Question
Which of these solutions has the highest freezing point?
A.
1.0 M molecular sucrose (C12H22O11)
B.
1.0 M ionic sodium chloride (NaCl)
C.
1.0 M ionic calcium bromide (CaBr2)
D.
1.0 M ionic aluminum bromide (AlBr3)
E.
1.0 M ionic potassium chloride (KCl)
A.
1.0 M molecular sucrose (C12H22O11)
B.
1.0 M ionic sodium chloride (NaCl)
C.
1.0 M ionic calcium bromide (CaBr2)
D.
1.0 M ionic aluminum bromide (AlBr3)
E.
1.0 M ionic potassium chloride (KCl)
Answers
GPT-4o mini
The freezing point depression, which indicates how much the freezing point of a solution is lowered, is proportional to the number of solute particles in the solution. The equation that describes this relationship is:
\[
\Delta T_f = i \cdot K_f \cdot m
\]
where:
- \(\Delta T_f\) is the change in freezing point,
- \(i\) is the van 't Hoff factor (the number of particles the solute dissociates into),
- \(K_f\) is the cryoscopic constant (which is a property of the solvent, typically water), and
- \(m\) is the molality of the solution.
To evaluate the freezing point of each solution, we need to consider the van 't Hoff factor (\(i\)) for each solute:
1. **Sucrose (C₁₂H₂₂O₁₁)**: It is a non-ionic, molecular solute, hence \(i = 1\).
2. **Sodium chloride (NaCl)**: It dissociates into 2 ions (Na⁺ and Cl⁻), so \(i = 2\).
3. **Calcium bromide (CaBr₂)**: This dissociates into 3 ions (Ca²⁺ and 2Br⁻), so \(i = 3\).
4. **Aluminum bromide (AlBr₃)**: This dissociates into 4 ions (Al³⁺ and 3Br⁻), so \(i = 4\).
5. **Potassium chloride (KCl)**: It dissociates into 2 ions (K⁺ and Cl⁻), so \(i = 2\).
Now we can summarize the effective concentration of particles in each solution:
- **Sucrose**: \(1.0 \, \text{M} \times 1 = 1.0 \, \text{particles}\)
- **NaCl**: \(1.0 \, \text{M} \times 2 = 2.0 \, \text{particles}\)
- **CaBr₂**: \(1.0 \, \text{M} \times 3 = 3.0 \, \text{particles}\)
- **AlBr₃**: \(1.0 \, \text{M} \times 4 = 4.0 \, \text{particles}\)
- **KCl**: \(1.0 \, \text{M} \times 2 = 2.0 \, \text{particles}\)
Since a higher number of particles results in a greater depression of the freezing point, the solution with the least number of total particles has the highest freezing point.
Among the options:
- Sucrose has the least impact on freezing point since it generates only 1 particle.
- NaCl, KCl both increase the freezing point lower than sucrose due to having 2 particles.
- CaBr₂ and AlBr₃ will have even lower freezing points due to having 3 and 4 particles, respectively.
Thus, the solution with the highest freezing point is:
**A. 1.0 M molecular sucrose (C₁₂H₂₂O₁₁)**.
\[
\Delta T_f = i \cdot K_f \cdot m
\]
where:
- \(\Delta T_f\) is the change in freezing point,
- \(i\) is the van 't Hoff factor (the number of particles the solute dissociates into),
- \(K_f\) is the cryoscopic constant (which is a property of the solvent, typically water), and
- \(m\) is the molality of the solution.
To evaluate the freezing point of each solution, we need to consider the van 't Hoff factor (\(i\)) for each solute:
1. **Sucrose (C₁₂H₂₂O₁₁)**: It is a non-ionic, molecular solute, hence \(i = 1\).
2. **Sodium chloride (NaCl)**: It dissociates into 2 ions (Na⁺ and Cl⁻), so \(i = 2\).
3. **Calcium bromide (CaBr₂)**: This dissociates into 3 ions (Ca²⁺ and 2Br⁻), so \(i = 3\).
4. **Aluminum bromide (AlBr₃)**: This dissociates into 4 ions (Al³⁺ and 3Br⁻), so \(i = 4\).
5. **Potassium chloride (KCl)**: It dissociates into 2 ions (K⁺ and Cl⁻), so \(i = 2\).
Now we can summarize the effective concentration of particles in each solution:
- **Sucrose**: \(1.0 \, \text{M} \times 1 = 1.0 \, \text{particles}\)
- **NaCl**: \(1.0 \, \text{M} \times 2 = 2.0 \, \text{particles}\)
- **CaBr₂**: \(1.0 \, \text{M} \times 3 = 3.0 \, \text{particles}\)
- **AlBr₃**: \(1.0 \, \text{M} \times 4 = 4.0 \, \text{particles}\)
- **KCl**: \(1.0 \, \text{M} \times 2 = 2.0 \, \text{particles}\)
Since a higher number of particles results in a greater depression of the freezing point, the solution with the least number of total particles has the highest freezing point.
Among the options:
- Sucrose has the least impact on freezing point since it generates only 1 particle.
- NaCl, KCl both increase the freezing point lower than sucrose due to having 2 particles.
- CaBr₂ and AlBr₃ will have even lower freezing points due to having 3 and 4 particles, respectively.
Thus, the solution with the highest freezing point is:
**A. 1.0 M molecular sucrose (C₁₂H₂₂O₁₁)**.