Osmotic pressure

Osmosis is the spontaneous flow of solvent molecules from a region of lower solute concentration (or pure solvent) toward a region of higher solute concentration, across a semipermeable membrane — one that lets small solvent molecules through but blocks the larger or hydrated solute particles. Left alone, this flow continues until the concentrations equalise or until an opposing pressure builds up.

Osmotic pressure ($\Pi$) is defined as the excess external pressure that must be applied on the solution side to just stop that net solvent flow. Van't Hoff noticed that dilute solutions obey a relation with exactly the form of the ideal gas equation:

where: $\Pi$ is osmotic pressure (atm or Pa), $M$ is the molarity of the solution ($\text{mol L}^{-1}$), $R$ is the gas constant ($0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}$, or $8.314\ \text{J K}^{-1}\text{mol}^{-1}$ in SI), and $T$ is the absolute temperature (K). The resemblance to $P = \dfrac{n}{V}RT$ is not a coincidence of algebra — it is treated as three van't Hoff law analogues. At constant $T$, $\Pi \propto 1/V$, so osmotic pressure rises directly with concentration (a Boyle's-law echo). At constant concentration, $\Pi \propto T$ (a Charles's-law echo). And solutions of equal molarity at the same temperature exert equal osmotic pressure regardless of what the solute actually is (an Avogadro's-law echo) — this last point is what makes $\Pi$ a true colligative property, depending on the number of solute particles, not their identity. The svg figure below fixes the picture: a membrane separates pure solvent from a solution, solvent crosses toward the solution side, and the applied piston pressure needed to halt that crossing is $\Pi$.

Molar mass from $\Pi$. Since molarity is $M = \dfrac{w_2/M_2}{V}$ (where $w_2$ is the mass of solute, $M_2$ its molar mass, and $V$ the solution volume in litres), substituting into $\Pi = MRT$ gives

Because $\Pi$ is comparatively large even in very dilute solutions — unlike the tiny $\Delta T_f$ or $\Delta T_b$ produced by the same solute — this route is the standard one for high-molar-mass, low-concentration solutes such as proteins and synthetic polymers, where freezing-point or boiling-point shifts would be too small to measure reliably.

Isotonic, hypertonic, hypotonic. Two solutions with equal osmotic pressure are isotonic — no net solvent flow occurs between them across a membrane. Of two unequal solutions, the one with the higher $\Pi$ is hypertonic, and the one with the lower $\Pi$ is hypotonic; solvent always flows from the hypotonic side toward the hypertonic side. This is why 0.9% saline, isotonic with blood plasma, is the fluid used in intravenous drips — anything markedly hypotonic would burst red blood cells by osmotic inflow, and anything markedly hypertonic would shrink them by osmotic outflow.

Reverse osmosis. If a pressure greater than $\Pi$ is applied on the concentrated side, the natural direction of flow reverses: pure solvent is forced from the concentrated solution through the membrane into the dilute side. This is the operating principle behind desalination plants and household water purifiers, at a definitional level — apparatus and engineering detail are outside this syllabus.

A solution containing 6 g of a non-volatile solute dissolved in enough water to make 500 mL of solution has an osmotic pressure of $1.5\ \text{atm}$ at 300 K. Calculate the molar mass of the solute. ($R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}$)

Step 1. Write down the working formula from the derivation above: $M_2 = \dfrac{w_2 RT}{\Pi V}$, with $w_2 = 6\ \text{g}$, $R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}$, $T = 300\ \text{K}$, $\Pi = 1.5\ \text{atm}$, and $V = 500\ \text{mL} = 0.500\ \text{L}$ — converting the volume to litres before substituting.

Step 2. Substitute the numbers: $M_2 = \dfrac{6 \times 0.0821 \times 300}{1.5 \times 0.500}$.

Step 3. Evaluate numerator and denominator separately, then divide: numerator $= 6 \times 0.0821 \times 300 = 147.78$; denominator $= 1.5 \times 0.500 = 0.75$; so $M_2 = \dfrac{147.78}{0.75} = 197.0\ \text{g mol}^{-1}$.

Final answer: the solute has a molar mass of $\boxed{197.0\ \text{g mol}^{-1}}$.

Competitive stretch

Two solutions, A (0.20 M glucose) and B (0.35 M glucose), sit on either side of a semipermeable membrane at the same temperature. Which is hypertonic, and which way does water flow?

Step 1. Since $\Pi = MRT$ at fixed $T$, $\Pi \propto M$, so the higher-molarity solution has the higher osmotic pressure — that is solution B.

Step 2. By definition, the solution with higher $\Pi$ is hypertonic, so B is hypertonic relative to A, and A is hypotonic relative to B.

Step 3. Solvent flows from the hypotonic side to the hypertonic side, so water flows from A into B, diluting B and concentrating A until the two osmotic pressures approach equality.

• Using molality instead of molarity in $\Pi = MRT$ — unlike the freezing-point and boiling-point formulas, the osmotic-pressure equation is built on molarity, not molality.
• Forgetting to convert volume to litres or temperature to kelvin before substituting — a volume left in mL or a temperature left in Celsius silently ruins the arithmetic.
• Confusing osmosis (solvent flow through a semipermeable membrane, driven by a concentration difference) with plain diffusion (movement of any species from high to low concentration, needing no membrane at all).
• Mixing up hypertonic and hypotonic — the solution with the higher osmotic pressure is hypertonic, and solvent flows toward it, not away from it.

Ask the tutor: curious why osmotic pressure gives a much more precise molar mass for a protein than freezing-point depression would — ask the AI tutor to compare the two methods numerically for the same dilute solute.