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Chapter 10: Problem 3

Assume that you place a freshwater plant into a saltwater solution and examine it under a microscope. What happens to the plant cells? What if you placed a saltwater plant in pure water? Explain. Draw pictures to illustrate your explanations.

Short Answer

Expert verified
When a freshwater plant is placed in a saltwater solution, the plant cells lose water through osmosis, causing them to become flaccid and shrink, a process called plasmolysis. This leads to the wilting and death of the plant. On the other hand, when a saltwater plant is placed in pure water, the plant cells gain water through osmosis and become turgid, expanding and pressing the cell membrane against the cell wall. This may ultimately lead to cell bursting and damage to the plant.

Step by step solution

01

Understanding osmosis

First, we need to understand the concept of osmosis. Osmosis is the movement of water molecules across a selectively permeable membrane from a region of lower solute concentration (higher water concentration) to a region of higher solute concentration (lower water concentration). This process continues until the solute concentrations on both sides of the membrane become equal. In the context of plant cells, the selectively permeable membrane is the cell membrane.
02

Freshwater plant in saltwater solution

When you place a freshwater plant into a saltwater solution, the plant cell will lose water through osmosis. This is because the saltwater solution outside the cell has a higher solute concentration (less water concentration) than the inside of the cell, causing the water in the cell to move outside through the selectively permeable membrane. As the cells lose water, they will become flaccid and shrink, a process called plasmolysis. This will ultimately lead to the wilting and death of the plant.
03

Illustration for freshwater plant in saltwater solution

To illustrate this process: 1) Draw a plant cell in its normal state with a large central vacuole filled with water. 2) Draw the same plant cell after being placed in the saltwater solution, with the central vacuole and cytoplasm shrunk due to water loss, and the cell membrane pulling away from the cell wall.
04

Saltwater plant in pure water

When you place a saltwater plant in pure water, the plant cell will gain water through osmosis. This is because the pure water outside the cell has a lower solute concentration (higher water concentration) than the inside of the cell, causing the water in the surroundings to move inside the cell through the selectively permeable membrane. The cell will continue to absorb water until it becomes turgid, expanding and pressing the cell membrane against the cell wall, which may ultimately lead to the bursting of the cells, damaging the plant.
05

Illustration for saltwater plant in pure water

To illustrate this process: 1) Draw a plant cell adapted to saltwater conditions with a smaller central vacuole and higher solute concentration. 2) Draw the same plant cell after being placed in pure water, with the central vacuole and cytoplasm expanded due to water gain. The cell membrane pressing against the cell wall, giving a stressed appearance. By understanding the cellular processes that occur in each case, including osmosis and the resulting plasmolysis or turgidity, we can explain and visualize the effects of placing a freshwater plant in a saltwater solution and a saltwater plant in pure water on the plant cells.

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Most popular questions from this chapter

The normal boiling point of diethyl ether is \(34.5^{\circ} \mathrm{C} .\) A solution containing a nonvolatile solute dissolved in diethyl ether has a vapor pressure of 698 torr at \(34.5^{\circ} \mathrm{C}\). What is the mole fraction of diethyl ether in this solution?

A solution is prepared by mixing 0.0300 mole of \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and 0.0500 mole of \(\mathrm{CH}_{2} \mathrm{Br}_{2}\) at \(25^{\circ} \mathrm{C}\). Assuming the solution is ideal, calculate the composition of the vapor (in terms of molefractions) at \(25^{\circ} \mathrm{C} .\) At \(25^{\circ} \mathrm{C},\) the vapor pressures of pure \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and pure \(\mathrm{CH}_{2} \mathrm{Br}_{2}\) are 133 and 11.4 torr, respectively.

For an acid or a base, when is the normality of a solution equal to the molarity of the solution and when are the two concentration units different?

In lab you need to prepare at least \(100 \mathrm{mL}\) of each of the following solutions. Explain how you would proceed using the given information. a. \(2.0 \mathrm{m} \mathrm{KCl}\) in water (density of \(\mathrm{H}_{2} \mathrm{O}=1.00 \mathrm{g} / \mathrm{cm}^{3}\) ) b. \(15 \%\) NaOH by mass in water \(\left(d=1.00 \mathrm{g} / \mathrm{cm}^{3}\right)\) c. \(25 \%\) NaOH by mass in \(\mathrm{CH}_{3} \mathrm{OH}\left(d=0.79 \mathrm{g} / \mathrm{cm}^{3}\right)\) d. 0.10 mole fraction of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) in water \(\left(d=1.00 \mathrm{g} / \mathrm{cm}^{3}\right)\)

The freezing-point depression of a 0.091-m solution of CsCl is \(0.320^{\circ} \mathrm{C} .\) The freezing-point depression of a \(0.091-m\) solution of \(\mathrm{CaCl}_{2}\) is \(0.440^{\circ} \mathrm{C}\). In which solution does ion association appear to be greater? Explain.
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