Osmosis and Osmotic Pressure — Dr. Georgios Malis

Experiment 01

Two solutions – one semipermeable membrane

Select the concentration and solute on each side. The ideal semipermeable membrane allows water, but not dissolved particles, to pass through.

Left-hand solution A 0,10 M

0 M0,60 M

Solute A

Right-hand solution B 0,30 M

0 M0,60 M

Solute B

Temperature 27 °C

What should you observe?

In a real system, the hydrostatic pressure difference resulting from different liquid levels can counterbalance the osmotic driving force. At dynamic equilibrium, water molecules continue moving in both directions, but there is no net transfer of water.

Semipermeable membrane

WaterDissolved particlesMembrane

The orange solute particles cannot pass through the membrane. Liquid levels change schematically over time.

Schematic depiction – not to actual molecular scaleTime: 0,0 s

2,46

Π_Α initial (atm)

7,38

Π_Β initial (atm)

Α → Β

Initial net H₂O flow

Π = i · c · R · T i: van’t Hoff factor (ideal number of dissolved particles per formula unit) · R = 0,082 L·atm·mol⁻¹·K⁻¹ · T in K

Illustrative volumes: Α 1,00 L · Β 1,00 LNet flow: Α → Β

Step-by-step explanation

1The membrane is permeable to the solvent (H₂O) and impermeable to the solute.

2Before a hydrostatic pressure difference develops, the initial net water flow is from the solution with lower osmotic pressure to that with higher osmotic pressure.

3The difference in liquid levels creates a hydrostatic pressure difference opposing the initial net flow.

Experiment 02

van’t Hoff’s law

Adjust molar concentration, temperature and solute type. Observe how osmotic pressure changes.

Molar concentration c 0,10 MTemperature θ 27 °C

Solute type

Important assumption

In the school-level model, Π = cRT applies to dilute nonelectrolyte solutions. For electrolyte solutions, we use the approximation Π = i·cRT, where the van’t Hoff factor i takes the ideal values 1, 2 or 3 depending on the solute.

Osmotic pressure calculation

Π = 0,10 · 0,082 · 300 = 2,46 atm

2,46

Π (atm)

0,10

i·c (osmol/L), ideal

300

Temperature T (K)

Π ∝ c·T, for constant i (ideal dilute-solution model)Dashed lines mark the selected value

Try this:

Increase concentration from 0.10 M to 0.20 M at constant temperature. How does Π change? Then hold concentration constant and change the solute from glucose to NaCl.

Calculations 01

Solution dilution

During dilution, the amount (mol) of solute remains constant. We assume constant temperature and additive volumes.

Initial concentration (M)

Initial volume (mL)

Water added (mL)

Temperature (°C)

c₁V₁ = c₂V₂

0,10

Final concentration (M)

4,92

Initial Π (atm)

2,46

Final Π (atm)

Calculations 02

Mixing solutions

Two solutions are mixed at 27 °C. We assume the solutes do not react with each other and that volumes are additive.

Concentration A (M)

Volume A (mL)

Concentration B (M)

Volume B (mL)

Type of solute A

Type of solute B

Π_final = [(i₁c₁V₁ + i₂c₂V₂)/(V₁ + V₂)]·R·T

0,20

Final osmotic concentration (osmol/L)

4,92

Final Π (atm)

1.000

Final volume (mL)

Calculations 03

Osmometry — Determining Mᵣ

For a dilute nonelectrolyte solution, we calculate molar mass M (g/mol) from mass m, volume V, and osmotic pressure Π. The relative molecular mass Mᵣ has the same numerical value but no unit.

Mass of solute (g)

Solution volume (mL)

Osmotic pressure (atm)

Temperature (°C)

M = m·R·T/(Π·V) (g/mol)

Calculations 04

Isotonicity

For two dilute solutions at the same temperature we have: Π₁ = Π₂ ⇔ i₁c₁ = i₂c₂ (ideal approximation).

Reference concentration (M)

Reference solute type

Required solution’s solute type

i₁·c₁ = i₂·c₂

Experiment 03

Osmosis and cells

Observe schematically how a red blood cell responds when the tonicity of the external solution changes.

Osmotic concentration of external solution 0,30 osmol/L

0,050,60

Reference intracellular osmotic concentration: 0.30 osmol/LSchematic model with solute particles unable to cross the membrane, at constant temperature

How do I identify the solution type?

The terms “hypertonic” and “hypotonic” are comparative and describe the external solution relative to the cell. Tonicity is mainly determined by solute particles that cannot cross the particular cell membrane.

Schematic representation of a red blood cell

Isotonic

Surroundings

↔

Net water movement

Stable

Change in shape / volume

Connection to biology

In a strongly hypotonic environment, a red blood cell may swell and eventually undergo hemolysis. In a hypertonic environment, it loses water and shrinks. This diagram is qualitative and does not replace the study of real cells.

Assessment

Quick quiz — Osmosis

Seven multiple-choice questions with instant scoring and explanations.

Revision summary for Greek national exams

1Osmosis: net solvent movement through a semipermeable membrane from lower Π toward higher Π before a balancing hydrostatic pressure develops.

2Osmotic pressure: the external pressure required to prevent net solvent flow through a semipermeable membrane at a given temperature.

3van’t Hoff’s law for a dilute molecular solution: Π = cRT = (n/V)RT.

4Ideal complete electrolyte dissociation: Π ≈ i·c·R·T. The total number of dissolved particles is the determining factor.

5At the same temperature, two ideal solutions are isotonic when their osmotic pressures are equal, even if their molar concentrations differ.

Pay attention to units: T/K ≈ 273 + θ/°C, V in L, c in mol/L, Π in atm and R = 0,082 L·atm·mol⁻¹·K⁻¹. The ideal electrolyte approximation does not describe real solutions exactly.