Determine the freezing point (in °C) of a solution that contains 204 g of CsH1206 dissolved in 234 mL of acetic acid (density = 1.05 g/mL). Pure acetic acid has a freezing point of 16.6°C and a K = 3.90°C/m.
Determine the freezing point (in °C) of a solution that contains 204 g of CsH1206 dissolved in 234 mL of acetic acid (density = 1.05 g/mL). Pure acetic acid has a freezing point of 16.6°C and a K = 3.90°C/m.
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter13: Solutions And Their Behavior
Section: Chapter Questions
Problem 107SCQ
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![**Title: Determining the Freezing Point of a Solution: A Practical Application**
**Objective:**
Learn how to determine the freezing point of a solution given specific solute and solvent properties.
**Problem Statement:**
Determine the freezing point (in °C) of a solution that contains 204 g of \( \text{C}_6\text{H}_{12}\text{O}_6 \) dissolved in 234 mL of acetic acid (density = 1.05 g/mL). Pure acetic acid has a freezing point of 16.6°C and a \( K_f \) (cryoscopic constant) = 3.90°C·kg/mol.
**Solution Approach:**
1. **Identify the Given Data:**
- Mass of solute (Glucose, \( \text{C}_6\text{H}_{12}\text{O}_6 \)): \( m = 204 \) g
- Volume of solvent (Acetic Acid): \( V = 234 \) mL
- Density of solvent: \( \rho = 1.05 \) g/mL
- Freezing point of pure acetic acid: \( T_f^0 = 16.6 \)°C
- Cryoscopic constant (\( K_f \)): 3.90°C·kg/mol
2. **Calculate Moles of Solute:**
- Molar mass of glucose (\( \text{C}_6\text{H}_{12}\text{O}_6 \)): 180 g/mol
\[
\text{Moles of glucose} = \frac{\text{mass}}{\text{molar mass}} = \frac{204 \text{ g}}{180 \text{ g/mol}} = 1.133 \text{ mol}
\]
3. **Calculate Mass of Solvent:**
- Convert volume to mass using the density:
\[
\text{Mass of acetic acid} = \text{density} \times \text{volume} = 1.05 \frac{\text{g}}{\text{mL}} \times 234 \text{ mL} = 245.7 \text{ g} = 0.2457 \text{ kg}
\]
4. **Determine Molality of the Solution:**
- Molality (\( m \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdddcaf98-1498-4874-b480-699455f18f4b%2F9d641781-8c4a-4ae3-a140-b8577486ed0e%2F7ngsfih_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Determining the Freezing Point of a Solution: A Practical Application**
**Objective:**
Learn how to determine the freezing point of a solution given specific solute and solvent properties.
**Problem Statement:**
Determine the freezing point (in °C) of a solution that contains 204 g of \( \text{C}_6\text{H}_{12}\text{O}_6 \) dissolved in 234 mL of acetic acid (density = 1.05 g/mL). Pure acetic acid has a freezing point of 16.6°C and a \( K_f \) (cryoscopic constant) = 3.90°C·kg/mol.
**Solution Approach:**
1. **Identify the Given Data:**
- Mass of solute (Glucose, \( \text{C}_6\text{H}_{12}\text{O}_6 \)): \( m = 204 \) g
- Volume of solvent (Acetic Acid): \( V = 234 \) mL
- Density of solvent: \( \rho = 1.05 \) g/mL
- Freezing point of pure acetic acid: \( T_f^0 = 16.6 \)°C
- Cryoscopic constant (\( K_f \)): 3.90°C·kg/mol
2. **Calculate Moles of Solute:**
- Molar mass of glucose (\( \text{C}_6\text{H}_{12}\text{O}_6 \)): 180 g/mol
\[
\text{Moles of glucose} = \frac{\text{mass}}{\text{molar mass}} = \frac{204 \text{ g}}{180 \text{ g/mol}} = 1.133 \text{ mol}
\]
3. **Calculate Mass of Solvent:**
- Convert volume to mass using the density:
\[
\text{Mass of acetic acid} = \text{density} \times \text{volume} = 1.05 \frac{\text{g}}{\text{mL}} \times 234 \text{ mL} = 245.7 \text{ g} = 0.2457 \text{ kg}
\]
4. **Determine Molality of the Solution:**
- Molality (\( m \
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