Osmotic pressure — determine molecular weight of solute using π = CRT

Question

2.5 g of a non-volatile, non-electrolyte solute is dissolved in 100 mL of water. The osmotic pressure of the solution at 27°C is found to be 2.46 atm. Determine the molecular weight of the solute.

(NCERT Class 12, Chapter 2)

Solution — Step by Step

For a dilute solution:

\pi = CRT = \frac{n}{V}RT = \frac{w}{MV}RT

where $\pi$ is osmotic pressure, $w$ is mass of solute, $M$ is molar mass, $V$ is volume in litres, $R = 0.0821$ L atm K $^{-1}$ mol $^{-1}$ , and $T$ is temperature in Kelvin.

$w = 2.5$ g, $V = 100$ mL $= 0.1$ L, $T = 27 + 273 = 300$ K, $\pi = 2.46$ atm.

M = \frac{wRT}{\pi V}

M = \frac{2.5 \times 0.0821 \times 300}{2.46 \times 0.1}

M = \frac{61.575}{0.246}

\boxed{M = 250.3 \text{ g/mol} \approx 250 \text{ g/mol}}

Why This Works

Osmotic pressure is a colligative property — it depends on the number of solute particles, not their identity. This makes it ideal for determining molecular weight, especially for large molecules (polymers, proteins) where other methods (boiling point elevation, freezing point depression) give very small, hard-to-measure changes.

The van’t Hoff equation $\pi = CRT$ has the same form as the ideal gas law ( $PV = nRT$ ). This is not a coincidence — both arise from the kinetic theory of particles in a dilute medium. The solute particles create pressure on the semipermeable membrane just as gas molecules create pressure on container walls.

Alternative Method

For electrolytes, the formula is modified by the van’t Hoff factor $i$ : $\pi = iCRT$ . For NaCl ( $i = 2$ ), the osmotic pressure is twice that of a non-electrolyte at the same concentration. If the problem mentions an electrolyte, always include $i$ .

Osmotic pressure is the most sensitive colligative property for molecular weight determination. Even a small number of moles produces a measurable osmotic pressure. This is why it is the preferred method for determining molecular weights of proteins and polymers — where masses are large but mole numbers are small.

Common Mistake

The two most common errors: (1) forgetting to convert temperature to Kelvin — using 27 instead of 300 gives an answer that is off by a factor of 10, and (2) using volume in mL instead of L. Always convert: $V$ must be in litres when using $R = 0.0821$ L atm K $^{-1}$ mol $^{-1}$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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