Question

root tissue has a water potential of -3.1 bars. the root tissue is placed it in a 0.1 m solution of sucrose at 20°c in an open beaker. from the statements given select the solute potential of the solution and the true statements about this scenario. net flow of water will be into the solution the solution is considered hypotonic compared to the cell -3.42 -0.17 net flow of water will be into the cell the solution is considered hypertonic compared to the cell -2.43

Explanation:

Step1: Calculate the solute potential of the solution

Given \(i = 1.0\) (for sucrose), \(C=0.1M\), \(R = 0.0831\) liter bars/mole K, \(T=(20 + 273)K=293K\)
Using the formula \(\Psi_{S}=-iCRT\)
Substitute the values: \(\Psi_{S}=-(1.0)\times(0.1)\times(0.0831)\times(293)\)

$$
LATEXBLOCK0
$$

Step2: Determine the direction of water flow

The water potential of the root tissue \(\Psi=-3.1\) bars. The water potential of the solution (in an open container \(\Psi_{p} = 0\), so \(\Psi=\Psi_{S}=- 2.43\) bars)
Water flows from an area of higher water potential to an area of lower water potential. Since \(-2.43>-3.1\), water will flow into the cell (root tissue)

Step3: Determine tonicity

A hypotonic solution has a higher water potential (less negative solute potential in this case, since \(\Psi=\Psi_{S}\) for the solution) compared to the cell. Here, \(\Psi_{solution}=-2.43\) and \(\Psi_{cell}=-3.1\). So the solution is hypotonic compared to the cell

Answer:

-2.43, net flow of water will be into the cell, the solution is considered hypotonic compared to the cell