Which of the following solutes will lower the freezing point of 1 L H_2O the most when 1 mole is dissolved in it?

• A. KNO_3
• B. NaCl
• C. CaCl_2
• D. C_6H_{12}O_6 (glucose)

Answer: (C)

Verified step by step guidance

1. Identify that the freezing point depression depends on the number of solute particles in solution, not just the amount of solute in moles. This is described by the formula for freezing point depression: \(\Delta T_f = i \cdot K_f \cdot m\), where \(i\) is the van't Hoff factor, \(K_f\) is the freezing point depression constant of the solvent, and \(m\) is the molality of the solution.
2. Determine the van't Hoff factor (\(i\)) for each solute, which represents the number of particles the compound dissociates into in solution: for example, glucose (\(C_6H_{12}O_6\)) does not dissociate, so \(i=1\); NaCl dissociates into Na\(^+\) and Cl\(^-\), so \(i=2\); KNO\(_3\) dissociates into K\(^+\) and NO\(_3^-\), so \(i=2\); CaCl\(_2\) dissociates into Ca\(^{2+}\) and 2 Cl\(^-\) ions, so \(i=3\).
3. Since 1 mole of each solute is dissolved in 1 L of water, and assuming the density of water is close to 1 kg/L, the molality \(m\) is approximately 1 mol/kg for all solutions, making \(m\) constant across all solutes.
4. Calculate the effective number of particles in solution by multiplying the moles of solute by the van't Hoff factor \(i\). The solute with the highest \(i\) will produce the greatest number of particles, leading to the largest freezing point depression.
5. Conclude that CaCl\(_2\), with the highest van't Hoff factor of 3, will lower the freezing point of water the most when 1 mole is dissolved, compared to the other solutes listed.