Question

What is the water potential of a .20 M solution of sucrose in an open beaker at 23 degrees Celsius?

Answer 1

Answer:

-4.92 bars

Explanation:

The water potential of a solution is the sum of the pressure potential and the solute potential of the solution. This can be mathematically expressed as;

Фw = Фp + Фs, where Фw = water potential, Фp = pressure potential, and Фs = solute potential.

For a solution that is in an open beaker, the Фp = 0. hence

Фw = -Фs

Also,

Фs = iCRT, where i = number of particles the molecule will make in water (1 for sucrose), C = molar concentration, R = pressure constant (usually 0.0831 liter bar/mole K), and T = temperature (in kelvin).

In this case, i = 1, C = 0.20 M, R = 0.0831, T = 273+23

Фs = 1 x 0.20 x 0.0831 x 296 = 4.92 bars

Therefore,

Фw = -4.92 bars