Arrange the following solutions according to decreasing order of osmotic pressure under similar condition of temperature and assuming complete dissociation.
I. 0.2 m KCl
II. 0.3 m MgSO\(_4\)
III. 0.1 m BaCl\(_2\)
IV. 0.5 m Al\(_2\)(SO\(_4\))\(_3\)
Show Hint
When calculating osmotic pressure, remember that the more ions a solute dissociates into, the higher the osmotic pressure, even if the molarity is lower.
Step 1: Understanding the Question:
Osmotic pressure (\( \pi \)) is a colligative property proportional to the total concentration of particles in solution (\( i \times C \)). Step 2: Key Formula or Approach:
Calculate \( \pi \propto i \times C \).
I. \( \text{KCl} \rightarrow \text{K}^+ + \text{Cl}^- \) (\( i = 2 \))
II. \( \text{MgSO}_4 \rightarrow \text{Mg}^{2+} + \text{SO}_4^{2-} \) (\( i = 2 \))
III. \( \text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\text{Cl}^- \) (\( i = 3 \))
IV. \( \text{Al}_2(\text{SO}_4)_3 \rightarrow 2\text{Al}^{3+} + 3\text{SO}_4^{2-} \) (\( i = 5 \)) Step 3: Detailed Explanation:
Calculate effective concentration (\( i \cdot m \)):
I. \( 2 \times 0.2 = 0.4 \)
II. \( 2 \times 0.3 = 0.6 \)
III. \( 3 \times 0.1 = 0.3 \)
IV. \( 5 \times 0.5 = 2.5 \)
Decreasing order of effective concentration: \( 2.5 (\text{IV})>0.6 (\text{II})>0.4 (\text{I})>0.3 (\text{III}) \). Step 4: Final Answer:
The decreasing order of osmotic pressure is \( \text{IV}>\text{II}>\text{I}>\text{III} \).