The boiling point of pure ethanol (C₂H₂OH, MM = 46.07 g/mol) is 78.4 °C. A solution is made using 100.0 g of ethanol (the solvent) and 24.1 g of glucose (MM = 180.16 g/mol). What is the boiling point of this solution, in °C? (Kb of ethanol is 1.22 °C/m)
The boiling point of pure ethanol (C₂H₂OH, MM = 46.07 g/mol) is 78.4 °C. A solution is made using 100.0 g of ethanol (the solvent) and 24.1 g of glucose (MM = 180.16 g/mol). What is the boiling point of this solution, in °C? (Kb of ethanol is 1.22 °C/m)
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![**Boiling Point Elevation of Ethanol Solution**
The boiling point of pure ethanol \((C_2H_5OH, \text{Molecular Mass} = 46.07 \ \text{g/mol})\) is 78.4°C. A solution is made by dissolving 24.1 g of glucose \((\text{Molecular Mass} = 180.16 \ \text{g/mol})\) in 100.0 g of ethanol (the solvent). What is the boiling point of this solution, in °C? The ebullioscopic constant \((K_b)\) of ethanol is 1.22 °C/m.
To find the boiling point of the solution, use the boiling point elevation formula:
\[ \Delta T_b = K_b \cdot m \]
where \( \Delta T_b \) is the boiling point elevation, \( K_b \) is the ebullioscopic constant, and \( m \) is the molality of the solution.
1. **Calculate the molality (m)**:
- Moles of glucose:
\[ \text{Moles of glucose} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}} = \frac{24.1 \ \text{g}}{180.16 \ \text{g/mol}} \approx 0.1338 \ \text{mol} \]
- Mass of ethanol (in kg):
\[ \text{Mass of ethanol} = 100.0 \ \text{g} = 0.100 \ \text{kg} \]
- Molality:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} = \frac{0.1338 \ \text{mol}}{0.100 \ \text{kg}} = 1.338 \ m \]
2. **Calculate the boiling point elevation (\(\Delta T_b\))**:
\[ \Delta T_b = K_b \cdot m = 1.22 \ \text{°C/m} \cdot 1.338 \ m \approx 1.632 \ \text{°C} \]
3. **Determine the new boiling point**:
\[ \text{Boiling point of solution} = \text](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbc3b72c7-eb99-406f-83fe-82fc41b2a8b5%2F9669c553-e676-4ed4-baa8-300ff98a6a96%2Fzqjsvpg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Boiling Point Elevation of Ethanol Solution**
The boiling point of pure ethanol \((C_2H_5OH, \text{Molecular Mass} = 46.07 \ \text{g/mol})\) is 78.4°C. A solution is made by dissolving 24.1 g of glucose \((\text{Molecular Mass} = 180.16 \ \text{g/mol})\) in 100.0 g of ethanol (the solvent). What is the boiling point of this solution, in °C? The ebullioscopic constant \((K_b)\) of ethanol is 1.22 °C/m.
To find the boiling point of the solution, use the boiling point elevation formula:
\[ \Delta T_b = K_b \cdot m \]
where \( \Delta T_b \) is the boiling point elevation, \( K_b \) is the ebullioscopic constant, and \( m \) is the molality of the solution.
1. **Calculate the molality (m)**:
- Moles of glucose:
\[ \text{Moles of glucose} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}} = \frac{24.1 \ \text{g}}{180.16 \ \text{g/mol}} \approx 0.1338 \ \text{mol} \]
- Mass of ethanol (in kg):
\[ \text{Mass of ethanol} = 100.0 \ \text{g} = 0.100 \ \text{kg} \]
- Molality:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} = \frac{0.1338 \ \text{mol}}{0.100 \ \text{kg}} = 1.338 \ m \]
2. **Calculate the boiling point elevation (\(\Delta T_b\))**:
\[ \Delta T_b = K_b \cdot m = 1.22 \ \text{°C/m} \cdot 1.338 \ m \approx 1.632 \ \text{°C} \]
3. **Determine the new boiling point**:
\[ \text{Boiling point of solution} = \text
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