What Osmosis and Osmotic Pressure Are
A semipermeable membrane (SPM) is a film whose pores allow small solvent molecules such as water to pass while hindering the passage of larger solute molecules. When such a membrane separates pure solvent from a solution, solvent molecules flow spontaneously through the membrane into the solution. This net flow of solvent across a semipermeable membrane is called osmosis.
The flow continues until equilibrium is reached. It can be stopped earlier by applying an extra pressure on the solution side. The exact pressure that just stops the inflow of solvent is the osmotic pressure ($\Pi$) of the solution. Equivalently, osmotic pressure is the excess pressure that must be applied to a solution to prevent the passage of solvent molecules into it through a semipermeable membrane. The key directional rule is that solvent always migrates from a region of lower solute concentration to one of higher solute concentration.
Because the inflow of solvent depends on the difference in solute concentration and not on the chemical nature of the solute, osmotic pressure is classified as a colligative property — it counts particles, not identities. This places it alongside relative lowering of vapour pressure, elevation of boiling point and depression of freezing point.
The Osmotic Pressure Equation
For dilute solutions, experiment shows that the osmotic pressure is directly proportional to the molar concentration (molarity) $C$ of the solution at a fixed temperature $T$. This gives the central relation:
$$ \Pi = C\,R\,T $$Writing the molarity as moles of solute $n_2$ per litre of solution volume $V$, the same relation becomes the more useful working form:
$$ \Pi = \frac{n_2}{V}\,R\,T \qquad\Longrightarrow\qquad \Pi\,V = n_2\,R\,T $$Here $R$ is the gas constant. The structural similarity to the ideal-gas equation $PV = nRT$ is deliberate and worth remembering, but $\Pi$ is a property of the solution, not of a gas. The value of $R$ chosen must match the pressure unit: $R = 0.083\ \text{L bar mol}^{-1}\text{K}^{-1}$ when $\Pi$ is in bar, or $R = 0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}$ when $\Pi$ is in atm.
| Symbol | Meaning | Common Units |
|---|---|---|
Π | Osmotic pressure | bar, atm, Pa |
C | Molar concentration (molarity) | mol L⁻¹ |
n₂ | Moles of solute | mol |
V | Volume of solution | L |
R | Gas constant | 0.083 L bar mol⁻¹ K⁻¹ |
T | Absolute temperature | K |
A plot of $\Pi$ against $C$ at constant $T$ is therefore a straight line through the origin whose slope equals $RT$. NEET has used precisely this graphical reading to back-calculate the temperature of measurement, so the slope-equals-$RT$ idea is examination-critical.
Molar Mass from Osmotic Pressure
If $w_2$ grams of a solute of molar mass $M_2$ are dissolved, then $n_2 = w_2 / M_2$, and substituting into $\Pi V = n_2 RT$ gives the molar-mass form:
$$ M_2 = \frac{w_2\,R\,T}{\Pi\,V} $$Knowing $w_2$, $T$, $\Pi$ and $V$, the molar mass follows directly. This is the single most important application of osmotic pressure: it is the standard route to the molar masses of proteins, polymers and other macromolecules. The method has decisive advantages over freezing-point or boiling-point methods — measurements are made near room temperature so heat-sensitive biomolecules do not decompose, molarity is used instead of molality, and crucially the osmotic pressure remains large and measurable even in very dilute solution, the only regime in which poorly soluble macromolecules can be studied.
200 cm³ of an aqueous solution of a protein contains 1.26 g of the protein. Its osmotic pressure at 300 K is $2.57 \times 10^{-3}$ bar. Find the molar mass of the protein. (NCERT Example 1.11)
Given $w_2 = 1.26\ \text{g}$, $V = 200\ \text{cm}^3 = 0.200\ \text{L}$, $T = 300\ \text{K}$, $\Pi = 2.57\times10^{-3}\ \text{bar}$, $R = 0.083\ \text{L bar mol}^{-1}\text{K}^{-1}$.
$$ M_2 = \frac{w_2 R T}{\Pi V} = \frac{1.26 \times 0.083 \times 300}{2.57\times10^{-3} \times 0.200} $$
$$ M_2 \approx 61{,}022\ \text{g mol}^{-1} $$
The large molar mass (≈ 61 kg mol⁻¹) is exactly the regime where only osmotic pressure gives a reliably measurable signal.
Calculate the osmotic pressure (in pascals) of a solution made by dissolving 1.0 g of a polymer of molar mass 185,000 in 450 mL of water at 37 °C. (NCERT Intext 1.12)
$n_2 = \dfrac{1.0}{185000} = 5.405\times10^{-6}\ \text{mol}$, $V = 0.450\ \text{L}$, $T = 310\ \text{K}$, $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$.
$$ \Pi = \frac{n_2 R T}{V} = \frac{5.405\times10^{-6} \times 8.314 \times 310}{0.450 \times 10^{-3}\ \text{m}^3} $$
$$ \Pi \approx 30.96\ \text{Pa} $$
Even this minuscule pressure (about 31 Pa) is measurable, which is why osmometry works for huge molar masses; a freezing-point depression for the same solution would be undetectably small.
See why small molecules are better handled by depression of freezing point while macromolecules need osmometry.
Isotonic, Hypertonic and Hypotonic Solutions
Two solutions that have the same osmotic pressure at a given temperature are called isotonic solutions. When isotonic solutions are separated by a semipermeable membrane, no net osmosis occurs between them because there is no driving concentration difference. The classic biological example is that the fluid inside a blood cell has an osmotic pressure equal to that of $0.9\%$ (mass/volume) $\ce{NaCl}$ — the normal saline solution that is therefore safe to inject intravenously.
| Solution type | Osmotic pressure vs. cell | Direction of water flow | Effect on cell |
|---|---|---|---|
| Isotonic (≈ 0.9% NaCl) | Equal | No net flow | Cell unchanged |
| Hypertonic (> 0.9% NaCl) | Higher | Out of the cell | Cell shrinks |
| Hypotonic (< 0.9% NaCl) | Lower | Into the cell | Cell swells |
Everyday phenomena follow the same logic. A raw mango in concentrated salt solution loses water by osmosis and shrivels into pickle; wilted flowers and a limp carrot revive in fresh water as water flows back into their cells. Preservation of meat by salting and of fruits by sugar works because the surrounding hypertonic medium draws water out of any bacterium, which then shrivels and dies. Excess dietary salt causes water retention in tissue spaces by osmosis, producing the swelling called edema.
Two traps in one topic
Trap 1 — best method for polymers. When a question asks which colligative property is most suitable for determining the molar mass of a protein or polymer, the answer is osmotic pressure, not freezing-point depression. Osmotic pressure stays measurably large in dilute solution and needs only room-temperature conditions.
Trap 2 — isotonic means equal Π, not equal concentration. Isotonic solutions share the same osmotic pressure. For two non-electrolytes that means equal molarity, but a 0.1 M $\ce{NaCl}$ solution ($i \approx 2$) is isotonic with roughly 0.2 M glucose ($i = 1$), not with 0.1 M glucose.
Macromolecule molar mass → osmotic pressure. Isotonic → equal $\Pi$, so always compare $iC$, never just $C$.
Reverse Osmosis and Water Purification
The direction of osmosis can be reversed. If a pressure larger than the osmotic pressure is applied to the solution side, pure solvent is forced out of the solution through the semipermeable membrane — the opposite of natural osmosis. This phenomenon is called reverse osmosis and is of great practical utility, most famously in the desalination of sea water.
The pressure required for reverse osmosis is quite high. A workable porous membrane is a film of cellulose acetate placed over a suitable support; it is permeable to water but impermeable to the impurities and ions present in sea water. A variety of polymer membranes are available for the purpose. Many countries now run desalination plants based on reverse osmosis to meet their potable-water requirements, and the same principle drives domestic RO water purifiers.
Electrolytes and the van't Hoff Factor
The relation $\Pi = CRT$ assumes the solute neither dissociates nor associates. When it does, the effective particle count changes and the van't Hoff factor $i$ is introduced, so the osmotic pressure becomes:
$$ \Pi = \frac{i\,n_2\,R\,T}{V} = i\,C\,R\,T $$For an electrolyte like $\ce{KCl}$ that dissociates into $\ce{K+}$ and $\ce{Cl-}$, $i$ is close to 2, so the osmotic pressure is nearly double that of an equimolar non-electrolyte. For an associating solute (such as benzoic acid dimerising in benzene) $i$ is less than one. This factor is why osmotic-pressure comparisons must always be made on $iC$, not $C$ alone, and it is the bridge to the wider treatment of abnormal molar masses.
A solution of 36 g glucose ($\ce{C6H12O6}$, M = 180) in 1 L at 300 K has $\Pi = 4.98$ bar. What molar concentration of an isotonic non-electrolyte solution at 300 K gives $\Pi = 1.52$ bar? (NCERT Intext 1.22)
Since $\Pi = CRT$ at fixed $T$, concentration is directly proportional to $\Pi$:
$$ C_1 = \frac{36}{180} = 0.20\ \text{mol L}^{-1}\ \text{(gives 4.98 bar)} $$
$$ \frac{C_2}{C_1} = \frac{\Pi_2}{\Pi_1} \;\Rightarrow\; C_2 = 0.20 \times \frac{1.52}{4.98} \approx 0.061\ \text{mol L}^{-1} $$
Any non-electrolyte solution at 0.061 M would be isotonic with a 1.52 bar reference, since equal $\Pi$ at equal $T$ means equal $C$ for non-electrolytes.
Osmotic Pressure in One Screen
- Osmosis: solvent flows across a semipermeable membrane from dilute to concentrated; $\Pi$ is the pressure that just stops it.
- $\Pi = CRT = \dfrac{n_2}{V}RT$; it is a colligative property (counts particles, not identity).
- Molar mass: $M_2 = \dfrac{w_2 RT}{\Pi V}$ — the best method for proteins and polymers (large signal, room temperature, very dilute).
- Isotonic = equal $\Pi$; hypertonic draws water out (cell shrinks); hypotonic pushes water in (cell swells); 0.9% NaCl is isotonic with blood.
- Reverse osmosis: apply $P > \Pi$ to push pure water out through cellulose acetate — used for desalination.
- For electrolytes, use $\Pi = iCRT$; KCl gives $i \approx 2$.