What is the freezing point depression of a solution containing 2 moles of NaCl in 1 kg of water? (Kf for water = 1.86 °C kg/mol)

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Q: What is the freezing point depression of a solution containing 2 moles of NaCl in 1 kg of water? (Kf for water = 1.86 °C kg/mol)

Solution: Freezing point depression = i * Kf * m = 3 * 1.86 * 2 = 11.16 °C.

Steps: 7

Step 1: Identify the formula for freezing point depression, which is: ΔTf = i * Kf * m.
Step 2: Determine the van 't Hoff factor (i) for NaCl. Since NaCl dissociates into 2 ions (Na+ and Cl-), i = 2.
Step 3: Identify the Kf value for water, which is given as 1.86 °C kg/mol.
Step 4: Calculate the molality (m) of the solution. Since there are 2 moles of NaCl in 1 kg of water, m = 2 moles/kg.
Step 5: Substitute the values into the formula: ΔTf = i * Kf * m = 2 * 1.86 * 2.
Step 6: Perform the multiplication: 2 * 1.86 = 3.72, then 3.72 * 2 = 7.44 °C.
Step 7: The freezing point depression is 7.44 °C.