Question

"A" cell is placed in hypertonic solution. After some time, its osmotic potential is measured as -0.5 MPa. Then its water potential would be:

Hint:

Water potential = Osmotic potential + Pressure potential.
Hypertonic solution → water leaves → pressure potential = 0.
Always assign negative sign to osmotic potential in hypertonic solutions.

Answer 1

1. Water potential \(\Psi_w = \Psi_s + \Psi_p\), where \(\Psi_s\) = osmotic potential, \(\Psi_p\) = pressure potential.
2. In hypertonic solution, cell loses water → turgor pressure drops → \(\Psi_p \approx 0\).
3. Thus, \(\Psi_w \approx \Psi_s = -0.5\) MPa.
4. Hence the water potential is (2) -0.5 MPa.