Homework Help is Here – Start Your Trial Now!
Science
Chemistry
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

Introductory Chemistry: An Active Learning Approach
Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN:9781305079250
Author:Mark S. Cracolice, Ed Peters
Publisher:Mark S. Cracolice, Ed Peters
Chapter16: Solutions
Section: Chapter Questions
Problem 14E
See similar textbooks
icon
Related questions
Question
### 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
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Solutions
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Introductory Chemistry: An Active Learning Approa…
Introductory Chemistry: An Active Learning Approa…
Chemistry
ISBN:
9781305079250
Author:
Mark S. Cracolice, Ed Peters
Publisher:
Cengage Learning
World of Chemistry, 3rd edition
World of Chemistry, 3rd edition
Chemistry
ISBN:
9781133109655
Author:
Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:
Brooks / Cole / Cengage Learning
Chemistry for Today: General, Organic, and Bioche…
Chemistry for Today: General, Organic, and Bioche…
Chemistry
ISBN:
9781305960060
Author:
Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher:
Cengage Learning
General, Organic, and Biological Chemistry
General, Organic, and Biological Chemistry
Chemistry
ISBN:
9781285853918
Author:
H. Stephen Stoker
Publisher:
Cengage Learning
Introductory Chemistry For Today
Introductory Chemistry For Today
Chemistry
ISBN:
9781285644561
Author:
Seager
Publisher:
Cengage
Chemistry: The Molecular Science
Chemistry: The Molecular Science
Chemistry
ISBN:
9781285199047
Author:
John W. Moore, Conrad L. Stanitski
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS