Respuesta :
Answer:
1. Solute potential= -2.43 bars
2. Solution to root tissue
Explanation:
The solute potential is the potential or concentration gradient exerted by the solute in a solution.
The mathematical expression of the solute potential is
= -iRCT, where i= ionization of molecule
C- concentration of a molecule
R - pressure constant (0.0831)
T- Temperature (kelvin)
According to the given question,
i=1 ( sucrose is non-polar)
R= 0.0831
T= 273+ 20= 293
C= 0.1
Putting the values in -iRCT
we will have solute potential = -2.43 bars (high water potential)
Since the potential in root tissue is -3.3 bars. (low water potential)
We know that water moves from high water potential to low water potential and the water will move from the solution to the root tissue that is from -2.43 bars to -3.3 bars
A) The value of Ψ ( water potential ) of the solution is : - 2.43 bars
B) The direction of the net flow of water would be : From the solution to the root tissue
Given data :
0.1 M solution of sucrose
Temperature = 20°C + 273 = 293 K
Initial value of solute potential ( Ψ ) = -3.3 bars
pressure constant ( R ) = 0.0831
Ionization of molecule ( For sucrose ) = 1
A) Calculate the value of water potential in the solution
Applying the expression used for determining the value of solute potential.
solute potential ( Ψ ) = - IRCT ------ ( 1 )
where ; I = 1, R = 0.0831 , C = 0.1 , T = 293
Insert the values into equation ( 1 )
( Ψ ) = - ( 1 * 0.0831 * 0.1 * 293 )
= - 2.43
B) Given that the solute potential in the solution is -2.43 which is higher than the solute potential found in the root tissue, therefore the water will move from the solution to the root tissue.
Hence we can conclude that; The value of Ψ ( water potential ) of the solution is; - 2.43 bars. The direction of the net flow of water would be : From the solution to the root tissue.
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