Question 19 of 30 What is the predicted change in the boiling point of water when 4.00 g of barium chloride (BaCl,) is dissolved in 2.00 kg of water? Kp of water 0.51°C/mol molar mass BaCl2 = 208.23 g/mol ivalue of BaCl2 = 3 !3! %3D A. -1.0°C B. 0.0016°C O C. 0.015°C D. -0.0049°C
Question 19 of 30 What is the predicted change in the boiling point of water when 4.00 g of barium chloride (BaCl,) is dissolved in 2.00 kg of water? Kp of water 0.51°C/mol molar mass BaCl2 = 208.23 g/mol ivalue of BaCl2 = 3 !3! %3D A. -1.0°C B. 0.0016°C O C. 0.015°C D. -0.0049°C
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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![### Pretest: Unit 1
#### Question 19 of 30
**Question:**
What is the predicted change in the boiling point of water when 4.00 g of barium chloride (BaCl₂) is dissolved in 2.00 kg of water?
**Given Data:**
- \( K_b \) of water = 0.51°C/mol
- Molar mass of BaCl₂ = 208.23 g/mol
- \( i \) value of BaCl₂ = 3
**Options:**
- A. -1.0°C
- B. 0.0016°C
- C. 0.015°C
- D. -0.0049°C
**Explanation:**
To calculate the boiling point elevation, we use the formula:
\[ \Delta T_b = i \times K_b \times m \]
where:
- \( \Delta T_b \) is the change in boiling point,
- \( i \) is the van't Hoff factor,
- \( K_b \) is the ebullioscopic constant,
- \( m \) is the molality of the solution.
The molality \( m \) is calculated as follows:
\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]
First, we need to find the moles of BaCl₂:
\[ \text{Moles of BaCl₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{4.00 \, \text{g}}{208.23 \, \text{g/mol}} = 0.0192 \, \text{mol} \]
Next, we calculate the molality:
\[ m = \frac{0.0192 \, \text{mol}}{2.00 \, \text{kg}} = 0.0096 \, \text{mol/kg} \]
Finally, we calculate the change in boiling point:
\[ \Delta T_b = 3 \times 0.51 \, \text{°C/mol} \times 0.0096 \, \text{mol/kg} = 0.0147 \, \text{°C} \]
**Answer:**
The closest option to our calculated value is:
- C. 0.015°C](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ea5bb80-8ede-42b5-bd36-211bb2d7d3bc%2F6c5f5242-3f6d-4a25-a785-3095b2876357%2Fzdraps6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Pretest: Unit 1
#### Question 19 of 30
**Question:**
What is the predicted change in the boiling point of water when 4.00 g of barium chloride (BaCl₂) is dissolved in 2.00 kg of water?
**Given Data:**
- \( K_b \) of water = 0.51°C/mol
- Molar mass of BaCl₂ = 208.23 g/mol
- \( i \) value of BaCl₂ = 3
**Options:**
- A. -1.0°C
- B. 0.0016°C
- C. 0.015°C
- D. -0.0049°C
**Explanation:**
To calculate the boiling point elevation, we use the formula:
\[ \Delta T_b = i \times K_b \times m \]
where:
- \( \Delta T_b \) is the change in boiling point,
- \( i \) is the van't Hoff factor,
- \( K_b \) is the ebullioscopic constant,
- \( m \) is the molality of the solution.
The molality \( m \) is calculated as follows:
\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]
First, we need to find the moles of BaCl₂:
\[ \text{Moles of BaCl₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{4.00 \, \text{g}}{208.23 \, \text{g/mol}} = 0.0192 \, \text{mol} \]
Next, we calculate the molality:
\[ m = \frac{0.0192 \, \text{mol}}{2.00 \, \text{kg}} = 0.0096 \, \text{mol/kg} \]
Finally, we calculate the change in boiling point:
\[ \Delta T_b = 3 \times 0.51 \, \text{°C/mol} \times 0.0096 \, \text{mol/kg} = 0.0147 \, \text{°C} \]
**Answer:**
The closest option to our calculated value is:
- C. 0.015°C
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