Solutions · Lesson 9 of 11
Osmosis, Osmotic Pressure and Reverse Osmosis
“Explain selective solvent movement and calculate osmotic pressure, solute molar mass and the pressure needed to reverse osmosis.”
• Explain osmosis using a membrane that passes solvent but retains solute. • Define osmotic pressure as an opposing pressure difference. • Calculate osmotic pressure and molar mass using consistent units. • Compare isotonic, hypertonic and hypotonic conditions and explain reverse osmosis.
Selective membranes make solvent movement visible
A raisin swells in water and a piece of raw mango shrivels in concentrated brine. These changes involve water crossing barriers that separate solutions of different composition. A semipermeable membrane in the ideal model passes solvent molecules but retains dissolved solute. Solvent can cross in both directions; osmosis is the net transfer caused by a difference in solvent escaping tendency.
At equal temperature and initially equal pressure, water flows from pure water or a more dilute solution toward a more concentrated solution of the same solvent. This increases the solution volume and dilutes it. In a thistle funnel with sugar solution separated from water by an appropriate membrane, the column rises until the resulting pressure difference opposes further net flow. The membrane is selective: it is not simply a barrier with holes that indiscriminately passes every small ion. Real membrane selectivity depends on chemistry as well as size.
Net solvent transfer through a membrane permeable to the solvent but retaining the relevant solute, driven by a solvent chemical-potential difference. With no opposing pressure and dilute solutions, net flow is toward the side with greater effective dissolved-particle concentration.
Osmotic pressure is the pressure that just prevents net flow
Apply additional pressure on the solution side. The pressure makes it less favourable for more solvent to enter. At one particular excess pressure, the two opposing tendencies balance and net flow stops. That minimum opposing pressure difference is osmotic pressure Π. It is not automatically the atmospheric pressure or the total pressure exerted by the liquid. If the applied excess is smaller than Π, normal osmosis can continue; if it exceeds Π sufficiently, the solvent-flow direction reverses.
At fixed temperature, doubling particle concentration doubles the predicted pressure. At fixed composition and volume, raising thermodynamic temperature increases Π. The equation resembles the ideal gas law because both depend on particle number density and temperature. This does not mean that the dissolved solute has become a gas; the solution remains a liquid.
| Pressure and volume units | Suitable R | Required concentration unit |
|---|---|---|
| bar and L | 0.08314 L bar mol⁻¹ K⁻¹ | mol L⁻¹ |
| atm and L | 0.082057 L atm mol⁻¹ K⁻¹ | mol L⁻¹ |
| Pa and m³ | 8.314 J mol⁻¹ K⁻¹ = 8.314 Pa m³ mol⁻¹ K⁻¹ | mol m⁻³ |
A litre is 10⁻³ m³; therefore 1 mol L⁻¹ is 1000 mol m⁻³. When using R=8.314 to obtain pascals, convert volume to cubic metres or concentration to mol m⁻³. Convert Celsius to kelvin by adding 273.15. A common mistake is to combine mol L⁻¹ with the SI gas constant and obtain a pressure too small by a factor of 1000.
Derive the molar-mass expression
Multiply Π=CRT by V, substitute C=n₂/V, then replace amount n₂ by w₂/M₂. Multiply by M₂ and divide by ΠV to isolate molar mass. Unit cancellation leaves mass per mole. Osmotic methods are useful for proteins and polymers because they can measure a detectable pressure in dilute solutions near room temperature, avoiding heating that could damage the molecules. A polydisperse polymer sample yields an effective number-average molar mass in the ideal limit, rather than proving every chain has identical length.
Problem
A 200 cm³ solution contains 1.26 g protein and has osmotic pressure 2.57 × 10⁻³ bar at 300 K. Find molar mass using R=0.08314 L bar mol⁻¹ K⁻¹.
- 1.Convert volume: 200 cm³=200 mL=0.200 L.
- 2.Use M=1.26×0.08314×300/[(2.57×10⁻³)×0.200]=61141.868 g mol⁻¹.
- 3.At the data precision, M≈6.11×10⁴ g mol⁻¹. A more rounded R=0.083 gives about 6.10×10⁴; use one constant consistently.
- 4.Large molar mass means few solute particles per gram, explaining the small measured pressure.
Problem
Dissolve 1.0 g polymer of molar mass 185000 g mol⁻¹ to make approximately 450 mL solution at 37 °C. Estimate its osmotic pressure in Pa.
- 1.Amount=1/185000=5.405405×10⁻⁶ mol. Temperature=310.15 K and volume=450×10⁻⁶=4.50×10⁻⁴ m³.
- 2.Π=nRT/V=(5.405405×10⁻⁶×8.314×310.15)/(4.50×10⁻⁴)=30.9740 Pa.
- 3.Report about 31 Pa. Using 310 K gives about 30.96 Pa, illustrating a harmless final rounding difference.
- 4.If 450 mL refers strictly to solvent rather than final solution, treating them as equal is a dilute-solution approximation.
Compare solutions at the same temperature
| Term | Relative effective osmotic pressure | Net water response of an ideal retained-solute cell model |
|---|---|---|
| Isotonic | Equal | No net osmotic volume change |
| Hypertonic outside | Greater outside | Water leaves and the cell shrinks |
| Hypotonic outside | Lower outside | Water enters and the cell swells |
Equal molarities are isotonic only if they generate the same effective retained-particle concentration. A salt and a sugar at equal analytical molarity need not do so, because the salt can form more than one ion per formula unit. Biological tonicity also depends on whether particular solutes can cross the cell membrane. Therefore equal total measured osmotic pressure and identical long-term cell-volume effects are not always interchangeable claims.
A 0.9% w/v sodium chloride solution is a familiar approximate isotonic reference for red blood cells. Higher retained-solute concentration outside tends to draw water out; lower concentration tends to draw water in. Wilted plant material can regain firmness when water enters its cells. Concentrated salt or sugar can inhibit many microbes by lowering available water and promoting water loss, but neither treatment guarantees that every organism is killed. Root water uptake involves osmotic effects, while long-distance transport also depends on other plant processes.
Reverse osmosis uses work to oppose the natural direction
If the applied pressure difference on the concentrated side exceeds the osmotic-pressure difference, water can be driven from that side through the membrane toward the less concentrated side. This is reverse osmosis. Desalination uses pressure and a supported selective membrane to obtain water with a much smaller dissolved-salt concentration. Cellulose acetate and other polymer membranes are examples. Real membranes have finite salt rejection and can foul; a schematic ideal membrane should not be interpreted as perfectly removing every possible impurity.
Problem
A solution has osmotic pressure 5 bar relative to pure solvent at the same temperature. Predict flow with applied solution-side excess pressures of 0, 5 and 8 bar.
- 1.At 0 bar excess, net solvent enters the solution by osmosis.
- 2.At 5 bar excess, opposing pressure balances the osmotic driving tendency: no net flow in the ideal static model.
- 3.At 8 bar excess, the pressure exceeds the osmotic requirement and drives solvent out of the solution: reverse osmosis.
- 4.Practical flowing systems require additional pressure to overcome hydraulic resistance.
Quick Quiz
Quiz
What crosses an ideal semipermeable membrane during osmosis?
Explanation: The model membrane passes solvent and retains the solute; net solvent transfer can change the solution volume.
Osmotic pressure is:
Explanation: It is an opposing pressure difference required to balance the solvent transfer tendency.
Which temperature is used in Π=CRT?
Explanation: The gas constant form requires thermodynamic temperature, so 37 °C corresponds to 310.15 K.
Under the ideal retained-solute model, a cell placed in a hypertonic surrounding solution tends to:
Explanation: The outside has greater effective osmotic pressure, so net water flows out.
Reverse osmosis requires which condition?
Explanation: Sufficient opposing pressure reverses net solvent flow; boiling is not required.
Practice Problems
- A glucose solution containing 36 g L⁻¹ has Π=4.98 bar at 300 K. What concentration gives 1.52 bar at the same temperature in the dilute proportional model?
- Calculate the osmotic pressure of a 0.0100 M non-electrolyte at 298.15 K with R=0.08314 L bar mol⁻¹ K⁻¹.
- Why is osmotic pressure often more convenient than a thermal shift for determining a protein’s molar mass?
- Two solutions have equal molarity: glucose with i=1 and an ideal fully dissociated 1:1 salt with i=2. Are they isotonic at the same temperature?
- Explain a thistle-funnel osmosis demonstration and the meaning of the rising liquid column.
- Does an 8 bar reverse-osmosis pressure guarantee flow reversal for a solution whose osmotic-pressure difference is 10 bar?
At constant T with the same independent solute, Π is proportional to concentration. New mass concentration=36×1.52/4.98=10.98795 g L⁻¹≈11.0 g L⁻¹. Using glucose 180 g mol⁻¹ gives 0.0610442 mol L⁻¹. The proportional calculation avoids introducing a different R from the reference measurement.
Key Takeaways
• Osmosis is net solvent movement through a selectively permeable membrane. • At equal pressure, solvent tends toward greater effective retained-solute concentration. • Osmotic pressure is an opposing excess pressure, not automatically total pressure. • Dilute ideal solutions obey Π=CRT with temperature in kelvin. • Use final solution volume and units compatible with the chosen gas constant. • Osmotic measurements can determine protein and polymer molar masses. • Biological cell response depends on membrane permeability as well as particle concentration. • Reverse osmosis uses sufficient applied pressure to drive solvent out of a concentrated solution.