Question

Plants that thrive in salt water must have internal solutions (inside the plant cells) that are isotonic with (have the same osmotic pressure as) the surrounding solution. A leaf of a saltwater plant is able to thrive in an aqueous salt solution (at ${25}^{\circ }\mathrm{C})$ that has a freezing point equal to $-{0.621}^{\circ }\mathrm{C}.$ You would like to use this information to calculate the osmotic pressure of the solution in the cell. a. In order to use the freezing-point depression to calculate osmotic pressure, what assumption must you make (in addition to ideal behavior of the solutions, which we will assume)? b. Under what conditions is the assumption (in part a) reasonable? c. Solve for the osmotic pressure (at ${25}^{\circ }\mathrm{C})$ of the solution in the plant cell. d. The plant leaf is placed in an aqueous salt solution (at ${25}^{\circ }\mathrm{C}$ ) that has a boiling point of ${102.0}^{\circ }\mathrm{C}.$ What will happen to the plant cells in the leaf?

Short answer

In addition to ideal behavior of the solutions, one must assume that the solute does not dissociate into ions in the solution. This assumption is reasonable under conditions where the solute does not ionize in the solution, or if ionization occurs, the extent of ionization is negligible. The osmotic pressure of the solution in the plant cell is approximately 8.169 atm. If the plant leaf is placed in a salt solution with a boiling point of 102.0°C, water from inside the plant cells will move out through osmosis, causing the plant cells to lose water and shrink, potentially leading to cell damage.

Step-by-step solution

a. Assumption for using freezing-point depression to calculate osmotic pressure

In addition to ideal behavior of the solutions, one must assume that the solute does not dissociate into ions in the solution. This is because the osmotic pressure is directly dependent on the number of solute particles in the solution, and dissociation into ions would change the number of particles.

b. Reasonable conditions for the assumption in part a

The assumption made in part a is reasonable under conditions where the solute does not ionize in the solution, or if ionization occurs, the extent of ionization is negligible. This can occur in solutions containing molecular solutes (i.e., non-ionic solutes) or solutes that do not ionize significantly in the particular solvent being used.

c. Calculating the osmotic pressure of the solution in the plant cell

First, we need to find the molality (m) of the solution using the formula for freezing-point depression: $$\mathrm{\Delta }{T}_f=K_f\cdot m$$ $\mathrm{\Delta }{T}_f$ is the freezing-point depression, which is the normal freezing point of the solvent minus the freezing point of the solution, and $K_f$ is the cryoscopic constant of the solvent (for water, it is 1.86 °C kg/mol). We have: $$\mathrm{\Delta }{T}_f=0-(-0.621)=0.621\text{degree}C$$ So to calculate the molality (m) of the solution, we rearrange the equation and solve for m: $$m=\frac{\mathrm{\Delta }{T}_f}{K_f}=\frac{0.621}{1.86}=0.334\frac{mol}{kg}$$ Now we will use the molality of the solution to calculate the osmotic pressure (Π) using the formula: $$\mathrm{\Pi }=MRT$$ M is the molarity (in mol/L), R is the ideal gas constant (0.0821 L atm/K mol), and T is the absolute temperature (in Kelvin). First, we need to convert the molality (m) to molarity (M) by multiplying it with the density of water (1 g/mL) since the density of a dilute solution is approximately equal to the density of the solvent. $$M=0.334\frac{mol}{kg}\times \frac{1kg}{1000g}\times \frac{1000ml}{1L}$$ $$M=0.334mol/L$$ Next, we convert the temperature from Celsius to Kelvin: $$T=25+273.15=298.15\text{ K}$$ Now we can calculate the osmotic pressure: $$\mathrm{\Pi }=MRT=(0.334\text{ mol/L})\cdot (0.0821\frac{L\cdot atm}{K\cdot mol})\cdot (298.15\text{ K})$$ $$\mathrm{\Pi }=8.169\text{ atm}$$ The osmotic pressure of the solution in the plant cell is approximately 8.169 atm.

d. Effect of the plant leaf being placed in a salt solution with a boiling point of 102.0°C

If the boiling point of the solution is higher than the natural boiling point of pure water, it means that the solution outside the plant cells has a higher concentration of solute particles than the one inside the cells. As a result, water from inside the plant cells will move out through osmosis to try and equilibrate the concentrations between the inside and outside of the cells. Consequently, the plant cells will lose water, causing them to shrink and potentially leading to cell damage.

Key concepts

Freezing-Point Depression and Osmotic Pressure

Freezing-point depression is a colligative property that occurs when the freezing point of a liquid is lowered by the addition of a solute. This phenomenon is often used to understand and calculate osmotic pressure, especially in biological systems like plant cells.
When a solute is added to a solvent such as water, the freezing point is lowered. This happens because solute particles disrupt the formation of the solid phase by interfering with the solvent's ability to form a crystalline structure.
To calculate osmotic pressure using freezing-point depression, you assume that the solute does not dissociate or ionize in the solution. The change in freezing point ($\mathrm{\Delta }{T}_f$) is proportional to the molality ($m$) of the solution, which helps in determining the concentration of solute particles.
In the case of plant cells, this concept helps in ensuring that the cells can maintain equal osmotic pressure with the surrounding saltwater, which stops the water from moving unfavorably across cell membranes.

Osmosis and Plant Cells

Understanding osmotic pressure is crucial for comprehending how plant cells interact with their environments, especially in saltwater conditions.
Plant cells use semipermeable membranes that allow water to flow in and out, balancing internal and external osmotic pressures. If the external environment has a higher solute concentration, water tends to leave the cell, causing it to shrink; if lower, water enters, leading to swelling.
For saltwater plants, the capability to maintain isotonic conditions is critical. They adapt by having internal solute concentrations that match their external surroundings, preventing undesirable water movement. This equilibrium ensures healthy cell function even when external conditions change.
Not achieving this balance can result in water loss or intake that is harmful to plant cells. Understanding how osmotic pressure differs with various solutes helps in predicting plant behavior in different water types.

Application of the Ideal Gas Law

The ideal gas law makes it easier to relate different properties of solutions engaged in osmotic processes. Although it primarily applies to gases, its principles aid in explaining osmotic pressure in solutions.
The formula $$\mathrm{\Pi }=MRT$$ uses elements from the ideal gas law to predict osmotic pressure.

• $\mathrm{\Pi }$ stands for osmotic pressure.
• $M$ represents molarity, or the number of moles of solute per liter of solution.
• $R$ is the ideal gas constant ($0.0821L\cdot atm/(K\cdot mol)$).
• $T$ is temperature in Kelvin.

By understanding how osmotic pressure is calculated using this formula, we appreciate why conditions of isotonicity are essential for plant life, particularly when exposed to saline environments.
Being familiar with these calculations supports a broader understanding of cellular behaviors, such as osmotic balance, an essential aspect for plant survival in changing environments.