Answer:
The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.
Explanation:
Water potential = Pressure potential + solute potential
[tex]P_w=P_p+P_s[/tex]
[tex]P_w=P_p+(-iCRT)[/tex]
We have :
C = 0.15 M, T = 273.15 K
i = 1
The water potential of a solution of 0.15 m sucrose= [tex]P_w[/tex]
[tex]P_p=0 bar [/tex] (At standard temperature)
[tex]P_s=-iCRT=-\times 1\times 8.314\times 10^{-2}bar L/mol K\times 273.15 K=-3.406 bar[/tex]
[tex]P_w=0 bar+(-3.406 ) bar[/tex]
The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.
