14. Certain compounds are known to “dimerize” in solution – this means that the intermolecular forces between them are strong enough to cause two molecules to assemble and behave as one object. Benzene melts at 5.50 °C and has a freezing point depression constant of 5.10 °C/m. Calculate the freezing point of a solution that contains 0.0500 mol of acetic acid, CH3COOH, in 125 g of benzene if acetic acid forms a dimer in this solvent. A) 4.48 °C B) -1.02 °C C) 3.46 °C D) 5.24 °C E) 6.01 °C

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### Question 14: Freezing Point Depression and Dimer Formation

#### Context:
Certain compounds are known to "dimerize" in solution. This means that the intermolecular forces between them are strong enough to cause two molecules to assemble and behave as one object. 

#### Provided Information:
- **Solvent:** Benzene
  - Melting Point: 5.50 °C
  - Freezing Point Depression Constant (\(K_f\)): 5.10 °C/m

- **Solute:** Acetic Acid (CH₃COOH)
  - Moles of Solute: 0.0500 mol
  - Mass of Solvent: 125 g of benzene

- **Condition:** Acetic acid forms a dimer in the solvent.

#### Problem:
Calculate the freezing point of a solution containing 0.0500 mol of acetic acid in 125 g of benzene.

#### Multiple Choices:
A) 4.48 °C  
B) -1.02 °C  
C) 3.46 °C  
D) 5.24 °C  
E) 6.01 °C  

#### Calculation Steps:
1. **Identify the molality** of the solution:
   \[ \text{Molality (m)} = \frac{\text{moles of solute}}{\text{kg of solvent}} \]

2. **Adjust for dimerization** assuming acetic acid forms a dimer which halves the effective number of solute particles.

3. **Use the freezing point depression formula**:
   \[ \Delta T_f = i \cdot K_f \cdot m \]
   Where \(i\) is the van't Hoff factor (dimerization implies \(i = 0.5\)).

4. **Determine the change in freezing point (\( \Delta T_f \))** and subtract it from the pure solvent's freezing point.

5. **Select the correct answer** from the given options.

---

This problem introduces students to concepts such as freezing point depression, molality, and the effects of molecular interactions like dimerization on colligative properties. Understanding these principles is fundamental in physical chemistry and solutions chemistry.
Transcribed Image Text:### Question 14: Freezing Point Depression and Dimer Formation #### Context: Certain compounds are known to "dimerize" in solution. This means that the intermolecular forces between them are strong enough to cause two molecules to assemble and behave as one object. #### Provided Information: - **Solvent:** Benzene - Melting Point: 5.50 °C - Freezing Point Depression Constant (\(K_f\)): 5.10 °C/m - **Solute:** Acetic Acid (CH₃COOH) - Moles of Solute: 0.0500 mol - Mass of Solvent: 125 g of benzene - **Condition:** Acetic acid forms a dimer in the solvent. #### Problem: Calculate the freezing point of a solution containing 0.0500 mol of acetic acid in 125 g of benzene. #### Multiple Choices: A) 4.48 °C B) -1.02 °C C) 3.46 °C D) 5.24 °C E) 6.01 °C #### Calculation Steps: 1. **Identify the molality** of the solution: \[ \text{Molality (m)} = \frac{\text{moles of solute}}{\text{kg of solvent}} \] 2. **Adjust for dimerization** assuming acetic acid forms a dimer which halves the effective number of solute particles. 3. **Use the freezing point depression formula**: \[ \Delta T_f = i \cdot K_f \cdot m \] Where \(i\) is the van't Hoff factor (dimerization implies \(i = 0.5\)). 4. **Determine the change in freezing point (\( \Delta T_f \))** and subtract it from the pure solvent's freezing point. 5. **Select the correct answer** from the given options. --- This problem introduces students to concepts such as freezing point depression, molality, and the effects of molecular interactions like dimerization on colligative properties. Understanding these principles is fundamental in physical chemistry and solutions chemistry.
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