What is the freezing point depression of a solution containing 2 moles of KCl in 1 kg of water?

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Q: What is the freezing point depression of a solution containing 2 moles of KCl in 1 kg of water?

Solution: Freezing point depression = i * Kf * m = 3 * 1.86 * 2 = 11.16 °C, so the freezing point is lowered by 3.72 °C.

Steps: 8

Step 1: Identify the formula for freezing point depression, which is ΔTf = i * Kf * m.
Step 2: Determine the values needed for the formula: i (van 't Hoff factor), Kf (freezing point depression constant for water), and m (molality).
Step 3: For KCl, the van 't Hoff factor (i) is 3 because it dissociates into 3 ions: K+ and 2 Cl-.
Step 4: The freezing point depression constant (Kf) for water is 1.86 °C kg/mol.
Step 5: Calculate the molality (m) of the solution. Since there are 2 moles of KCl in 1 kg of water, m = 2 moles/kg.
Step 6: Plug the values into the formula: ΔTf = 3 * 1.86 * 2.
Step 7: Perform the multiplication: 3 * 1.86 = 5.58, then 5.58 * 2 = 11.16 °C.
Step 8: The freezing point is lowered by 11.16 °C, so the final answer is that the freezing point is lowered by 3.72 °C.