The freezing point of water H₂O is 0.00 °C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in water is glucose. How many grams of glucose, C6H12O6 (180.2 g/mol), must be dissolved in 218.0 grams of water to reduce the freezing point by 0.350 °C? Refer to the table for the necessary boiling or freezing point constant. Solvent Formula Kb (°C/m) Kf(°C/m) Water H₂O 0.512 Ethanol CH3 CH₂OH 1.22 Chloroform CHC13 3.67 Benzene C6H6 2.53 Diethyl ether CH3 CH₂ OCH₂ CH3 2.02 Mass= g 1.86 1.99 5.12
The freezing point of water H₂O is 0.00 °C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in water is glucose. How many grams of glucose, C6H12O6 (180.2 g/mol), must be dissolved in 218.0 grams of water to reduce the freezing point by 0.350 °C? Refer to the table for the necessary boiling or freezing point constant. Solvent Formula Kb (°C/m) Kf(°C/m) Water H₂O 0.512 Ethanol CH3 CH₂OH 1.22 Chloroform CHC13 3.67 Benzene C6H6 2.53 Diethyl ether CH3 CH₂ OCH₂ CH3 2.02 Mass= g 1.86 1.99 5.12
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
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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![**Boiling Point Elevation/Freezing Point Depression**
\[ \Delta T = m \, K \]
- **For freezing point depression:**
\[ \Delta T = T(\text{pure solvent}) - T(\text{solution}) \]
- **For boiling point elevation:**
\[ \Delta T = T(\text{solution}) - T(\text{pure solvent}) \]
Where:
- \( m = \) (# moles solute / Kg solvent)
- \( K_b = \) boiling point elevation constant
- \( K_f = \) freezing point depression constant
\( K_b \) and \( K_f \) depend only on the SOLVENT. Below are some common values. Use these values for the calculations that follow.
| Solvent | Formula | \( K_b (°C / m) \) | \( K_f (°C / m) \) |
|--------------|----------------|------------------|------------------|
| Water | H\(_2\)O | 0.512 | 1.86 |
| Ethanol | CH\(_3\)CH\(_2\)OH | 1.22 | 1.99 |
| Chloroform | CHCl\(_3\) | 3.67 | - |
| Benzene | C\(_6\)H\(_6\) | 2.53 | 5.12 |
| Diethyl ether| CH\(_3\)CH\(_2\)OCH\(_2\)CH\(_3\) | 2.02 | - |](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5e64e050-9cb6-467e-8132-ec94d7497579%2F99202ff5-67a1-4f42-9834-a90e11c4f45a%2Fcr9stsj_processed.png&w=3840&q=75)
Transcribed Image Text:**Boiling Point Elevation/Freezing Point Depression**
\[ \Delta T = m \, K \]
- **For freezing point depression:**
\[ \Delta T = T(\text{pure solvent}) - T(\text{solution}) \]
- **For boiling point elevation:**
\[ \Delta T = T(\text{solution}) - T(\text{pure solvent}) \]
Where:
- \( m = \) (# moles solute / Kg solvent)
- \( K_b = \) boiling point elevation constant
- \( K_f = \) freezing point depression constant
\( K_b \) and \( K_f \) depend only on the SOLVENT. Below are some common values. Use these values for the calculations that follow.
| Solvent | Formula | \( K_b (°C / m) \) | \( K_f (°C / m) \) |
|--------------|----------------|------------------|------------------|
| Water | H\(_2\)O | 0.512 | 1.86 |
| Ethanol | CH\(_3\)CH\(_2\)OH | 1.22 | 1.99 |
| Chloroform | CHCl\(_3\) | 3.67 | - |
| Benzene | C\(_6\)H\(_6\) | 2.53 | 5.12 |
| Diethyl ether| CH\(_3\)CH\(_2\)OCH\(_2\)CH\(_3\) | 2.02 | - |
![### Freezing Point Depression Example
The freezing point of water \( \text{H}_2\text{O} \) is 0.00 °C at 1 atmosphere. A nonvolatile, nonelectrolyte solute that dissolves in water is glucose.
**Problem:**
How many grams of glucose, \( \text{C}_6\text{H}_{12}\text{O}_6 \) (180.2 g/mol), must be dissolved in 218.0 grams of water to reduce the freezing point by 0.350 °C? Refer to the table for the necessary boiling or freezing point constant.
**Table: Freezing and Boiling Point Constants**
The table below shows various solvents along with their chemical formulas and constants:
| Solvent | Formula | \( K_b \) (°C/m) | \( K_f \) (°C/m) |
|---------------|----------------------|------------------|------------------|
| Water | \( \text{H}_2\text{O} \) | 0.512 | 1.86 |
| Ethanol | \( \text{CH}_3\text{CH}_2\text{OH} \) | 1.22 | 1.99 |
| Chloroform | \( \text{CHCl}_3 \) | 3.67 | |
| Benzene | \( \text{C}_6\text{H}_6 \) | 2.53 | 5.12 |
| Diethyl ether | \( \text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3 \) | 2.02 | |
### Calculate
To calculate the mass of glucose needed, use the freezing point depression formula:
\[
\Delta T_f = i \times K_f \times m
\]
Where:
- \( \Delta T_f \) is the change in freezing point (0.350 °C)
- \( i \) is the van't Hoff factor (1 for glucose since it doesn’t dissociate)
- \( K_f \) is the freezing point depression constant for water (1.86 °C/m)
- \( m \) is the molality of the solution
**Calculate the molality (\( m \)):**
Re](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5e64e050-9cb6-467e-8132-ec94d7497579%2F99202ff5-67a1-4f42-9834-a90e11c4f45a%2Fzjxdbsm_processed.png&w=3840&q=75)
Transcribed Image Text:### Freezing Point Depression Example
The freezing point of water \( \text{H}_2\text{O} \) is 0.00 °C at 1 atmosphere. A nonvolatile, nonelectrolyte solute that dissolves in water is glucose.
**Problem:**
How many grams of glucose, \( \text{C}_6\text{H}_{12}\text{O}_6 \) (180.2 g/mol), must be dissolved in 218.0 grams of water to reduce the freezing point by 0.350 °C? Refer to the table for the necessary boiling or freezing point constant.
**Table: Freezing and Boiling Point Constants**
The table below shows various solvents along with their chemical formulas and constants:
| Solvent | Formula | \( K_b \) (°C/m) | \( K_f \) (°C/m) |
|---------------|----------------------|------------------|------------------|
| Water | \( \text{H}_2\text{O} \) | 0.512 | 1.86 |
| Ethanol | \( \text{CH}_3\text{CH}_2\text{OH} \) | 1.22 | 1.99 |
| Chloroform | \( \text{CHCl}_3 \) | 3.67 | |
| Benzene | \( \text{C}_6\text{H}_6 \) | 2.53 | 5.12 |
| Diethyl ether | \( \text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3 \) | 2.02 | |
### Calculate
To calculate the mass of glucose needed, use the freezing point depression formula:
\[
\Delta T_f = i \times K_f \times m
\]
Where:
- \( \Delta T_f \) is the change in freezing point (0.350 °C)
- \( i \) is the van't Hoff factor (1 for glucose since it doesn’t dissociate)
- \( K_f \) is the freezing point depression constant for water (1.86 °C/m)
- \( m \) is the molality of the solution
**Calculate the molality (\( m \)):**
Re
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