When 151. g of benzamide C7H7NO are dissolved in 1200. g of a certain mystery liquid X, the freezing point of the solution is 8.0 °C lower than the freezing point of pure X. On the other hand, when 151. g of iron(III) chloride are dissolved in the same mass of X, the freezing point of the solution is 20.5°C lower than the freezing point of pure X. Calculate the van’t Hoff factor for iron(III) chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits.
The Correct Answer and Explanation is:
To calculate the van’t Hoff factor (i) for iron(III) chloride (FeCl₃) in solvent X, we’ll use freezing point depression: ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot m
Where:
- ΔTf\Delta T_f is the freezing point depression,
- ii is the van’t Hoff factor,
- KfK_f is the cryoscopic constant (same for both solutions since the solvent is the same),
- mm is the molality of the solution.
Step 1: Use benzamide to determine KfK_f
Given:
- Mass of benzamide = 151 g
- Molar mass of benzamide (C₇H₇NO) ≈ 121.14 g/mol
- Mass of solvent X = 1200. g = 1.200 kg
- Freezing point depression = 8.0°C
- Benzamide is a non-electrolyte, so i=1i = 1
Molality: m=151121.14÷1.200=1.039 mol/kgm = \frac{151}{121.14} \div 1.200 = 1.039 \text{ mol/kg}
Using the formula: 8.0=1⋅Kf⋅1.039⇒Kf=8.01.039≈7.7 °C\cdotpkg/mol8.0 = 1 \cdot K_f \cdot 1.039 \Rightarrow K_f = \frac{8.0}{1.039} ≈ 7.7 \text{ °C·kg/mol}
Step 2: Use FeCl₃ data to find ii
Given:
- Mass of FeCl₃ = 151 g
- Molar mass of FeCl₃ ≈ 162.20 g/mol
- Mass of solvent X = 1.200 kg
- Freezing point depression = 20.5°C
Molality: m=151162.20÷1.200=0.7749 mol/kgm = \frac{151}{162.20} \div 1.200 = 0.7749 \text{ mol/kg}
Now solve for ii: 20.5=i⋅7.7⋅0.7749⇒i=20.57.7⋅0.7749≈20.55.9667≈3.420.5 = i \cdot 7.7 \cdot 0.7749 \Rightarrow i = \frac{20.5}{7.7 \cdot 0.7749} ≈ \frac{20.5}{5.9667} ≈ 3.4
Final Answer:
i=3.4 (unitless)\boxed{i = 3.4\ (\text{unitless})}
Explanation
The van’t Hoff factor (i) represents the number of particles into which a compound dissociates in solution. For non-electrolytes like benzamide, which don’t dissociate, i = 1. For ionic compounds like FeCl₃, which dissociate into ions, i can be more than 1. The theoretical maximum for FeCl₃ is 4, since it dissociates into Fe³⁺ and 3 Cl⁻ ions.
To calculate i experimentally, we compare the freezing point depression (ΔTf\Delta T_f) of a known non-electrolyte and the ionic compound in the same solvent. The formula ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot m links the depression to molality and the van’t Hoff factor. We first used benzamide’s data to find the cryoscopic constant KfK_f of solvent X, since i = 1 for benzamide. Then, using the same solvent and mass, we determined the molality of the FeCl₃ solution and plugged everything into the formula to solve for i.
We found the experimental van’t Hoff factor for FeCl₃ to be 3.4, slightly lower than the ideal value of 4. This suggests incomplete dissociation or ion-pairing in the solvent, common in non-aqueous or less polar solvents. The difference highlights real-world behavior versus ideal dissociation assumptions.
