5. The water potential in a dilute aqueous solution depends on the pressure P an

Question

5. The water potential in a dilute aqueous solution depends on the pressure P and the solute concentration solte as follows: where ??120 is the chemical potential of pure water (ie., atx,01 te-0) at the reference pressure PO, and 18 mL/mol is the volume per mole of water. Here, the chemical potential is given per mole of H,o rather than per molecule A) Show that the H20 in a salt solution can never be at chemical equilibrium with pure water at the same temperature and pressure. In which liquid will the H2O molecules accumulate? B) Show that a salt solution can be at chemical equilibrium with pure water at the same temperature if the two liquids are held at different pressures, and find the required pressure difference (called the "osmotic pressure") as a function of solute concentration. C) Cytoplasm has a concentration ofxsolute 0.005. Find the osmotic pressure of cytoplasm relative to pure water at 300 K.

Explanation / Answer

A) the chemical potential of pure water is = u*H2O

For salt solution with concetration Y, the chemical potential is uH2O = u*H2O - RT Y + (P + P0) V0

from the above equation, at a salt concentration of Y with a same temperature and pressure just as pure water, the chemical potential is given by u*H2O - Y.

Hence at zero salt concetration will the chemical potential be equal to pure water's.

Because pure water always has higher chemical potential, so water will flow from pure water (higner potential) to water which contatins solutes (lower potential). Hence water will accumulate in water containing the solute

B) uH2O can be equal to u*H2O but only if P changes. Now we calculate the P at which uH2O + u*H2O

From the formula given, uH2O = u*H2O - RT X + ( P + Po) Vo

And we know uH2O = u*H2O and Vo = 18,

so putting these values in the formula, we get

u*H2O = u*H2O - RT X + (P + Po) 18

therefore,

Rt X = (P + Po) 18,

P + Po = RT X/18

So, P = (RT X)/18 - Po

So the pressure required is  P = (RT X)/18 - Po

C) xsolute = 0.005 and RT = 300k

So, uH2O = u*H2O - 300 * 0.005 + (P -Po) * 18

Hence the osmotic pressure is = u*H2O -1.5 + 18(P + Po)

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.