Calculate osmotic pressure and understand reverse osmosis.

Osmotic Pressure — π = iCRT

Question

A 0.1 M solution of glucose (non-electrolyte) and a 0.1 M solution of NaCl are prepared at 27°C. Calculate the osmotic pressure of each. (R = 0.0831 L·bar·mol⁻¹·K⁻¹)

Solution — Step by Step

The osmotic pressure formula is:

\pi = iCRT

where $i$ is the van’t Hoff factor (number of particles the solute breaks into), $C$ is molarity, $R$ is the gas constant, and $T$ is temperature in Kelvin.

For glucose: $i = 1$ (it doesn’t ionise). For NaCl: $i = 2$ (NaCl → Na⁺ + Cl⁻).

T = 27 + 273 = 300 \text{ K}

Always convert before plugging in. R is in L·bar units, so our answer will be in bar directly.

\pi_{\text{glucose}} = 1 \times 0.1 \times 0.0831 \times 300

\pi_{\text{glucose}} = 2.493 \text{ bar}

\pi_{\text{NaCl}} = 2 \times 0.1 \times 0.0831 \times 300

\pi_{\text{NaCl}} = 4.986 \text{ bar} \approx 4.99 \text{ bar}

NaCl gives exactly double the osmotic pressure because it produces twice as many particles in solution.

Final Answers:

Osmotic pressure of glucose solution = 2.49 bar
Osmotic pressure of NaCl solution = 4.99 bar

Why This Works

Osmotic pressure is a colligative property — it depends only on the number of solute particles, not on what they are. This is why we multiply by $i$ before anything else. A 0.1 M NaCl solution behaves like a 0.2 M solution of particles.

The formula $\pi = iCRT$ looks identical to the ideal gas law $PV = nRT$ — and that’s not a coincidence. Van’t Hoff showed that solute particles in dilute solution behave much like gas molecules, exerting a “pressure” through the semipermeable membrane.

This is why seawater is so difficult to desalinate via reverse osmosis — the osmotic pressure of seawater (~27 bar) must be exceeded to push water back through the membrane. That’s a direct application of this formula and has appeared in JEE Main 2023 as a conceptual question.

Alternative Method — Checking with Isotonic Solutions

If two solutions have the same osmotic pressure, they’re isotonic. We can use this to quickly verify our work or solve comparison problems.

For two solutions to be isotonic:

i_1 C_1 = i_2 C_2

So a 0.1 M NaCl solution ( $i = 2$ , effective concentration = 0.2 M) is isotonic with a 0.2 M glucose solution ( $i = 1$ , effective concentration = 0.2 M).

This shortcut is extremely useful in NEET MCQs where you’re asked which pair of solutions is isotonic — you never need to calculate $\pi$ explicitly, just compare $iC$ values.

In NEET and CBSE boards, isotonic pair questions appear almost every year. The trick: just compare $iC$ for each solution. Whichever pair has equal $iC$ is the isotonic pair. No calculator needed.

Common Mistake

The most common error is forgetting the van’t Hoff factor $i$ for electrolytes — or worse, using $i = 2$ for weak electrolytes like acetic acid (CH₃COOH) when they’re only partially dissociated. Acetic acid in water gives $i$ between 1 and 2, depending on degree of dissociation $\alpha$ : use $i = 1 + \alpha$ for a binary electrolyte. For strong electrolytes like NaCl in dilute solution, $i = 2$ is safe. For questions specifying “0.1 M NaCl (assume complete dissociation)”, always use $i = 2$ .

Also watch the units of $R$ . If the question gives $R = 8.314 \text{ J·mol}^{-1}\text{·K}^{-1}$ , your answer comes out in Pascals, not bar. The CBSE 2024 board exam used $R = 0.0831 \text{ L·bar·mol}^{-1}\text{·K}^{-1}$ to keep the answer clean — that’s the value to memorise for boards.

Question

Solution — Step by Step

Why This Works

Alternative Method — Checking with Isotonic Solutions

Common Mistake

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