consistently copy the symbols for units (mosmol, l, mosM) in each of the tables helps them memorize the units and prevents them from mistaking the amount for the concentrations of solute.

The complexity of these box problem assignments can be taken to the next level by creating a variety of real-life scenarios in which a person is subjected, for instance, to sweat loss from increased physical activity followed by the inappropriate choice of a rehydration drink. In addition to following the usual set of instructions for filling in the charts, students demonstrate the ability to calculate the osmotic concentration of a solution and deduce how the wrong choice of sport drink can lead to a disturbance in osmotic homeostasis (Fig. 3).

When a cell and a solution have the same concentration of solutes that are considered?

Fig. 3.This is the example of a box problem assignment that helps students understand the importance of osmolarity and tonicity within the context of systemic homeostasis. This particular task illustrates how the inappropriate choice of a rehydration drink following sweat loss from increased physical activity may disturb osmotic homeostasis. Students follow the steps laid out in the different parts of Table 2. For simplicity, sweat was assumed to include only nonpenetrating solute lost from the ECF. In reality, sweat is made mostly of sodium with smaller components of chloride, potassium, lactate, and urea. This question could be made more complex by including these substances in the question stem.


Glucose as a solute.

The examples above used NaCl as the prototype nonpenetrating solute and urea as the freely penetrating solute. But what about glucose, clinically known as dextrose? Many IV solutions contain glucose, either as the sole solute (for example, 5% dextrose in water, or D5W) or in combination with NaCl (for example, 5% dextrose in half-normal saline). As it turns out, glucose as a solute must be treated as a special case for the purpose of working problems.

Students will state that glucose is a penetrating solute because they know that it enters cells. However, it is not freely penetrating because, in most cells, glucose that enters is immediately phosphorylated by hexokinase to glucose-6-phosphate (G-6-P). Phosphorylated compounds are not able to leave the cell, so the glucose that enters adds to the cell’s pool of nonpenetrating solutes when it becomes G-6-P.

In a normal person given a load of glucose by mouth or IV, over a period of several hours 100% of the administered load will enter the cells. In addition, through normal aerobic metabolism, much of the administered glucose will be processed from G-6-P to CO2 and water. Thus the net result of giving a pure glucose IV over time is the same as giving plain water, making glucose IV solutions hypotonic, even if they are isosmotic. Failure to recognize hypotonicity and the metabolism of glucose to water can result in iatrogenic hyponatremia when D5W is used as a routine maintenance solution (8).

When administering a solution containing glucose in the box problems described above, we use a “freeze frame” approach. Students are asked to calculate changes in ECF and ICF volumes and osmolarities at the time when X% of glucose has entered the cell. They must recognize that 100 − X% of the glucose has remained in the ECF, and they must distribute the appropriate amounts of glucose into each compartment to do their calculations.

One variant of glucose as a solute involves the distribution of glucose in a patient with type 1 diabetes mellitus, where absence of insulin prevents glucose uptake into most cells. In these situations, glucose remains in the blood and effectively becomes a nonpenetrating solute. A similar situation occurs in hyperglycemic hyperosmolar states (11). These two scenarios help students link changes in tonicity to physiologically significant disruptions of normal function.

Extending the concept.

The box problems described above are a simple example of the principle of mass balance: take what was in the body at the start, then add and subtract solutes and volume to find what is in the body now. The box problem approach can be extended to include calculating the concentration of specific components of the different compartments. For example, a patient comes in dehydrated with an elevated plasma potassium concentration. If you replace his volume loss with an IV of isotonic NaCl, what will happen to his plasma K+ concentration? Or a patient comes in with acute hyponatremia from water intoxication. How will administering a small volume of hypertonic saline alter both her plasma Na+ concentration and the distribution of water between the ECF and ICF?

Answering these questions requires using the box problem strategy with the addition of a set of boxes that are specific to the ion (Table 8). In the ion-specific tables, students calculate the amount of ion in the ECF from the plasma ion concentration and the ECF volume. This is a good place to reinforce that we can substitute plasma ion concentration as a surrogate for ECF ion concentration (unless significant amounts of ion are bound to plasma proteins, as is the case for Ca2+).

Table 8. Calculating plasma ion changes

Initial ECF ConditionsECF Conditions Following 1 liter of 100 mosM NaCl and 300 mosM Urea
Amount of K+ (meq) 4.5 meq/l × 10 liters = 45 45
Volume (liters) 10 (from Table 2) 10.6 (from Table 7)
Plasma [K+] (meq/l) 4.5 4.2

The K+ example shown in Table 8 is based on the subject being given 1 liter of 100 mosM NaCl and 300 mosM urea from the set of Tables 2–7 calculations. Students should recognize that the amount of K+ in the ECF does not change post-IV. But there is a new ECF volume after giving the IV, so the plasma K+ concentration will change.

In the example of hyponatremia treated with a bolus of hypertonic saline, after students work out the effects of the saline IV on the body, they use a second set of Na+ boxes to add Na+ from the IV to the ECF Na+ amount. The revised Na+ amount divided by the post-IV ECF volume yields the new ECF Na+ concentration.

When a cell is in a solution where the concentration of solute is the same?

If the solute concentration outside the cell is the same as inside the cell, and the solutes cannot cross the membrane, then that solution is isotonic to the cell.

What is the solution called when its concentration is the same as concentration inside the cell a hypertonic solution C hypotonic solution B isotonic solution d dilute solution?

Or, isotonic solution is the solution that has the same concentration of solvent in both surrounding solution and solution inside the cell. There is no movement of solvent as there is an equilibrium between the surrounding solution and inside of the cell.

What is the solution called when its concentration is the same as concentration inside the cell 1 point?

A solution is isotonic when its effective osmole concentration is the same as that of another solution. In biology, the solutions on either side of a cell membrane are isotonic if the concentration of solutes outside the cell is equal to the concentration of solutes inside the cell.

When a cell is in a solution that has the same concentration of water and solute the cell is said to be in an?

In this type of solution, where you have the same concentration of solute in the solution, as you do inside the cell, we would call this an isotonic. This is an isotonic solution. Isotonic solution. The prefix, iso, refers to things that are the same.