On this page
Osmotic pressure questions in JEE and NEET often come down to one rearranged formula, M = wRT/πV, but the numbers involved are awkward: tiny pressures, huge molar masses and several possible units for R. This post shows where the formula comes from, works one full example with a polymer, explains why chemists use osmotic pressure rather than freezing point for big molecules, and lists the slips that cost marks.
Where does this come from in NCERT, and is it on the syllabus?
The NCERT Class 12 chapter Solutions (Reprint 2026-27) defines osmotic pressure as the excess pressure that must be applied to a solution to stop solvent passing into it through a semipermeable membrane. For dilute solutions it is proportional to molarity: π = CRT. Writing C = n₂/V and n₂ = w₂/M₂ gives
M₂ = w₂RT/(πV).
The chapter says this method "is widely used to determine molar masses of proteins, polymers and other macromolecules", and its own Example 1.11 finds a protein's molar mass of about 61,000 g/mol.
The JEE (Main) 2026 syllabus lists osmotic pressure and "determination of molecular mass using colligative properties", and the NEET (UG) 2026 syllabus lists the same. The JEE (Advanced) 2026 brochure lists osmotic pressure and the van 't Hoff factor. Check the syllabus for your exam year.
Worked example: the molar mass of a polymer
A chemist dissolves 2.00 g of a polymer in water to make 250 mL of solution. At 27 °C the osmotic pressure is 4.98 × 10⁻³ bar (sample numbers). Find the molar mass.
Convert: V = 0.250 L, T = 300 K, π = 4.98 × 10⁻³ bar.
Choose R to match the units: 0.0831 L bar mol⁻¹ K⁻¹. NCERT rounds this to 0.083.
Substitute: M = (2.00 × 0.0831 × 300)/(4.98 × 10⁻³ × 0.250) = 49.86/1.245 × 10⁻³ ≈ 4.0 × 10⁴ g/mol.
Check in SI units: π = 498 Pa, V = 2.50 × 10⁻⁴ m³, w = 2.00 × 10⁻³ kg, R = 8.314 J mol⁻¹ K⁻¹. M = (2.00 × 10⁻³ × 8.314 × 300)/(498 × 2.50 × 10⁻⁴) ≈ 40.1 kg/mol, the same answer.
Why not use freezing point instead?
Take the same solution. Its concentration is 2.00/40,000/0.250 = 2.0 × 10⁻⁴ mol/L, and for a dilute aqueous solution the molality is almost the same. With NCERT's value Kf = 1.86 K kg mol⁻¹ for water:
ΔTf = 1.86 × 2.0 × 10⁻⁴ ≈ 3.7 × 10⁻⁴ K.
That is far too small to measure with a school thermometer. The osmotic pressure, 498 Pa, would hold up a column of water about 5 cm high (h = π/ρg = 498/(1000 × 9.81) ≈ 0.051 m), which you can see. NCERT gives the reasons in one paragraph: osmotic pressure is measured near room temperature, it uses molarity, and "its magnitude is large even for very dilute solutions". Proteins also tend to be unstable at higher temperatures, which rules out boiling point methods.
What happens with a salt?
Ions count as separate particles, so the formula becomes π = iCRT, where i is the van 't Hoff factor. NCERT notes that van 't Hoff introduced i in 1880, and he received the first Nobel Prize in Chemistry in 1901, cited in part for discovering the laws of osmotic pressure in solutions.
Example: 0.585 g of NaCl in 1.00 L of solution at 300 K. Assuming full dissociation (i = 2), π = 2 × 0.0100 × 0.0831 × 300 ≈ 0.499 bar. If you forget i and feed that pressure into M = wRT/πV, you get 0.585 × 0.0831 × 300/(0.499 × 1.00) ≈ 29 g/mol, half the true 58.5 g/mol. NCERT's table gives i = 1.94 for NaCl at 0.01 m, so the real pressure is a little lower than the ideal estimate.
How does this connect to isotonic solutions?
NCERT says the fluid inside blood cells has the same osmotic pressure as 0.9% (mass/volume) sodium chloride, called normal saline. Work out an ideal estimate of that pressure at 37 °C (310 K):
0.9 g per 100 mL is 9.0 g/L, or 9.0/58.44 = 0.154 mol/L of NaCl.
With i = 2, the particle concentration is 0.308 mol/L.
π ≈ 0.308 × 0.0831 × 310 ≈ 7.9 bar.
A glucose solution with the same osmotic pressure would need 0.308 mol/L of glucose, about 55 g/L. This is a classroom estimate only, using ideal behaviour.
Which mistakes cost marks?
Using mL for V when R is in litre units.
Celsius for T. 27 °C is 300 K.
Mismatched R and pressure units. Convert first, then substitute.
Ignoring i for electrolytes, which halves the apparent molar mass for NaCl.
Using molality in π = CRT. Osmotic pressure uses molarity, one of NCERT's stated advantages.
For the other three colligative properties and the van 't Hoff factor in detail, see our colligative properties guide.
A practice routine
Redo NCERT Example 1.11 without the book, then repeat it in SI units.
Solve NCERT Exercise 1.22, which compares two glucose solutions at 300 K.
Find the freezing point depression for the protein in Example 1.11 and compare it with the osmotic pressure.
Take one electrolyte question and solve it twice, with i = 1 and with the ideal i, and explain the difference.
If you keep getting the units wrong, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, redo the polymer example with π given as 4.92 × 10⁻³ atm and check that you get the same molar mass.
