Use the lecture notes and chapter readings to answeelowing questions. Show all w

Question

Use the lecture notes and chapter readings to answeelowing questions. Show all work for full credit. 1. Two arms of a U-tube are separated by a semipermeable membrane. In the left arm is a sucrose solution with a solute (or osmotic) potential, s, of-0.5 MPa, and the right arm is a different sucrose solution with a s of-25 MPa. a. Which arm of the tube has a higher concentration of sucrose? b. What is the molar concentration of sucrose on both sides of the membrane if the temperature is 20 °C? c. In which direction does water movement occur? d. Describe two ways water movement could be stopped. 2. A single living cell with a pressure potential ('s) of 0.4 MPa and a solute potential ('s) of -0.9 MPa is dropped into a large open vessel of pure water (at sea level at 20 C). After equilibration, the cell has a Wp of 0.8 MPa. What was the solute potential of the cell after equilibration? Explain your answer. (i) The total water potential (pw) of an algal cell is-1.7 MPa and s within the cell is -2.6 MPa. The algal cell is placed in a large volume of 0.95 Molar sucrose solution at 20 °C. 3. a. Will the water potential, Vw, of the algal cell change? b. If so, at equilibrium, what will be the new Pw of the algal cell? If 's does not change, what will be the new 'p? (ii) After the algal cell had equilibrated to the 0.95 Molar sucrose solution, it was then transferred to a large volume of a 0.4 Molar NaCl solution at 30 °C a. What will be the new w of the algal cell after equilibration? b. If Y's does not change, what will be the new Wp?

Explanation / Answer

1a. The right-hand side has more concentration of the sucrose.

Explanation:

The osmotic potential of pure water is 0.

The osmotic potential becomes negative with the increase in the concentration of water. Higher the concentration of a solute, more negative is the osmotic potential.

The osmotic potential in the left arm is -0.5MPa, and

The osmotic potential in the right arm is -2.5MPa.

The right-hand side has more negative water potential, thus the right-hand side has more concentration of the sucrose.

b). The molar concentration of sucrose can be calculated as follows:

According to Hoff’s equation

Water Potential = cRT

Where c is the concentration of the solution

R is Gas constant = 0.082 liter-atmospheres per degree-mole

T is temperature = 21oC = 294.15 K

Therefore, the sucrose concentration at the left side is

Concentration = 0.5/ (0.082x294.15)

C= 0.02073 M or 20.73 mM

The sucrose concentration at the right side is

Concentration = 2.5/ (0.082x294.15)

C= 0.10367 M or 103.67 mol

c). The water would move from higher concentration to lower concentration i.e., from the right side to the left side.

d). The water movement could be stopped

(i) If water potential is made equal on both sides by either increasing the sucrose concentration on the left side or by adding water to the right side.

(ii) replacing the semipermeable membrane by an impermeable membrane.

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.