Question

Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?

• A. \(C_6H_{12}O_6\)
• B. \(Al_2(SO_4)_3\)
• C. \(K_2SO_4\)
• D. \(KCl\)

Answer 1

The Correct Option is B

To determine which 0.10 m aqueous solution will exhibit the largest freezing point depression, we need to understand the concept of colligative properties. Freezing point depression (\(\Delta T_f\)) is a colligative property, which depends on the number of solute particles in a solution and not on their identity.

The formula for freezing point depression is given by:

\(\Delta T_f = i \cdot K_f \cdot m\)

where:

• \(\Delta T_f\) is the freezing point depression.
• i is the van't Hoff factor, which represents the number of particles into which a compound dissociates in solution.
• K_f is the freezing point depression constant of the solvent (for water, it's a known constant).
• m is the molality of the solution.

For equal molality solutions, the key factor affecting freezing point depression is the van't Hoff factor i:

• \(C_6H_{12}O_6\) (glucose) doesn't ionize; i = 1.
• \(Al_2(SO_4)_3\) dissociates into 2 Al^{3+} ions and 3 SO_4^{2-} ions; i = 2 + 3 = 5.
• \(K_2SO_4\) dissociates into 2 K^+\) and 1 SO_4^{2-} ion; i = 2 + 1 = 3.
• \(KCl\) dissociates into 1 K^+ and 1 Cl^-\); i = 1 + 1 = 2.

Comparing these values, \(Al_2(SO_4)_3\) with i = 5 has the highest van't Hoff factor, indicating the maximum number of particles produced upon dissociation.

Since the freezing point depression is directly proportional to the van't Hoff factor i, the solution of \(Al_2(SO_4)_3\) will exhibit the largest freezing point depression.

Therefore, the correct answer is:
\(Al_2(SO_4)_3\)