Osmotic Pressure Chemistry Tutorial - AUS-e-TUTE

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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 lower solute concentration to the side of higher solute concentration in a process known as osmosis. So the dilute solution becomes more concentrated over time, and at the same time, the concentrated solution becomes more dilute.
• 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 (c) and its temperature in Kelvin (T).

  Π ∝ cT

• van't Hoff Equation: Π = cRT

  Π = osmotic pressure c = 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.

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Example: Osmotic Pressure Calculation for a Nonelectrolyte Solution

Osmotic Pressure Problem:

Calculate the osmotic pressure exhibited by a 0.10 mol L-1 sucrose solution at 20oC.

Osmotic Pressure Problem Solution:

1. For pressure in atmospheres (atm)

   Π = cRT c = 0.10 mol L-1 R = 0.0821 L atm K-1mol-1 (from data sheet) T = 20oC = 20 + 273 = 293 K

   Π = 0.10 x 0.0821 × 293 = 2.4 atm

2. For pressure in kilopascals (kPa)

   Π = cRT c = 0.10 mol L-1 R = 8.314 J K-1mol-1 (from data sheet) T = 20oC = 20 + 273 = 293 K

   Π = 0.10 × 8.314 x 293 = 243.6 kPa

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Example: Osmotic Pressure Calculation for an Electrolyte Solution

Osmotic Pressure Problem:

Calculate the osmotic pressure exhibited by a 0.42 mol L-1 KOH solution at 30oC.

Osmotic Pressure Problem Solution:

1. For pressure in atmospheres (atm): Π = cRT Since KOH → K+(aq) + OH-(aq) c(K+(aq)) = 0.42 mol L-1 c(OH-(aq)) = 0.42 mol L-1 c(solute) = 0.42 + 0.42 = 0.84 mol L-1 R = 0.0821 L atm K-1mol-1 (from data sheet) T = 30oC = 30 + 273 = 303 K

   Π = 0.84 × 0.0821 × 303 = 20.9 atm

2. For pressure in kilopascals (kPa) Π = cRT Since KOH → K+(aq) + OH-(aq) c(K+(aq)) = 0.42 mol L-1 c(OH-(aq)) = 0.42 mol L-1 c(solute) = 0.42 + 0.42 = 0.84 mol L-1 R = 8.314 J K-1mol-1 (from data sheet) T = 30oC = 30 + 273 = 303 K

   Π = 0.84 × 8.314 × 303 = 2116 kPa

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Example: Calculation of Molar Mass Using Osmotic Pressure

Osmotic Pressure Problem:

0.500 g hemoglobin was dissolved in enough water to make 100.0 mL of solution. At 25oC the osmotic pressure was found to be 1.78 × 10-3 atm. Calculate the molar mass of the hemoglobin.

Osmotic Pressure Problem Solution:

1. Calculate the molarity, c, of the solution: c = Π ÷ RT Π = 1.78 × 10-3 atm R = 0.0821 L atm K-1 mol-1 (from data sheet) T = 25oC = 25 + 273 = 298 K c = 1.78 × 10-3 ÷ (0.0821 × 298) = 7.28 × 10-5 mol L-1
2. Calculate the moles, n, of hemoglobin present in solution: n = c × V c = 7.28 × 10-5 mol L-1 V = 100.0 mL = 100.0 × 10-3 L n = 7.28 × 10-5 × 100.0 × 10-3 = 7.28 × 10-6 mol
3. Calculate the molar mass of the hemoglobin: molar mass = mass ÷ moles mass = 0.500 g moles = 7.28 × 10-6 mol molar mass = 0.500 ÷ (7.28 × 10-6) = 68 681 g mol-1

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