Question

Arrange the following solutions in the correct order of their osmotic pressures:
(A) 0.1 M NaCl
(B) 0.1 M Urea
(C) 0.1 M BaCl\(_2\)

• A. \(B > C > A\)
• B. \(C > B > A\)
• C. \(C > A > B\)
• D. \(A > B > C\)

Hint:

Higher van’t Hoff factor \(i\) means higher osmotic pressure for same molarity solution.

Answer 1

The Correct Option is C

Step 1: Osmotic pressure dependence.
π = iMRT, so at identical molarity and temperature, π scales strictly with the van't Hoff factor i.

Step 2: Dissociation into ions.
C (BaCl₂) gives Ba²⁺ + 2Cl⁻ → i=3; A (NaCl) gives Na⁺ + Cl⁻ → i=2; B (Urea) does not dissociate → i=1.

Step 3: Ordering by i.
Larger i produces greater osmotic pressure: BaCl₂ (3)>NaCl (2)>Urea (1).

Step 4: Final sequence.
Therefore, C>A>B.

Step 5: Rationale.
Electrolytes generate more solute particles, enhancing colligative effects relative to non-electrolytes.