Osmosis Flashcards
What is osmosis?
Osmosis is the diffusion of water molecules across a partially permeable membrane, from an area of higher water potential (i.e. higher concentration of water molecules) to an area of lower water potential (i.e. lower concentration of water molecules).
What is water potential?
Water potential is the potential (likelihood) of water molecules to diffuse out of or into a solution.
Pure water has the highest water potential. All solutions have a lower water potential than pure water.
What is are two solutions with the same water potential known as?
If two solutions have the same water potential, they’re said to be isotonic.
What 3 factors does osmosis depend on?
(same as diffusion)
The water potential gradient the higher the water potential gradient, the faster the - rate of osmosis. As osmosis takes place, the difference in water potential on either side of the membrane decreases, so the rate of osmosis levels off over time.
2) The thickness of the exchange surface the thinner the exchange surface, the faster the rate of osmosis. -
3) The surface area of the exchange surface the larger the surface area, the faster the rate of osmosis.
How will dissolving salt into a beaker of pure water affect the water potential of the liquid in the beaker?
Adding salt to pure water would lower the water potential of the liquid in the beaker. Pure water has the highest water potential. All solutions have a lower water potential than pure water.
A student makes five serial dilutions of sucrose solution and carries out an experiment in order to identify the water potential of potato cells. Describe a method the student could have used to produce the results shown in the graph.
E.g. use a cork borer to cut potatoes into identically sized chips. Divide the chips into groups of three and measure the mass of each group using a mass balance. Place one group into each of the five sucrose solutions. Leave the chips in the solutions for at least 20 minutes. Remove the chips and pat dry gently with a paper towel. Weigh each group again and record the results. Calculate the % change in mass for each group. Use the results to make a calibration curve, showing % change in mass against sucrose concentration.
Describe how you could make 5 serial diultions of a sucrose solution, starting with an initial sucrose concentration of 2 M and diluting each solution by a factor of 2.
- Add 10 cm³ of the initial 2 M sucrose solution to the first test tube and 5 cm³ of distilled water to the other four test tubes.
- Then, using a pipette, draw 5 cm³ of the solution from the first test tube, add it to the distilled water in the second test tube and mix the solution thoroughly. You now have 10 cm solution that’s half as concentrated as the solution in the first test tube (it’s 1 M).
- Repeat this process three more times to create solutions of 0.5 M, 0.25 M and 0.125 M.
How can you make solutions of different concerntrations by finding the scale factor?
For example, if you want to make 15 cm³ of 0.4 M sucrose solution…
1) Start with a solution of a known concentration, e.g. 1 M.
2) Find the scale factor by dividing the concentration of this solution by the concentration of the solution you want to make. So in this case the scale factor = 1 M÷ 0.4 M = 2.5.
3) This means that the solution you want to make is 2.5 times weaker than the one you have. To make the solution 2.5 times weaker, use 2.5 times less of it, i.e. 15 cm³ ÷ 2.5 = 6 cm³. Transfer this amount to a clean test tube.
4) Top up the test tube with distilled water to get the volume you want to make. In this case you want to make 15 cm³ of solution, so you need to add: 15 - 6 = 9 cm³ of distilled water.
a. Using the graph, estimate the concentration of sucrose solution that would have the same water potential as the potato cells.
b. How could the student use this estimation to find out the water potential of the potato cells?
c. Suggest how the student could repeat the experiment to produce a more accurate estimate of the cells’ water potential.
a. E.g. 0.3 mol dm³. (The point at which the curve crosses the x-axis, where the % change in mass is 0, is the point at which the water potential of the sucrose solution is the same as the water potential of the potato cells.)
b. Look up the water potential for that concentration of sucrose solution in a textbook or online.
c. E.g. the student could repeat the experiment with new sucrose solutions with concentrations between 0.25 and 0.5 mol dm-³. This would allow the student to make a calibration curve with a higher resolution and therefore find the water potential of the potato cells to a greater degree of accuracy.
Give one way in which a specialised cell might be adapted to increase the rate at which it absorbs water by osmosis.
E.g. it might have a larger surface area (which could be created through folds or projections in the cell-surface membrane).
The diagram on the right shows the exterior and interior of a plant cell vacuole. In which direction will the net movement of water molecules be?
The net movement of water molecules will be from the vacuole interior to the vacuole exterior. This is because there is a higher water potential inside the vacuole than outside the vacuole, so water molecules will diffuse out of the vacuole.
Two solutions of differing concentration are separated by a partially permeable membrane. Explain why the rate of osmosis between the solutions levels off over time.
Water molecules will diffuse across the membrane from the solution with the higher water potential to the solution with the lower water potential. As this diffusion takes place, the difference in water potential between the two sides of the membrane will decrease until it reaches an equilibrium (the water potential on both sides is equal). As the water potential gradient decreases, so will the rate of osmosis.
A student has 10 cm³ of a 4 mol dm-³ sucrose solution. Describe how the student could make 8 cm³ of a 3.2 mol dm-³ sucrose solution.
Work out the scale factor, which in this case = 4 mol dm³ ÷ 3.2 mol dm³ =1.25. This means that the solution the student wants to make is 1.25 times weaker than the one they have. To make the solution 1.25 times weaker, they should use 1.25 times less than the desired volume. In this case, that’s 8 cm³ ÷ 1.25 = 6.4 cm³. Transfer this amount to a clean test tube. Top up the test tube with distilled water to get the desired volume (8 – 6.4 = 1.6 cm³ of distilled water).
