s = solute potential p = pressure potential = water potential Assume i = 1 A pie

Question

s = solute potential

p = pressure potential

= water potential

Assume i = 1

A piece of plant tissue was incubated in 0.3 molal mannitol at 25°C. After 1 hour, the tissue's mass was the same as it was before it was incubated. After incubation in 0.4 molal mannitol, the tissue's mass was lowered by 5%. After correction for intercellular water, the tissues initial s was-0.9 MPa. Using the same assumptions as we did in the lab, calculate the following (no graphs are required) a. . Wo and for 0.4 molal mannitol solution b. initial V of the plant tissue c. initial Vp of the plant tissue d, equilibrium Y. . Wo of the plant tissue in 0.4 molal mannitol

Explanation / Answer

Water potential = pressure potential + solute potential Water potential = 0 - 0.9Mpa Initial water potential = -0.9 Mpa Initial pressure potential = water potential - solute potential Initial P. Potential = -0.9 -0.9 = - 1.8 Mpa solute potential for 0.3 molal mannitol solution is = -iCRT (1) i = Ioniation constatnt for manniol C = Molar concentration of plant tissue R = gas constant T = temperature Substituting all the values in equation (1) solute potential = 1 x 0.4 x 8.314 k pa L / mol K x 298 = 991. 03 = 0.9 x 103 kpa = 0.9Mkpa

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