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 cell is placed in a saltwater solution, water moves out of the cell due to osmosis, causing the cell to shrink (plasmolysis). This results in a loss of turgor pressure and may lead to wilting and cell death. On the other hand, when a saltwater plant cell is placed in pure water, water moves into the cell, causing it to expand and become turgid. If too much water enters, the cell membrane may rupture, causing cell lysis.

Step by step solution

01

Understand Osmosis

Osmosis is the process wherein water molecules move from an area of higher concentration to lower concentration through a semipermeable membrane. In the case of plant cells, the semipermeable membrane is the cell membrane which allows water to move in and out of the cell.
02

Freshwater plant in saltwater solution

When a freshwater plant cell is placed in a saltwater solution, the concentration of water is higher inside the cell than in the surrounding environment. Due to osmosis, water will start moving out of the cell, causing it to shrink. This shrinking of the cell is called plasmolysis. The cell will lose its turgor pressure and may lead to the plant cells wilting and eventually dying, if not removed from the saltwater solution. To illustrate this, draw a plant cell before placing it in saltwater with a distinct vacuole and cytoplasm. Then, draw the cell after being in saltwater with the cytoplasm shrinking and the cell membrane detaching from the cell wall.
03

Saltwater plant in pure water

When a saltwater plant cell is placed in pure water, the concentration of water is lower inside the cell than in the surrounding environment. Due to osmosis, water will start moving into the cell. This causes the cell to expand and the vacuole to increase in size. The cell will become more turgid, which is normal for healthy plant cells for maintaining their structure. However, if too much water enters the cell, it could lead to the rupture of the cell membrane and cause cell lysis, or the breaking down of the cell. To illustrate this, draw a saltwater plant cell before placing it in pure water with a smaller vacuole and turgid cytoplasm. Then, draw the cell after being in pure water with an increased vacuole size and an increased turgidity, or the cell membrane ruptured due to cell lysis, if too much water entered. In conclusion, placing a freshwater plant in a saltwater solution will cause its cells to lose water and shrink due to plasmolysis, while placing a saltwater plant in pure water will cause its cells to take in water and either become turgid or rupture due to cell lysis. Illustrations of these changes can help visualize the effects of osmosis on plant cells in different water environments.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which of the following will have the lowest total vapor pressure at \(25^{\circ} \mathrm{C} ?\) a. pure water (vapor pressure \(=23.8\) torr at \(25^{\circ} \mathrm{C}\) ) b. a solution of glucose in water with \(\chi_{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}=0.01\) c. a solution of sodium chloride in water with \(\chi_{\mathrm{NaCl}}=0.01\) d. a solution of methanol in water with \(\chi_{\mathrm{CH}_{3} \mathrm{OH}}=0.2\) (Consider the vapor pressure of both methanol \(\left[143\right.\) torr at \(\left.25^{\circ} \mathrm{C}\right]\) and water.)

How would you prepare \(1.0 \mathrm{~L}\) of an aqueous solution of sodium chloride having an osmotic pressure of \(15 \mathrm{~atm}\) at \(22^{\circ} \mathrm{C} ?\) Assume sodium chloride exists as \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions in solution.

An aqueous solution containing \(0.250 \mathrm{~mol} \mathrm{Q}\), a strong electrolyte, in \(5.00 \times 10^{2} \mathrm{~g}\) water freezes at \(-2.79^{\circ} \mathrm{C}\). What is the van't Hoff factor for Q? The molal freezing-point depression constant for water is \(1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\). What is the formula of \(\mathrm{Q}\) if it is \(38.68 \%\) chlorine by mass and there are twice as many anions as cations in one formula unit of \(\mathrm{Q}\) ?

Rubbing alcohol contains 585 g isopropanol \(\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\right)\) per liter (aqueous solution). Calculate the molarity.

At \(25^{\circ} \mathrm{C}\), the vapor in equilibrium with a solution containing carbon disulfide and acetonitrile has a total pressure of 263 torr and is \(85.5\) mole percent carbon disulfide. What is the mole fraction of carbon disulfide in the solution? At \(25^{\circ} \mathrm{C}\), the vapor pressure of carbon disulfide is 375 torr. Assume the solution and vapor exhibit ideal behavior.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free