5. What is the boiling point of a 1.42molal MgS solution? (Ko of water = 0.512°C/molality) 6. What is the freezing point of a 1.83molal C,H14 solution? (Krof water 1.86°C/molality)
5. What is the boiling point of a 1.42molal MgS solution? (Ko of water = 0.512°C/molality) 6. What is the freezing point of a 1.83molal C,H14 solution? (Krof water 1.86°C/molality)
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Molality and mole fraction problem can someone help me with those two
![### Example Problems on Colligative Properties
**5. What is the boiling point of a 1.42 molal MgS solution?**
Given:
- Molality (m) of MgS solution: 1.42 molal
- Boiling point elevation constant (\( K_b \)) of water: 0.512°C/molality
**Solution:**
The boiling point elevation (\( \Delta T_b \)) is calculated using the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \( i \) is the van't Hoff factor (number of particles the solute breaks into)
- \( K_b \) is the boiling point elevation constant
- \( m \) is the molality of the solution
For MgS, the van't Hoff factor (\( i \)) is 2 since MgS dissociates into Mg²⁺ and S²⁻ ions.
\[ \Delta T_b = 2 \cdot 0.512°C/molality \cdot 1.42 molal \]
\[ \Delta T_b = 2 \cdot 0.512 \cdot 1.42 \]
\[ \Delta T_b = 1.45264°C \]
The boiling point of pure water is 100°C. Therefore, the boiling point of the MgS solution is:
\[ 100°C + 1.45264°C = 101.45264°C \]
**6. What is the freezing point of a 1.83 molal C₆H₁₄ solution?**
Given:
- Molality (m) of C₆H₁₄ solution: 1.83 molal
- Freezing point depression constant (\( K_f \)) of water: 1.86°C/molality
**Solution:**
The freezing point depression (\( \Delta T_f \)) is calculated using the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( i \) is the van't Hoff factor (number of particles the solute breaks into)
- \( K_f \) is the freezing point depression constant
- \( m \) is the molality of the solution
For C₆H₁₄ (hexane), the van't Hoff factor (\( i \)) is 1 since it](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdd4f4500-c170-4a7d-84e4-ab98d31e3953%2F05b73988-fec4-470d-9021-ddc0638294b5%2Fxsnbt4f_processed.png&w=3840&q=75)
Transcribed Image Text:### Example Problems on Colligative Properties
**5. What is the boiling point of a 1.42 molal MgS solution?**
Given:
- Molality (m) of MgS solution: 1.42 molal
- Boiling point elevation constant (\( K_b \)) of water: 0.512°C/molality
**Solution:**
The boiling point elevation (\( \Delta T_b \)) is calculated using the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \( i \) is the van't Hoff factor (number of particles the solute breaks into)
- \( K_b \) is the boiling point elevation constant
- \( m \) is the molality of the solution
For MgS, the van't Hoff factor (\( i \)) is 2 since MgS dissociates into Mg²⁺ and S²⁻ ions.
\[ \Delta T_b = 2 \cdot 0.512°C/molality \cdot 1.42 molal \]
\[ \Delta T_b = 2 \cdot 0.512 \cdot 1.42 \]
\[ \Delta T_b = 1.45264°C \]
The boiling point of pure water is 100°C. Therefore, the boiling point of the MgS solution is:
\[ 100°C + 1.45264°C = 101.45264°C \]
**6. What is the freezing point of a 1.83 molal C₆H₁₄ solution?**
Given:
- Molality (m) of C₆H₁₄ solution: 1.83 molal
- Freezing point depression constant (\( K_f \)) of water: 1.86°C/molality
**Solution:**
The freezing point depression (\( \Delta T_f \)) is calculated using the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( i \) is the van't Hoff factor (number of particles the solute breaks into)
- \( K_f \) is the freezing point depression constant
- \( m \) is the molality of the solution
For C₆H₁₄ (hexane), the van't Hoff factor (\( i \)) is 1 since it
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