The method to calculate osmolarity is a basic concept of Biology that students should be aware of. The maintenance of osmolarity is essential for the body because it ensures the survival of cells. If cells are placed in a solution of an osmolarity much higher than the cell’s cytoplasm, the cell would lose water and die. This is why fluid homeostasis in the body functions to balance the intake and loss of water from the cells.
This article will explain how to calculate osmolarity using some example questions that will improve your understanding of the topic.
Osmolarity formula
Osmolarity is the total concentration of osmotically active solutes per unit volume of solution. Osmotically active solute refers to the dissociated ions or solutes in a solution that affect the movement of water across a selectively permeable membrane (the cell membrane) via osmosis. A formula to explain osmolarity is:
Osmolarity = Moles of solutes1 L of solution ×no. of ions
Osmolarity equation
A simple equation for calculating the osmolarity of a solution is derived from the molarity of the solution such that:
Osmolarity = Molarity × n
- n is the number of particles or ions present in the solution. For instance a solution of NaCl will dissociate into two ions that are Na+ and Cl-.
How to measure Osmolarity
Steps required to measure the Osmolarity of a solution:
- Calculate the moles of a solute dissolved in a solution.
- Now calculate the Molarity of the solution.
- Multiple the Molarity of the solution by the no. of ionizable particles in the solution.
Now let’s have a look at some background information related to the concept of osmolarity:
- Osmosis
Osmosis refers to the net movement of a solvent such as water, across a selectively permeable membrane from a region of lower solute concentration to a region of higher solute concentration.
Osmosis occurs as a result of the difference in concentration of solutes dissolved in water. When a substance dissolves in water it forms intermolecular bonds with the polar molecules of water. The formation of these bonds restricts the movement of water. This explains why an area having a higher solute concentration would have lesser free water molecules while an area having a lower solute concentration would have more water molecules free to move. This allows the movement of water molecules from areas of lower solute concentration to areas of higher solute concentration. In other words, water molecules tend to move from a region of low osmolarity towards regions of higher osmolarity.
- Osmosis in living cells
Osmosis can occur in all cells due to the very small size of water molecules, thus, water can pass through the phospholipid cell membranes even though it is hydrophobic. Some cells such as kidney cells carry aquaporins which are water channels in the cell membrane that greatly increase its permeability to water.
- Types of solutions
The terms used to describe the relative osmolarities between different solutions are hyperosmotic, hypoosmotic, and isoosmotic. For example, consider two solutions having different osmolarities. If a given solution has higher osmolarity compared to another solution, It is called hyperosmotic. When a solution has lower osmolarity than another solution, it is called hypoosmotic. Lastly, in case the two solutions have the same osmolarity, they are called isoosmotic.
A diagram depicting the effect of hypertonic, isotonic, and hypotonic solutions on animal and plant cells.
When placed in the hypertonic solution, both plant cell and animal cell lost water to the surrounding resulting in shriveled Red blood cells (RBCs) and plamolysed plant cells. Cells appear normal when placed in the isotonic solution. In the hypotonic solution, water from the solution moved into the cells. As a result, RBCs burst while plant cells become turgid due to extra protection provided by the cell wall.
Example 1: Steps to measure the osmolarity of a plant tissue
The best way to understand the steps of measuring osmolarity is by solving an example. For instance, let’s look at the protocol for calculating the osmolarity of a potato cell’s cytoplasm:
- Prepare 6 beakers carrying 40 ml of sucrose solution at different concentrations.
- Cut out potato cores, each having the same dimensions.
- Now weigh each potato core on a weighing balance and note down the mass in a table next to the corresponding concentration of sucrose.
- Place the potato cores in their respective beakers and let them soak for 24 hours.
- During this time, potato cores placed in hypertonic solutions will lose water, and hence the weight of potatoes will decrease. While potato cores placed in hypotonic solutions will gain water and the weight of potato will increase. This is based on the fact that water moves from an area having lower osmolarity(lower solute concentration) to an area having higher osmolarity (high solute concentration).
- The potato core placed in an isotonic solution will have no change in mass. This is because the concentration or molarity of the sucrose solution is the same as the concentration of solute or molarity inside the potato cells. This will give us the molarity of cytoplasm inside the potato cells, allowing us to calculate osmolarity from it.
- 24 hrs later measure the masses again and record them.
- Data analysis
| Sucrose concentration (M) | Mass of potato core in g (before) | Mass of potato core in g (after) | Difference in mass (g) | Percentage change in mass (%) | Average percentage change in mass (%) |
| 0.0 | 2.2 | 3.0 | 0.8 | 36 | |
| 0.2 | 2.3 | 2.8 | 0.5 | 22 | |
| 0.4 | 2.4 | 2.76 | 0.36 | ||
| 0.8 | 2.4 | 2.66 | 0.26 |
This table depicts the kind of columns you will need to create in excel for plotting a graph.
- Calculate the percentage change in mass of each potato core.
- Percentage change in mass = new mass-old massold mass × 100
- The experiment should be repeated three times for each sucrose concentration to get accurate results.
- The average percentage change in mass will be plotted against the sucrose concentration.
- From the linear trend line graph, you can obtain the molarity of sucrose concentration at which there is no percent mass change in the potato core. This will be the sucrose concentration at which the graph is cutting the x-axis such that the percentage mass change is zero.
- This osmolarity of the potato cytoplasm in this example will be the same as the molarity of the sucrose concentration at which the mass change percentage is zero. This is because there is only one dissolved particle in a sucrose solution i.e. sucrose. Thus, there is no need for further calculations.
Practice Question
One way to investigate the concentration of solutes within plant tissues is by placing the plant tissue in varying concentrations of NaCl dissolved in water. Potato samples were immersed in solutions of different concentrations of NaCl. After a while, the percentage changes in the mass of potatoes were recorded to create a graph of change in mass percentage against varying NaCl solution concentrations. The following graph was plotted:
Graph extracted from https://www.ibdocuments.xyz/IB%20QUESTIONBANKS/4.%20Fourth%20Edition/questionbank.ibo.org/en/teachers/00000/questionbanks/43-dp-biology/questions/126107.html
(i) What is the estimated osmolarity of the potato tissue?
Ans: Molarity of NaCl solution at which there is zero percent change in mass = 0.28 moles dm-3. No change in mass of the potato indicates that the solution was isotonic which means that the concentration inside the cells was equal to the concentration of the NaCl solution. Thus, concentration or molarity inside the plant tissue was also = 0.28 moles dm-3. Since the solution is carrying 2 ions i.e. Na+ and Cl-, we will multiply the value of molarity by 2.
Osmolarity = 0.28 × 2 = 0.56 osmoles dm-3.
(ii) Identify which part of the graph represents samples measured in a hypotonic solution.
The graph above the x-axis, showing a positive percentage change in mass, represents samples measured in a hypotonic solution.