Osmotic pressure measurements are commonly used to determine the molecular weigh

Question

Osmotic pressure measurements are commonly used to determine the molecular weights of proteins and polymers. An osmometer is used to measure the equilibrium pressure difference between the pure solvent and the solvent containing macromolecules, which are too large to pass through the membrane. The pressure difference AP is the osmotic pressure is equal to pgh where is the solution density and h is the difference in the liquid heights. If the solute a. RTc concentration is small, show that solute . Here, Csolute Is the concentration of solute the solute in units of mass per unit volume and Msolute is the molecular weight of the solute. b. Relatively small concentrations of macromolecules produce easily measurable osmotic pressure differences. Consider 1 gm of a protein of molecular weight 60,000 Da (same as 60,000 gm/mol) dissolved in 100 mL of water and placed in an osmometer. Show that the osmotic pressure difference will be 4.13×10-3 bar Use R = 8.3 14x1 0-5 bar-m/mol-K. A rough rule of thumb in polymer solution theory is that a 4 molar aqueous polymer solution will have an osmotic pressure of approximately 100 bar. Is this c. water rule of thumb in approximate agreement with the expression water A 4 molar solution is 4 moles per liter of solution, which approximately contains 55 moles of water. First, show that the mole fraction of water xwater will be 0.9322. Then use R 8.314x10-5 bar-m'/mol-K, T-298.15 K, vwater- 18.015x10*m'/mol for the remainder of vour calculation

Explanation / Answer

given osmotic pressure = RTCsolute/M

1 gm of solute is dissovled in 100ml, converting the ml to m3 gives 100ml =0.0001m3

R= 8.314*10-5 barm3/mol.K, T = 25 deg.c= 25+273= 298K

C solute= 1/0.0001 g/m3= 10000 g/mm3

M= 60000 g/mole,

Osmotic pressure = 8.314*10-5*298*1000/60000= 0.004129 bar

2. concentration 4 moles/Liter of solution

volume of water= 1000ml, assuming the density of water =1 g/ml, mass of water= 1000ml*1g/ml= 1000gm

moles of water= mass of water/ 18= 1000/18= 55.5 moles, total moles of solution = 4+55.5= 59.5

mole fraction water = 55.5/59.5 =0.932

osmotic pressure =- 8.314*10-5* 298*ln(0.932)/18.015*10-3)=95.71 bar. This approximation is correct.

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

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