Calculate Kb of the solvent given the normal boiling point of the 2.7m solution is 35.6 C. The normal boiling point of the solvent is 32.1°C. The solute is potassium iodide in a polar sOvene
Calculate Kb of the solvent given the normal boiling point of the 2.7m solution is 35.6 C. The normal boiling point of the solvent is 32.1°C. The solute is potassium iodide in a polar sOvene
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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![**Question 13:**
Calculate \( K_b \) of the solvent given the normal boiling point of the 2.7 m solution is 35.6°C. The normal boiling point of the solvent is 32.1°C. The solute is potassium iodide in a polar solvent.
---
In this problem, we are asked to calculate the boiling point elevation constant (\( K_b \)) for a solution. The boiling point elevation can be determined using the formula:
\[
\Delta T_b = K_b \cdot m
\]
Where:
- \(\Delta T_b\) is the boiling point elevation.
- \( K_b \) is the ebullioscopic constant.
- \( m \) is the molality of the solution.
**Given:**
- Normal boiling point of the solvent = 32.1°C
- Boiling point of the solution = 35.6°C
- Molality (\( m \)) = 2.7 m
**Steps to Solve:**
1. Calculate \(\Delta T_b\) by finding the difference between the boiling point of the solution and the normal boiling point of the solvent:
\[
\Delta T_b = 35.6°C - 32.1°C = 3.5°C
\]
2. Rearrange the formula to solve for \( K_b \):
\[
K_b = \frac{\Delta T_b}{m} = \frac{3.5°C}{2.7 \, \text{m}}
\]
3. Calculate \( K_b \).
This calculation will give you the ebullioscopic constant \( K_b \) for the solvent, which describes how much the boiling point increases per molal concentration of the solute.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7d2b8eef-4fbe-4a55-b286-a596fe8c553e%2F6c2b5398-a597-4df9-b1ce-cf565927c482%2Fnjb9eeh_processed.png&w=3840&q=75)
Transcribed Image Text:**Question 13:**
Calculate \( K_b \) of the solvent given the normal boiling point of the 2.7 m solution is 35.6°C. The normal boiling point of the solvent is 32.1°C. The solute is potassium iodide in a polar solvent.
---
In this problem, we are asked to calculate the boiling point elevation constant (\( K_b \)) for a solution. The boiling point elevation can be determined using the formula:
\[
\Delta T_b = K_b \cdot m
\]
Where:
- \(\Delta T_b\) is the boiling point elevation.
- \( K_b \) is the ebullioscopic constant.
- \( m \) is the molality of the solution.
**Given:**
- Normal boiling point of the solvent = 32.1°C
- Boiling point of the solution = 35.6°C
- Molality (\( m \)) = 2.7 m
**Steps to Solve:**
1. Calculate \(\Delta T_b\) by finding the difference between the boiling point of the solution and the normal boiling point of the solvent:
\[
\Delta T_b = 35.6°C - 32.1°C = 3.5°C
\]
2. Rearrange the formula to solve for \( K_b \):
\[
K_b = \frac{\Delta T_b}{m} = \frac{3.5°C}{2.7 \, \text{m}}
\]
3. Calculate \( K_b \).
This calculation will give you the ebullioscopic constant \( K_b \) for the solvent, which describes how much the boiling point increases per molal concentration of the solute.
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