Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.18 m Ni(NO3)2 A. Lowest freezing point 2. 0.25 m ZNSO4 B. Second lowest freezing point 3. 0.20 m CoCl2 C. Third lowest freezing point 4. 0.56 m Sucrose(nonelectrolyte) D. Highest freezing point
Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.18 m Ni(NO3)2 A. Lowest freezing point 2. 0.25 m ZNSO4 B. Second lowest freezing point 3. 0.20 m CoCl2 C. Third lowest freezing point 4. 0.56 m Sucrose(nonelectrolyte) D. Highest freezing point
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Freezing Point Depression Activity**
Match the following aqueous solutions with the appropriate letter from the column on the right.
**Solutions:**
1. 0.18 m Ni(NO₃)₂
2. 0.25 m ZnSO₄
3. 0.20 m CoCl₂
4. 0.56 m Sucrose (nonelectrolyte)
**Categories:**
A. Lowest freezing point
B. Second lowest freezing point
C. Third lowest freezing point
D. Highest freezing point
---
**Explanation:**
Freezing point depression depends on the molality of the solution and the number of particles the compound dissociates into when dissolved in water. Electrolytes, which dissociate into ions, will cause a greater freezing point depression than nonelectrolytes at the same molality.
- **Ni(NO₃)₂** dissociates into 3 ions (Ni²⁺ and 2 NO₃⁻).
- **ZnSO₄** dissociates into 2 ions (Zn²⁺ and SO₄²⁻).
- **CoCl₂** dissociates into 3 ions (Co²⁺ and 2 Cl⁻).
- **Sucrose** does not dissociate, so it remains 1 molecule in solution.
Using the van 't Hoff factor (i), which represents the number of particles into which a solute dissociates, we can predict:
1. 0.18 m Ni(NO₃)₂ (i = 3)
2. 0.25 m ZnSO₄ (i = 2)
3. 0.20 m CoCl₂ (i = 3)
4. 0.56 m Sucrose (i = 1)
Multiplying molality by the van 't Hoff factor:
- **0.18 m Ni(NO₃)₂** : 0.18 * 3 = 0.54
- **0.25 m ZnSO₄** : 0.25 * 2 = 0.50
- **0.20 m CoCl₂** : 0.20 * 3 = 0.60
- **0.56 m Sucrose (nonelectrolyte)** : 0.56 * 1 = 0.56
Thus, the solutions in order from lowest to highest freezing](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fedee5435-55e0-4ec8-aecd-101c845dd6aa%2F688c99ec-f31e-4edf-a4f6-d29053a7cff1%2Fczpws5d_processed.png&w=3840&q=75)
Transcribed Image Text:**Freezing Point Depression Activity**
Match the following aqueous solutions with the appropriate letter from the column on the right.
**Solutions:**
1. 0.18 m Ni(NO₃)₂
2. 0.25 m ZnSO₄
3. 0.20 m CoCl₂
4. 0.56 m Sucrose (nonelectrolyte)
**Categories:**
A. Lowest freezing point
B. Second lowest freezing point
C. Third lowest freezing point
D. Highest freezing point
---
**Explanation:**
Freezing point depression depends on the molality of the solution and the number of particles the compound dissociates into when dissolved in water. Electrolytes, which dissociate into ions, will cause a greater freezing point depression than nonelectrolytes at the same molality.
- **Ni(NO₃)₂** dissociates into 3 ions (Ni²⁺ and 2 NO₃⁻).
- **ZnSO₄** dissociates into 2 ions (Zn²⁺ and SO₄²⁻).
- **CoCl₂** dissociates into 3 ions (Co²⁺ and 2 Cl⁻).
- **Sucrose** does not dissociate, so it remains 1 molecule in solution.
Using the van 't Hoff factor (i), which represents the number of particles into which a solute dissociates, we can predict:
1. 0.18 m Ni(NO₃)₂ (i = 3)
2. 0.25 m ZnSO₄ (i = 2)
3. 0.20 m CoCl₂ (i = 3)
4. 0.56 m Sucrose (i = 1)
Multiplying molality by the van 't Hoff factor:
- **0.18 m Ni(NO₃)₂** : 0.18 * 3 = 0.54
- **0.25 m ZnSO₄** : 0.25 * 2 = 0.50
- **0.20 m CoCl₂** : 0.20 * 3 = 0.60
- **0.56 m Sucrose (nonelectrolyte)** : 0.56 * 1 = 0.56
Thus, the solutions in order from lowest to highest freezing
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