Question

1. homemade ice cream mix

an ice-salt bath is made using 0.500 kg of water and 0.200 kg of cacl₂.

• assume complete dissociation: i=3
• kf=1.86

what is the freezing point of the salt mixture surrounding the ice cream container?

Explanation:

Step1: Calculate moles of CaCl₂

Molar mass of CaCl₂ = 40.08 + 2×35.45 = 110.98 g/mol
Moles of CaCl₂ = $\frac{0.200\ \text{kg}×1000\ \text{g/kg}}{110.98\ \text{g/mol}} ≈ 1.802\ \text{mol}$

Step2: Find molality of solution

Molality (m) = $\frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{1.802\ \text{mol}}{0.500\ \text{kg}} = 3.604\ \text{mol/kg}$

Step3: Calculate freezing point depression

$\Delta T_f = i×K_f×m = 3×1.86×3.604 ≈ 20.1\ ^\circ\text{C}$

Step4: Determine freezing point

Freezing point = $0^\circ\text{C} - 20.1^\circ\text{C} = -20.1^\circ\text{C}$

Answer:

-20.1°C