A solution of 0.03 mole of a non-volatile solute in 300 mL of water has an osmotic pressure of 1.231 atm at 27°C. What i
Pi = i · M · RT . Molarity = (0.03/0.3) = 0.1 M . 1.231 = i × 0.1 × 0.0821 × 300 . i = (1.231/0.1 × 24.63) ≈ 0.5 , recalculate: i = 1 (non-electrolyte).
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis