1 You place a plant cell in an open beaker of 05 M sucrose T

1.
You place a plant cell in an open beaker of 0.5 M sucrose. The solution has a s of -2.0 MPa. The plant cell has a s of -2.0 MPa and a p of +1.0.
Reference: water
Consider the cell above. You allow the cell to come to equilibrium in the sucrose solution. What will be the p of the cell?

A.
+2 MPa
B.
0 MPa
C.
+1 MPa
D.
-1 MPa
E.
-0.5 MPa

2.
You place a plant cell in an open beaker of 0.5 M sucrose. The solution has a s of -2.0 MPa. The plant cell has a s of -2.0 MPa and a p of +1.0.
Reference: water
After this cell has reached equilibrium in the solution in the previous question, you transfer it to an open beaker of pure water. What will happen immediately after placing the cell in water?

A.
The cell will burst.
B.
Nothing will happen.
C.
Water will flow into the cell.
D.
Water will flow out of the cell.

Solution

1. To measure the p of cell we have to calculate system pressure by following formula.

system = p + s =1+ (-2) = 1-2 =-1 MPa

But the solution of 0.5M sucrose has s’ -2.0 MPa. Therefore, the required pressure of system pressure is -2.0 MPa to meet the equilibrium pressure.

The amount of pressure needs = system – s’ =(-1) – (-2) = -1 + 2 = 1Mpa

Now the equilibrium p of cell = p - 1 Mpa = 1 - 1 = 0 MPa

The option B is correct.

2. When cell will place in the beaker of pure water the water will move inside the cell. Water has potential 0Mpa which is higher than equilibrium potential -2MPa. Therefore, water will move from higher to lower pressure for overcoming the potential difference. The option C is correct.