Question

calculate the water potential of a solution of 0.15m sucrose. the solution is at standard temperature.

Answer 1

Answer:

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

Explanation:

Water potential = Pressure potential + solute potential

$P_w=P_p+P_s$

$P_w=P_p+(-iCRT)$

We have :

C = 0.15 M, T = 273.15 K

i = 1

The water potential of a solution of 0.15 m sucrose= $P_w$

$P_p=0 bar$ (At standard temperature)

$P_s=-iCRT=-\times 1\times 8.314\times 10^{-2}bar L/mol K\times 273.15 K=-3.406 bar$

$P_w=0 bar+(-3.406 ) bar$

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.