Question

questions 1 - 3
a selectively permeable membrane separates the solutions in the arms of the u - tube shown. the membrane is permeable to water and to substance 1 but not to substance 2. forty grams of substance 1 and 20 grams of substance 2 have been added to the water on side a of the u - tube. twenty grams of substance 1 and 40 grams of substance 2 have been added to the water on side b of the u - tube. assume that after a period of time, the solutions on either side of the membrane have reached equilibrium.

1. how many grams of substance 1 will be in solution on side a of the u - tube? how many grams of substance 1 will be in solution on side b? explain.
2. how many grams of substance 2 will be in solution on side a of the u - tube? how many grams of substance 2 will be in solution on side b. explain.
3. explain what has happened to the water level in the u - tube.

questions 4 - 6
the cell membrane of red blood cells is permeable to water but not to sodium chloride, nacl. suppose that you have three flasks:

• flask x contains a solution that is 0.5 percent nacl.
• flask y contains a solution that is 0.9 percent nacl.
• flask z contains a solution that is 1.5 percent nacl.

to each flask, you add red blood cells, which contain a solution that is 0.9 percent nacl.

1. predict what will happen to the red blood cells in flask x.
2. predict what will happen to the red blood cells in flask y.
3. predict what will happen to the red blood cells in flask z.

Explanation:

Step1: Analyze substance 1 diffusion

Since the membrane is permeable to substance 1, at equilibrium, the concentration of substance 1 will be the same on both sides. The total amount of substance 1 is \(40 + 20=60\) grams. So each side will have \(\frac{60}{2}=30\) grams.

Step2: Analyze substance 2 diffusion

The membrane is not permeable to substance 2. So the amount of substance 2 on side A will remain 20 grams and on side B will remain 40 grams.

Step3: Analyze water - level change in U - tube

The initial solute concentration on side A is \(\frac{40 + 20}{V_A}\) (where \(V_A\) is the volume of water on side A initially) and on side B is \(\frac{20+40}{V_B}\). Since the membrane is permeable to water, water will move from the side with lower solute concentration to the side with higher solute concentration. But since the total amount of solute is the same on both sides (\(60\) grams of solute in total on each side considering both substances), at equilibrium, the water level will be the same on both sides of the U - tube.

Step4: Analyze red - blood cells in flask X

Flask X has a \(0.5\%\) NaCl solution and red - blood cells have a \(0.9\%\) NaCl solution. The solution in flask X is hypotonic compared to the red - blood cells. Water will move into the red - blood cells, causing them to swell and possibly burst (hemolysis).

Step5: Analyze red - blood cells in flask Y

Flask Y has a \(0.9\%\) NaCl solution, which is isotonic to the red - blood cells. There will be no net movement of water, and the red - blood cells will maintain their normal shape.

Step6: Analyze red - blood cells in flask Z

Flask Z has a \(1.5\%\) NaCl solution, which is hypertonic compared to the red - blood cells. Water will move out of the red - blood cells, causing them to shrink (crenation).

Answer:

1. Side A: 30 grams, Side B: 30 grams. Because the membrane is permeable to substance 1 and at equilibrium, the concentration of substance 1 will be equal on both sides.
2. Side A: 20 grams, Side B: 40 grams. Because the membrane is not permeable to substance 2.
3. The water level will be the same on both sides. Since the total amount of solute is equal on both sides at equilibrium.
4. The red - blood cells will swell and may burst (hemolysis) as the solution in flask X is hypotonic.
5. The red - blood cells will maintain their normal shape as the solution in flask Y is isotonic.
6. The red - blood cells will shrink (crenation) as the solution in flask Z is hypertonic.