Assuming complete dissociation, calculate the freezing point of a solution prepared by dissolving 6g of Na2SO4 · 10H2O.
The Correct Answer and Explanation is:
To calculate the freezing point of a solution prepared by dissolving 6 g of sodium sulfate decahydrate (Na₂SO₄·10H₂O), we use the freezing point depression formula:ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot mΔTf=i⋅Kf⋅m
Where:
- ΔTf\Delta T_fΔTf is the freezing point depression
- iii is the van’t Hoff factor (number of particles the solute dissociates into)
- KfK_fKf is the freezing point depression constant of water (1.86 °C·kg/mol)
- mmm is the molality of the solution
Step 1: Calculate molar mass of Na₂SO₄·10H₂O
- Na = 22.99 × 2 = 45.98 g/mol
- S = 32.07 g/mol
- O₄ = 16.00 × 4 = 64.00 g/mol
- 10H₂O = 10 × (2×1.008 + 16.00) = 10 × 18.016 = 180.16 g/mol
Total molar mass = 45.98 + 32.07 + 64.00 + 180.16 = 322.21 g/mol
Step 2: Find moles of solute
Moles of Na₂SO₄\cdotp10H₂O=6 g322.21 g/mol=0.01862 mol\text{Moles of Na₂SO₄·10H₂O} = \frac{6\text{ g}}{322.21\text{ g/mol}} = 0.01862 \text{ mol}Moles of Na₂SO₄\cdotp10H₂O=322.21 g/mol6 g=0.01862 mol
Step 3: Assume solution is made in 100 g (0.1 kg) of water
Molality m=0.018620.1=0.1862 mol/kgm = \frac{0.01862}{0.1} = 0.1862 \, \text{mol/kg}m=0.10.01862=0.1862mol/kg
Step 4: Van’t Hoff factor iii
Na₂SO₄ dissociates completely into 2 Na⁺ and 1 SO₄²⁻i=3i = 3i=3
Step 5: Calculate ΔTf\Delta T_fΔTf
\Delta T_f = 3 \cdot 1.86 \cdot 0.1862 = 1.039\,^\circ \text{C}
Step 6: Freezing point of water = 0 °C
Freezing point of solution =0−1.039=−1.04∘C0 – 1.039 = \boxed{-1.04^\circ \text{C}}0−1.039=−1.04∘C
Explanation
The freezing point depression is a colligative property, meaning it depends on the number of solute particles in a solution and not their identity. In this problem, we’re dissolving 6 grams of sodium sulfate decahydrate (Na₂SO₄·10H₂O) in water. This hydrated salt fully dissociates in water into three ions per formula unit: two sodium ions (Na⁺) and one sulfate ion (SO₄²⁻), giving it a van’t Hoff factor of 3. This factor is essential in determining how much the freezing point will be lowered.
First, we calculate the molar mass of the hydrate. Each molecule includes the base salt Na₂SO₄ plus ten water molecules. The full molar mass is approximately 322.21 g/mol. Using this, we determine the number of moles of the substance in the solution by dividing 6 grams by the molar mass, giving about 0.01862 moles. Assuming the salt is dissolved in 100 grams of water (a common assumption unless otherwise stated), we calculate molality by dividing the moles of solute by kilograms of solvent. This gives a molality of 0.1862 mol/kg.
Next, applying the freezing point depression formula, we multiply the van’t Hoff factor (3), the molality (0.1862 mol/kg), and the cryoscopic constant of water (1.86 °C·kg/mol). This yields a freezing point depression of 1.039 °C. Finally, subtracting this from the normal freezing point of water (0 °C) gives a final freezing point of approximately -1.04 °C.
This shows how ionic compounds can significantly depress the freezing point due to the number of particles formed in solution.
