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# Osmotic Pressure

## Key Concepts

• Osmotic pressure arises when two solutions of different concentrations, or a pure solvent and a solution, are separated by a semipermeable membrane. Molecules such as solvent molecules that can pass through the membrane will migrate from the side of higher concentration to the side of lower concentration in a process known as osmosis.

• The pressure required to stop osmosis is called the osmotic pressure.

• In dilute solutions, osmotic pressure (Π) is directly proportional to the molarity of the solution and its temperature in Kelvin.

• van't Hoff Equation: Π = MRT
Π = osmotic pressure
M = molarity = moles ÷ volume(L)
R = ideal gas constant
T = temperature (K)

• Solvent can be removed from a solution using a pressure greater than the osmotic pressure. This is known as reverse osmosis.

## Example: Osmotic Pressure Calculation for a Nonelectrolyte Solution

Calculate the osmotic pressure exhibited by a 0.10M sucrose solution at 20oC.
 Π = MRT M = 0.10M R = 0.0821 L atm K-1mol-1 T = 20oC = 20 + 273 = 293K Π = 0.10 x 0.0821 x 293 = 2.4 atm Π = MRT M = 0.10M R = 8.314 J K-1 mol-1 T = 20oC = 20 + 273 = 293K Π = 0.10 x 8.314 x 293 = 243.6 kPa

## Example: Osmotic Pressure Calculation for an Electrolyte Solution

Calculate the osmotic pressure exhibited by a 0.42M KOH solution at 30oC.
 Π = MRT Since KOH → K+(aq) + OH-(aq) M(K+(aq)) = 0.42mol/L M(OH-(aq)) = 0.42mol/L M(solute) = 0.42 + 0.42 = 0.84mol/L R = 0.0821 L atm K-1mol-1 T = 30oC = 30 + 273 = 303K Π = 0.84 x 0.0821 x 303 = 20.9 atm Π = MRT Since KOH → K+(aq) + OH-(aq) M(K+(aq)) = 0.42mol/L M(OH-(aq)) = 0.42mol/L M(solute) = 0.42 + 0.42 = 0.84mol/L R = 8.314 J K-1 mol-1 T = 30oC = 30 + 273 = 303K Π = 0.84 x 8.314 x 303 = 2116 kPa

## Example: Calculation of Molecular Mass (Formula Weight) Using Osmotic Pressure

0.500g hemoglobin was dissolved in enough water to make 100.0mL of solution.
At 25oC the osmotic pressure was found to be 1.78 x 10-3 atm.
Calculate the molecular mass (formula weight) of the hemoglobin.

1. Calculate the molarity, M, of the solution:
M = Π ÷ RT
Π = 1.78 x 10-3 atm
R = 0.0821 L atm K-1mol-1
T = 25oC = 25 + 273 = 298K
M = 1.78 x 10-3 ÷ 0.0821 x 298 = 7.28 x 10-5mol/L

2. Calculate the moles, n, of hemoglobin present in solution:
n = M x V
M = 7.28 x 10-5mol/L
V = 100.0mL = 100.0 x 10-3L
n = 7.28 x 10-5 x 100.0 x 10-3 = 7.28 x 10-6mol

3. Calculate the molecular mass, MM, of the hemoglobin:
MM = mass ÷ n
mass = 0.500g
n = 7.28 ÷ 10-6mol
MM = 0.500 ÷ 7.28 x 10-6 = 68,681 g/mol
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