Biol Water Potential Answer Key

Image Analysis Visual Elements: Title: "WATER POTENTIAL PROBLEMS" Header: "Name" Diagrams: A simple diagram representing a plant cell and a beaker of pure water. A diagram of a dialysis bag in a beaker of sucrose solution. Mathematical equations: Numerous equations related to water potential, solute potential, and pressure potential. Text: Questions and answers related to water potential concepts. Transcription of Text: What is the solute potential, Ψs, of a 1.0M sugar solution at 20 degrees Celsius under standard atmospheric conditions? Ψs = -iCRT = -(1)(0.0831 L.atm/mol.K)(1.0 mol/L)(293 K) = -24.1 bars Zucchini cores are measured and determined to have a sucrose concentration of 0.6 M. Calculate the solute potential, Ψs, of these cells. (Temperature is same as question #1.) Will water go in or out of the plant cell? Ψs = -iCRT = -(1)(0.6 mol/L)(0.0831 L.atm/mol.K)(293 K) = -14.6 bars Water will move out of the plant cell.   If solute potential in the plant cell is -6.35 bars and pressure potential is 0, what is the water potential of the plant cell? What does this indicate in terms of water movement? If it is -6.35 bars, water will move in. If it is less than -6.35 bars (for example), water will move out. If solute potential in the plant cell in question 3 is -6.25 bars and pressure potential is 6.25 bars, what is the water potential of the plant cell? What does this indicate in terms of water movement? Ψw = Ψs + Ψp = -6.25 bars + 6.25 bars = 0 bars Equal amounts move in and out.   A dialysis bag containing 0.1% sucrose is placed in a beaker containing 0.4% sucrose. The beaker is open to the atmosphere.   Initial Values: Beaker Content: 0.4% Sucrose Solution Dialysis Bag: 0.1% Sucrose Solution What is the pressure potential, Ψp, of the system? Ψp = 0 bars (open to atmosphere) What is the water potential of the dialysis bag? Ψw = Ψs + Ψp = -1 bars + 0 bars = -1 bars Water will move out of the dialysis bag. If a potato is allowed to dehydrate by sitting in the open air, would the water potential of the potato cells decrease or increase? Why? Decrease, because less water = lower solute potential. When it is dehydrated, the solutes are more concentrated = lower solute potential.   What is the water potential for a sucrose solution in an open container that is 0.1M? (temperature of 22°C) Ψw = Ψs + Ψp = -2.45 bars + 0 bars = -2.45 bars What is the solute potential for a NaCl solution that is 0.5M? (temperature of 10°C) Ψs = -iCRT = -(2)(0.5 mol/L)(0.0831 L.atm/mol.K)(283 K) = -23.5 bars A plant cell has a solute potential of -3.0 bars and a pressure potential of 1.0 bar. What is its water potential? If this cell is placed in a solution with a water potential of -5.0 bar. What will happen to this cell? Ψw = Ψs + Ψp = -3.0 bars + 1.0 bar = -2.0 bars Water will leave the cell.   Image Analysis Visual Elements: Continuation of previous image: The image appears to be a continuation of the previous image, containing more questions and answers related to water potential. Mathematical equations: Additional equations related to water potential calculations. Text: Questions and answers related to water potential concepts. Transcription of Text: A plant cell has a solute potential of -2.0 and a pressure potential of 0.0. What is its water potential? Show your work. It is placed in a solution with a water potential of -1.0. What happens to this plant cell? Ψw = -2.0 + 0 = -2.0 Water will go into the cell.   If a cell's Ψp is 2 bars and its Ψs is -3.5 bars, what is the resulting Ψw? Ψw = -3.5 bars + 2 bars = -1.5 bars If a cell from question #11 is placed in a beaker of sugar water with Ψw = 4.0 bars, in which direction will the net flow of water be? Water leaves the cell. The original cell from question #11 is placed in a beaker of sugar water with Ψw = -1.5 bars. In which direction will the net flow of water be? No net flow of water. The value for Ψw in root tissue was found to be -3.3 bars. If you take the root tissue and place it in a 0.1 M solution of sucrose at 20°C in an open beaker, what is the Ψw of the solution, and in which direction would the net flow of water be? Solution: Ψw = -iCRT = -(1)(0.1 mol/L)(0.0831 L.atm/mol.K)(293 K) = -2.4 bars Water will flow into the root.   NaCl dissociates into 2 particles in water: Na+ and Cl-. If the solution in question 14 contained 0.1M NaCl instead of 0.1M sucrose, what is the Ψw of the solution, and in which direction would the net flow of water be? Ψw = -iCRT = -(2)(0.1 mol/L)(0.0831 L.atm/mol.K)(293 K) = -4.9 bars   What is the solute potential, Ψs, of a 3.5M sugar solution at 22 degrees Celsius in an open beaker? Ψs = -iCRT = -(1)(3.5 mol/L)(0.0831 L.atm/mol.K)(295 K) = -85.8 bars At 20°C, a cell containing 0.6M glucose is in equilibrium with its surrounding solution containing 0.5M glucose in an open container. What is the cell's Ψw? Cells Ψs = -14.6 bars Ψp = 2.6 bars Ψw = -14.6 + 2.6 = -12 bars