Calculate osmotic pressure of 0.1M glucose solution at 27°C

Question

Calculate the osmotic pressure of a 0.1 M aqueous solution of glucose at 27°C. (R = 0.0821 L·atm·mol⁻¹·K⁻¹)

Solution — Step by Step

The osmotic pressure formula is analogous to the ideal gas law:

\pi = iCRT

Where:

$\pi$ = osmotic pressure (atm)
$i$ = van ‘t Hoff factor (number of solute particles per formula unit)
$C$ = molar concentration (mol/L)
$R$ = gas constant = 0.0821 L·atm·mol⁻¹·K⁻¹
$T$ = temperature in Kelvin

Glucose ( $C_6H_{12}O_6$ ) is a non-electrolyte — it does NOT dissociate in solution. One molecule of glucose remains as one particle in solution.

Therefore, $i = 1$ for glucose.

(Compare with NaCl where $i = 2$ , since it gives Na⁺ and Cl⁻.)

T = 27°C + 273 = 300 \text{ K}

\pi = iCRT = 1 \times 0.1 \times 0.0821 \times 300

\pi = 1 \times 0.1 \times 24.63

\pi = 2.463 \text{ atm}

Why This Works

Osmotic pressure is a colligative property — it depends on the number of solute particles, not their nature. The equation $\pi = iCRT$ comes from the van ‘t Hoff equation, which is mathematically identical to the ideal gas law. This analogy is not accidental: both relate to the tendency of particles to occupy space and exert pressure.

Osmotic pressure is the pressure needed to prevent osmosis (net flow of water across a semipermeable membrane from low to high solute concentration). Glucose solutions have high osmotic pressure because even at 0.1 M, the number of dissolved molecules is substantial.

At 25°C, the osmotic pressure of a 1 M non-electrolyte solution is approximately 24.5 atm — a useful benchmark. At 27°C (300 K), it’s $1 \times 1 \times 0.0821 \times 300 = 24.63$ atm. For a 0.1 M solution, it’s one-tenth of that: 2.463 atm. Remembering this benchmark helps you sanity-check numerical answers.

Extension — Converting to Other Units

1 atm = 101.325 kPa, so:

\pi = 2.463 \text{ atm} \times 101.325 \text{ kPa/atm} = 249.6 \text{ kPa} \approx 250 \text{ kPa}

Also: 1 atm ≈ 760 mmHg, so $\pi \approx 2.463 \times 760 \approx 1872$ mmHg — this is quite large, much greater than atmospheric pressure!

Common Mistake

The most common error is forgetting to convert temperature from Celsius to Kelvin. Using $T = 27$ instead of $T = 300$ gives $\pi = 0.222$ atm — an answer that is about 11 times too small. In all gas law and colligative property calculations, temperature must be in Kelvin. 27°C → 300 K is the standard “nice” temperature in JEE and CBSE problems, so recognise it immediately.

Question

Solution — Step by Step

Why This Works

Extension — Converting to Other Units

Common Mistake

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