Which of the following solutes, when dissolved in 1 kg of water, will lower the freezing point to -3.72 b0C? (Assume complete dissociation and use K_f = 1.86 b0C kg/mol for water.)

• A. 1.0 mol CaCl_2
• B. 2.0 mol C_6H_{12}O_6 (glucose)
• C. 2.0 mol NaCl
• D. 1.0 mol NaCl

Answer: (C)

Verified step by step guidance

1. Identify the freezing point depression formula: \(\Delta T_f = i \cdot K_f \cdot m\), where \(\Delta T_f\) is the freezing point depression, \(i\) is the van't Hoff factor (number of particles the solute dissociates into), \(K_f\) is the freezing point depression constant, and \(m\) is the molality of the solution.
2. Calculate the molality \(m\) for each solute, which is moles of solute per kilogram of solvent. Since the solvent is 1 kg of water, the molality equals the number of moles given for each solute.
3. Determine the van't Hoff factor \(i\) for each solute assuming complete dissociation: for \(CaCl_2\), \(i=3\) (1 Ca\(^{2+}\) and 2 Cl\(^-\)), for glucose \(C_6H_{12}O_6\), \(i=1\) (no dissociation), and for \(NaCl\), \(i=2\) (1 Na\(^+\) and 1 Cl\(^-\)).
4. Calculate the expected freezing point depression \(\Delta T_f\) for each solute using the formula \(\Delta T_f = i \cdot K_f \cdot m\) with \(K_f = 1.86\ ^\circ C \cdot kg/mol\) and the molality from step 2.
5. Compare the calculated \(\Delta T_f\) values to the target freezing point depression of 3.72 \(^\circ C\) (since the freezing point is lowered to -3.72 \(^\circ C\) from 0 \(^\circ C\)) to determine which solute matches this value.