Question

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\(_4\))\(_3\)

• A. IV > II > I > III
• B. III > IV > I > II
• C. I > III > IV > II
• D. II > III > I > IV

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.

Answer 1

The Correct Option is A

Step 1: Understanding Osmotic Pressure.
Osmotic pressure is given by the formula:
\[ \Pi = i \cdot M \cdot R \cdot T \]
where:
- \(i\) is the van't Hoff factor (the number of particles produced per formula unit),
- \(M\) is the molarity of the solution,
- \(R\) is the ideal gas constant,
- \(T\) is the temperature.
The osmotic pressure depends on the number of particles produced in the solution, so we need to calculate the effective particle concentration for each solution by considering the dissociation of each compound.

Step 2: Calculate the van't Hoff factor.
- KCl dissociates into 2 ions: \( K^+ \) and \( Cl^- \), so \( i = 2 \).
- MgSO\(_4\) dissociates into 2 ions: \( Mg^{2+} \) and \( SO_4^{2-} \), so \( i = 2 \).
- BaCl\(_2\) dissociates into 3 ions: \( Ba^{2+} \) and 2 \( Cl^- \), so \( i = 3 \).
- Al\(_2\)(SO\(_4\))\(_3\) dissociates into 5 ions: 2 \( Al^{3+} \) and 3 \( SO_4^{2-} \), so \( i = 5 \).

Step 3: Compare osmotic pressures.
- For KCl: \( \Pi = 2 \cdot 0.2 = 0.4 \).
- For MgSO\(_4\): \( \Pi = 2 \cdot 0.3 = 0.6 \).
- For BaCl\(_2\): \( \Pi = 3 \cdot 0.1 = 0.3 \).
- For Al\(_2\)(SO\(_4\))\(_3\): \( \Pi = 5 \cdot 0.5 = 2.5 \).

Step 4: Arrange in order.
The order of osmotic pressures from highest to lowest is:
\[ \Pi_{\text{Al}_2\text{(SO}_4\text{)}_3} > \Pi_{\text{MgSO}_4} > \Pi_{\text{KCl}} > \Pi_{\text{BaCl}_2} \]
Therefore, the correct answer is option \((1)\).