Question

Match the solutions to the descriptions of the freezing points.

a. One mole of the ionic compound Na3PO4 dissolved in 1000 g H2O
b. One mole of the ionic compound CuSO4 dissolved in 1000 g H2O
c. One mole of the nonelectrolyte C6H12O6 dissolved in 1000 g H2O

1. Highest freezing point
2. Intermediate freezing point
3. Lowest freezing point

Answer 1

1c

2b

3a

Step-by-step explanation:

Step 1: Write the general expression to calculate the freezing point depression

The freezing point depression is a colligative property, that can be calculated using the following expression.

ΔT = i × Kc × b

where,

i: van 't Hoff factor (number of ion particles per formula unit)

Kc: cryoscopic constant (Kc for water: 1.86 °C.Kg/mol)

b: molality (moles of solute per kilogram of solvent)

All the solution have the same Kc and the same b (1 mol/1 kg = 1 m), so ΔTf variation will depend on i

Step 2: Calculate the freezing point of the Na₃PO₄ solution

Na₃PO₄ has 4 ions (3 Na⁺ and 1 PO₄³⁻), so i = 4.

ΔT = i × Kc × b = 4 × 1.86 °C.Kg/mol × 1 mol/kg = 7.44 °C

T = 0°C - 7.44 °C = -7.44 °C

Step 3: Calculate the freezing point of the CuSO₄ solution

CuSO₄ has 2 ions (1 Cu²⁺ and 1 SO₄²⁻), so i = 2.

ΔT = i × Kc × b = 2 × 1.86 °C.Kg/mol × 1 mol/kg = 3.72 °C

T = 0°C - 3.72 °C = -3.72 °C

Step 4: Calculate the freezing point of the C₆H₁₂O₆ solution

C₆H₁₂O₆ is a nonelectrolyte (it doesn't ionize), so i = 1

ΔT = i × Kc × b = 1 × 1.86 °C.Kg/mol × 1 mol/kg = 1.86 °C

T = 0°C - 3.72 °C = -1.86 °C