The normal boiling point of a certain liquid X is 111.60 °C, but when 0.13 kg of potassium bromide (KBr) are dissolved in 900. g of X the solution boils at 113.6 °C instead. Use this information to calculate the molal boiling point elevation constant K, of X. Round your answer to 2 significant digits. °C-kg K, mol
The normal boiling point of a certain liquid X is 111.60 °C, but when 0.13 kg of potassium bromide (KBr) are dissolved in 900. g of X the solution boils at 113.6 °C instead. Use this information to calculate the molal boiling point elevation constant K, of X. Round your answer to 2 significant digits. °C-kg K, mol
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![**Boiling Point Elevation: Calculation of \( K_b \) for Liquid \( X \)**
The normal boiling point of a certain liquid \( X \) is 111.60 °C, but when 0.13 kg of potassium bromide (KBr) are dissolved in 900. g of \( X \), the solution boils at 113.6 °C instead. Use this information to calculate the molal boiling point elevation constant \( K_b \) of \( X \).
**Instructions:**
1. Report your answer to 2 significant digits.
**Given Data:**
- **Normal boiling point of liquid \( X \):** 111.60 °C
- **Boiling point of the solution:** 113.6 °C
- **Mass of potassium bromide (KBr) dissolved:** 0.13 kg
- **Mass of liquid \( X \):** 900. g (0.900 kg)
**Formula for Molal Boiling Point Elevation:**
\[ \Delta T_b = K_b \cdot m \]
Where:
- \(\Delta T_b\) is the boiling point elevation.
- \(K_b\) is the molal boiling point elevation constant.
- \(m\) is the molality of the solution (moles of solute per kilogram of solvent).
### Diagram Explanation:
#### Equation Box:
\[ K_b = \frac{ \Delta T_b }{ \text{molality} } \]
#### Input Box:
- A box indicating where to enter the value for \( K_b \).
- A checkbox indicating multiplication by 10.
- Buttons to submit (\(x\)), reset (\(\looparrowleft\)), and get help (\(?\)).
### Calculation Steps:
1. **Calculate Boiling Point Elevation:**
\[ \Delta T_b = 113.6\,°C - 111.60\,°C \]
\[ \Delta T_b = 2.00\,°C \]
2. **Calculate Molality (m):**
Step 1: Calculate the moles of KBr.
\[ \text{Molar mass of KBr} = 39.1 \,(\text{K}) + 79.9 \,(\text{Br}) = 119 \,\text{g/mol} \]
\[ \text{Mass of KBr} = 0.13 \,\text](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1067c048-27d4-462b-825c-b686bc07047b%2Fe49247f7-44df-4c55-81f1-501757840995%2Fve3i4xy.png&w=3840&q=75)
Transcribed Image Text:**Boiling Point Elevation: Calculation of \( K_b \) for Liquid \( X \)**
The normal boiling point of a certain liquid \( X \) is 111.60 °C, but when 0.13 kg of potassium bromide (KBr) are dissolved in 900. g of \( X \), the solution boils at 113.6 °C instead. Use this information to calculate the molal boiling point elevation constant \( K_b \) of \( X \).
**Instructions:**
1. Report your answer to 2 significant digits.
**Given Data:**
- **Normal boiling point of liquid \( X \):** 111.60 °C
- **Boiling point of the solution:** 113.6 °C
- **Mass of potassium bromide (KBr) dissolved:** 0.13 kg
- **Mass of liquid \( X \):** 900. g (0.900 kg)
**Formula for Molal Boiling Point Elevation:**
\[ \Delta T_b = K_b \cdot m \]
Where:
- \(\Delta T_b\) is the boiling point elevation.
- \(K_b\) is the molal boiling point elevation constant.
- \(m\) is the molality of the solution (moles of solute per kilogram of solvent).
### Diagram Explanation:
#### Equation Box:
\[ K_b = \frac{ \Delta T_b }{ \text{molality} } \]
#### Input Box:
- A box indicating where to enter the value for \( K_b \).
- A checkbox indicating multiplication by 10.
- Buttons to submit (\(x\)), reset (\(\looparrowleft\)), and get help (\(?\)).
### Calculation Steps:
1. **Calculate Boiling Point Elevation:**
\[ \Delta T_b = 113.6\,°C - 111.60\,°C \]
\[ \Delta T_b = 2.00\,°C \]
2. **Calculate Molality (m):**
Step 1: Calculate the moles of KBr.
\[ \text{Molar mass of KBr} = 39.1 \,(\text{K}) + 79.9 \,(\text{Br}) = 119 \,\text{g/mol} \]
\[ \text{Mass of KBr} = 0.13 \,\text
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