Osmotic pressure of 0.1 M Glucose at 27°C is π_g. For 0.1 M NaCl at same T, the osmotic pressure is approximately:
Aπ_g
B0
C2π_g (assuming full dissociation, i=2)
Dπ_g/2
Answer & Solution
Correct answer: C. 2π_g (assuming full dissociation, i=2)
NaCl dissociates into Na⁺ + Cl⁻, doubling particle count. So π_NaCl ≈ 2 × π_glucose ≈ 4.92 atm. Glucose stays as molecules, i = 1.
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