Arrange 0.1 M aqueous solution of the following in decreasing order of their osmotic pressure
(A) Ammonium phosphate
(B) Glucose
(C) Sodium chloride
(D) Potassium sulphate
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (D), (C), (B)
Osmotic pressure is a colligative property: it depends on the number of solute particles in the solution, not on their nature. For an electrolyte, the van’t Hoff factor (i) counts the particles formed from one formula unit:
\( \displaystyle \pi=i\,CRT \)
All four solutions have the same molar concentration (0.1 M) at the same temperature, so π depends only on i. Assuming complete dissociation of the salts:
\( \displaystyle \ce{(NH4)3PO4 -> 3NH4^+ + PO4^3-}\quad i=4 \)
\( \displaystyle \ce{K2SO4 -> 2K^+ + SO4^2-}\quad i=3 \)
\( \displaystyle \ce{NaCl -> Na^+ + Cl^-}\quad i=2 \)
\( \displaystyle \text{Glucose does not dissociate}\quad i=1 \)
Matching the given compounds:
(A) Ammonium phosphate, \( \ce{(NH4)3PO4} \): i = 4
(D) Potassium sulphate, \( \ce{K2SO4} \): i = 3
(C) Sodium chloride, NaCl: i = 2
(B) Glucose, \( \ce{C6H12O6} \): i = 1
Shortcut: Ammonium phosphate (A) gives the most ions and glucose (B) gives none. Only Option (2) starts with (A) and ends with (B).