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: Understanding osmotic pressure concept.
Osmotic pressure is given by: \[ \pi = iMRT \] where \(i\) is van’t Hoff factor, \(M\) is molarity, \(R\) is gas constant, and \(T\) is temperature. Since all solutions have same concentration (0.1 M) and same temperature, osmotic pressure depends only on van’t Hoff factor \(i\).

Step 2: Determining van’t Hoff factor.
For each solute: - \(C = BaCl_2 \rightarrow Ba^{2+} + 2Cl^- \Rightarrow i = 3\)
- \(A = NaCl \rightarrow Na^+ + Cl^- \Rightarrow i = 2\)
- \(B = Urea \rightarrow no dissociation \Rightarrow i = 1\)

Step 3: Comparing osmotic pressures.
Since \(\pi \propto i\), higher \(i\) means higher osmotic pressure: \[ BaCl_2 (3) > NaCl (2) > Urea (1) \]

Step 4: Writing final order.
Thus: \[ C > A > B \]

Step 5: Final justification.
Electrolytes produce more particles in solution, increasing colligative properties like osmotic pressure compared to non-electrolytes.
Final Answer: \[ \boxed{C > A > B} \]