The freezing point of water is 0.00°C at 1 atmosphere. How many grams of barium acetate (255.4 g/mol), must be dissolved in 262.0 grams of water to reduce the freezing point by 0.400°C ? Refer to the table for the necessary boiling or freezing point constant.
The freezing point of water is 0.00°C at 1 atmosphere. How many grams of barium acetate (255.4 g/mol), must be dissolved in 262.0 grams of water to reduce the freezing point by 0.400°C ? Refer to the table for the necessary boiling or freezing point constant.
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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### Freezing Point Depression Calculation - Educational Module
#### Question:
The freezing point of water is **0.00°C** at 1 atmosphere.
**Problem Statement:**
How many grams of *barium acetate* (255.4 g/mol) must be dissolved in **262.0 grams of water** to reduce the freezing point by **0.400°C**? Refer to the table for the necessary boiling or freezing point constant.
#### Table of Boiling and Freezing Point Constants:
| Solvent | Formula | \( K_b \) (°C/m) | \( K_f \) (°C/m) |
|--------------|-----------------|------------------|------------------|
| Water | \( H_2O \) | 0.512 | 1.86 |
| Ethanol | \( CH_3CH_2OH \)| 1.22 | 1.99 |
| Chloroform | \( CHCl_3 \) | 3.67 | --- |
| Benzene | \( C_6H_6 \) | 2.53 | 5.12 |
| Diethyl ether| \( CH_3CH_2OCH_2CH_3 \) | 2.02 | --- |
#### Required Information:
- Molar mass of barium acetate: **255.4 g/mol**
- Mass of water: **262.0 grams**
- Desired decrease in freezing point: **0.400°C**
- Freezing point depression constant for water \((K_f)\): **1.86 °C/m**
#### Calculation:
To determine the amount of barium acetate required, apply the freezing point depression formula:
\[ \Delta T_f = K_f \cdot m \]
Where:
\[ \Delta T_f \] = Change in freezing point (°C)
\[ K_f \] = Freezing point depression constant (°C/m)
\[ m \] = Molality (moles of solute / kg of solvent)
1. **Calculate Molality (\( m \))**:
\[ m = \frac{\Delta T_f}{K_f} = \frac{0.400°C}{1.86 °C/m} = 0.215 m \]
2. **Calculate Moles of Solute**:
\[ \text{Moles of B](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F60358686-21d1-41ff-a93a-9ba59c7b6516%2F8ff00a41-b1ba-4fc8-ac34-6daa497b285d%2Fk9urrq_reoriented.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Freezing Point Depression Calculation - Educational Module
#### Question:
The freezing point of water is **0.00°C** at 1 atmosphere.
**Problem Statement:**
How many grams of *barium acetate* (255.4 g/mol) must be dissolved in **262.0 grams of water** to reduce the freezing point by **0.400°C**? Refer to the table for the necessary boiling or freezing point constant.
#### Table of Boiling and Freezing Point Constants:
| Solvent | Formula | \( K_b \) (°C/m) | \( K_f \) (°C/m) |
|--------------|-----------------|------------------|------------------|
| Water | \( H_2O \) | 0.512 | 1.86 |
| Ethanol | \( CH_3CH_2OH \)| 1.22 | 1.99 |
| Chloroform | \( CHCl_3 \) | 3.67 | --- |
| Benzene | \( C_6H_6 \) | 2.53 | 5.12 |
| Diethyl ether| \( CH_3CH_2OCH_2CH_3 \) | 2.02 | --- |
#### Required Information:
- Molar mass of barium acetate: **255.4 g/mol**
- Mass of water: **262.0 grams**
- Desired decrease in freezing point: **0.400°C**
- Freezing point depression constant for water \((K_f)\): **1.86 °C/m**
#### Calculation:
To determine the amount of barium acetate required, apply the freezing point depression formula:
\[ \Delta T_f = K_f \cdot m \]
Where:
\[ \Delta T_f \] = Change in freezing point (°C)
\[ K_f \] = Freezing point depression constant (°C/m)
\[ m \] = Molality (moles of solute / kg of solvent)
1. **Calculate Molality (\( m \))**:
\[ m = \frac{\Delta T_f}{K_f} = \frac{0.400°C}{1.86 °C/m} = 0.215 m \]
2. **Calculate Moles of Solute**:
\[ \text{Moles of B
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