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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
Chemistry
10th Edition
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
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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
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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