Water has a molal freezing point constant of 1.86°C kg/mol. Calculate the freezing point of a solution containing 34.335 grams of calcium chloride (CaCl2) dissolved in 163.896 grams of water, assuming complete dissociation.

• A. -7.44°C
• B. -1.86°C
• C. -3.72°C
• D. -5.58°C

Answer: (D)

Verified step by step guidance

1. First, calculate the molar mass of calcium chloride (CaCl2). The molar mass is the sum of the atomic masses of calcium (Ca), chlorine (Cl), and chlorine (Cl) again. Use the periodic table to find these values: Ca = 40.08 g/mol, Cl = 35.45 g/mol. Therefore, the molar mass of CaCl2 is 40.08 + 35.45 + 35.45 g/mol.
2. Next, determine the number of moles of CaCl2 in the solution. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Substitute the mass of CaCl2 (34.335 g) and the molar mass calculated in the previous step to find the moles.
3. Calculate the molality of the solution. Molality (m) is defined as moles of solute per kilogram of solvent. Use the formula: \( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \). Convert the mass of water from grams to kilograms (163.896 g = 0.163896 kg) and use the moles of CaCl2 from the previous step.
4. Consider the van't Hoff factor (i) for CaCl2, which dissociates into three ions: one Ca²⁺ and two Cl⁻ ions. Therefore, \( i = 3 \). Use the formula for freezing point depression: \( \Delta T_f = i \cdot K_f \cdot m \), where \( K_f \) is the molal freezing point depression constant (1.86°C kg/mol). Substitute the values for \( i \), \( K_f \), and molality to find \( \Delta T_f \).
5. Finally, calculate the new freezing point of the solution. The freezing point depression \( \Delta T_f \) is subtracted from the normal freezing point of water (0°C) to find the freezing point of the solution. Use the formula: \( \text{Freezing point} = 0°C - \Delta T_f \).