For the equimolal solutions, increasing order of osmotic pressure is
- 1 m \(C_6H_5COOH\)
- 1 m \(BaCl_2\)
- 1 m \(KCl\)
- 1 m glucose
- 1 m \(H_3PO_4\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A < D < C < B < E
Osmotic pressure is a colligative property: \( \pi = iCRT \). For solutions of the same concentration at the same temperature, π depends only on the van’t Hoff factor i, the number of particles each formula unit gives in solution. Association lowers i below 1; dissociation raises it above 1.
The question uses the limiting values of i:
A. 1 m C6H5COOH — benzoic acid associates into dimers through hydrogen bonding, as carboxylic acids do in a non-polar solvent such as benzene, \( \ce{2C6H5COOH <=> (C6H5COOH)2} \); i is about 0.5.
D. 1 m glucose — a non-electrolyte; i = 1.
C. 1 m KCl — \( \ce{KCl -> K+ + Cl-} \); i = 2.
B. 1 m BaCl2 — \( \ce{BaCl2 -> Ba^2+ + 2Cl-} \); i = 3.
E. 1 m H3PO4 — taken as fully dissociated, \( \ce{H3PO4 -> 3H+ + PO4^3-} \); i = 4.
Shortcut: for equimolal solutions the order of any colligative property is the order of i; there is no need to calculate π.
NCERT (2026-27 edition):
Abnormal Molar Masses – “In case of association, value of i is less than unity while for dissociation it is greater than unity. For example, the value of i for aqueous KCl solution is close to 2, while the value for ethanoic acid in benzene is nearly 0.5.”