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1. Using the freezing point of the solutions and the freezing point of the pure solvent, calculate the molar mass of the solute for runs 2-5 in part A. Since you know the identity of the solute, you also calculate a % error for each run.

1. Using the freezing point of the solutions and the freezing point of the pure solvent, calculate the molar mass of the solute for runs 2-5 in part A. Since you know the identity of the solute, you also calculate a % error for each run.

Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter12: Solutions
Section: Chapter Questions
Problem 12.98QE
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### Data Tables: #### Part A: The table below summarizes the freezing points of various solutions compared to pure water. This data is useful for understanding how different solutes and their concentrations affect the freezing point of water. | **Run** | **Amount of Water (g)** | **Amount of Solute (g)** | **Freezing Point (°C)** | |---------|--------------------------|-------------------------|-------------------------| | 1 | 100 | 0 | 0.00 | | 2 | 100 | 10.0 g Sucrose | -0.49 | | 3 | 100 | 1.7 g NaCl | -1.07 | | 4 | 100 | 7.4 g NaCl | -4.17 | | 5 | 100 | 3.2 g CaCl2 | -1.57 | #### Table B: This additional table provides data on the freezing points of water solutions with different amounts of solute 'Sx'. This helps in comparing the effects of varying solute amounts of the same substance on the freezing point. | **Run** | **Amount of Water (g)** | **Amount of Sx (g)** | **Freezing Point (°C)** | |---------|--------------------------|-----------------------|-------------------------| | 1 | 100 | 5.0 g | -0.41 | | 2 | 100 | 10.0 g | -0.55 | ### Analysis: - **Pure Water** (Run 1 in Part A): The baseline freezing point is 0.00°C. - **Impact of Sucrose** (Run 2 in Part A): Adding 10.0 g of sucrose to 100 g of water lowers the freezing point to -0.49°C. - **Impact of NaCl**: - 1.7 g of NaCl (Run 3 in Part A) lowers the freezing point to -1.07°C. - 7.4 g of NaCl (Run 4 in Part A) results in a significant freezing point depression to -4.17°C. This illustrates the colligative property effect where more solute leads to a greater freezing point depression. - **Impact of Ca
Transcribed Image Text:### Data Tables: #### Part A: The table below summarizes the freezing points of various solutions compared to pure water. This data is useful for understanding how different solutes and their concentrations affect the freezing point of water. | **Run** | **Amount of Water (g)** | **Amount of Solute (g)** | **Freezing Point (°C)** | |---------|--------------------------|-------------------------|-------------------------| | 1 | 100 | 0 | 0.00 | | 2 | 100 | 10.0 g Sucrose | -0.49 | | 3 | 100 | 1.7 g NaCl | -1.07 | | 4 | 100 | 7.4 g NaCl | -4.17 | | 5 | 100 | 3.2 g CaCl2 | -1.57 | #### Table B: This additional table provides data on the freezing points of water solutions with different amounts of solute 'Sx'. This helps in comparing the effects of varying solute amounts of the same substance on the freezing point. | **Run** | **Amount of Water (g)** | **Amount of Sx (g)** | **Freezing Point (°C)** | |---------|--------------------------|-----------------------|-------------------------| | 1 | 100 | 5.0 g | -0.41 | | 2 | 100 | 10.0 g | -0.55 | ### Analysis: - **Pure Water** (Run 1 in Part A): The baseline freezing point is 0.00°C. - **Impact of Sucrose** (Run 2 in Part A): Adding 10.0 g of sucrose to 100 g of water lowers the freezing point to -0.49°C. - **Impact of NaCl**: - 1.7 g of NaCl (Run 3 in Part A) lowers the freezing point to -1.07°C. - 7.4 g of NaCl (Run 4 in Part A) results in a significant freezing point depression to -4.17°C. This illustrates the colligative property effect where more solute leads to a greater freezing point depression. - **Impact of Ca
**Calculations:** **Part A** 1. Using the freezing point of the solutions and the freezing point of the pure solvent, calculate the molar mass of the solute for runs 2-5 in part A. Since you know the identity of the solute, you also calculate a % error for each run. **Part B** 2. Determine the molar mass of S<sub>x</sub> from your two simulation runs in part B. Average the molar mass and then determine what the correct elemental formula is for S<sub>x</sub> (x is of course an integer). No % error calculation for this part. *Show all calculations. That’s it.*
Transcribed Image Text:**Calculations:** **Part A** 1. Using the freezing point of the solutions and the freezing point of the pure solvent, calculate the molar mass of the solute for runs 2-5 in part A. Since you know the identity of the solute, you also calculate a % error for each run. **Part B** 2. Determine the molar mass of S<sub>x</sub> from your two simulation runs in part B. Average the molar mass and then determine what the correct elemental formula is for S<sub>x</sub> (x is of course an integer). No % error calculation for this part. *Show all calculations. That’s it.*
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