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Which of the following two solutions has the lowest freezing point. Solution A or Solution B. For simplicity, assume each solution has a density of 1.0 g/mL Show all of your work clearly, then justify your choice (solution A or solution B) using full sentences. Solution A: 10 mass percent solution of sodium chloride in water Solution B: 10 mass percent solution of ammonium sulfate in water

Which of the following two solutions has the lowest freezing point. Solution A or Solution B. For simplicity, assume each solution has a density of 1.0 g/mL Show all of your work clearly, then justify your choice (solution A or solution B) using full sentences. Solution A: 10 mass percent solution of sodium chloride in water Solution B: 10 mass percent solution of ammonium sulfate in water

Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Chapter14: Mixtures And Solutions
Section: Chapter Questions
Problem 112A
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### Determining the Solution with the Lowest Freezing Point **Question:** Which of the following two solutions has the lowest freezing point, Solution A or Solution B? For simplicity, assume each solution has a density of 1.0 g/mL. You are required to show all of your work clearly and then justify your choice (Solution A or Solution B) using full sentences. **Solutions Provided:** - **Solution A:** 10 mass percent solution of sodium chloride in water - **Solution B:** 10 mass percent solution of ammonium sulfate in water ### Steps to Determine Freezing Point Depression 1. **Calculate Molarity of each Solution:** - For NaCl: - Molar mass of NaCl = 58.44 g/mol - 10 mass percent solution means 10 g of NaCl in 100 g of solution. - Given the density is 1.0 g/mL, 100 g solution is 100 mL or 0.1 L. - Moles of NaCl in 10 g = 10 g / 58.44 g/mol = 0.171 mol. - Molarity (M) = 0.171 mol / 0.1 L = 1.71 M. - For (NH₄)₂SO₄: - Molar mass of (NH₄)₂SO₄ = 132.14 g/mol - 10 mass percent solution means 10 g of (NH₄)₂SO₄ in 100 g of solution. - Given the density is 1.0 g/mL, 100 g of solution is 100 mL or 0.1 L. - Moles of (NH₄)₂SO₄ in 10 g = 10 g / 132.14 g/mol = 0.076 mol. - Molarity (M) = 0.076 mol / 0.1 L = 0.76 M. 2. **Account for Dissociation in Water:** - NaCl dissociates into 2 ions (Na⁺ and Cl⁻). - (NH₄)₂SO₄ dissociates into 3 ions (2 NH₄⁺ and 1 SO₄²⁻). 3. **Calculate the Van't Hoff Factor
Transcribed Image Text:### Determining the Solution with the Lowest Freezing Point **Question:** Which of the following two solutions has the lowest freezing point, Solution A or Solution B? For simplicity, assume each solution has a density of 1.0 g/mL. You are required to show all of your work clearly and then justify your choice (Solution A or Solution B) using full sentences. **Solutions Provided:** - **Solution A:** 10 mass percent solution of sodium chloride in water - **Solution B:** 10 mass percent solution of ammonium sulfate in water ### Steps to Determine Freezing Point Depression 1. **Calculate Molarity of each Solution:** - For NaCl: - Molar mass of NaCl = 58.44 g/mol - 10 mass percent solution means 10 g of NaCl in 100 g of solution. - Given the density is 1.0 g/mL, 100 g solution is 100 mL or 0.1 L. - Moles of NaCl in 10 g = 10 g / 58.44 g/mol = 0.171 mol. - Molarity (M) = 0.171 mol / 0.1 L = 1.71 M. - For (NH₄)₂SO₄: - Molar mass of (NH₄)₂SO₄ = 132.14 g/mol - 10 mass percent solution means 10 g of (NH₄)₂SO₄ in 100 g of solution. - Given the density is 1.0 g/mL, 100 g of solution is 100 mL or 0.1 L. - Moles of (NH₄)₂SO₄ in 10 g = 10 g / 132.14 g/mol = 0.076 mol. - Molarity (M) = 0.076 mol / 0.1 L = 0.76 M. 2. **Account for Dissociation in Water:** - NaCl dissociates into 2 ions (Na⁺ and Cl⁻). - (NH₄)₂SO₄ dissociates into 3 ions (2 NH₄⁺ and 1 SO₄²⁻). 3. **Calculate the Van't Hoff Factor
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