Below is a diagram of two solutions separated by a water-permeable membrane. Label the tonicity of each solution relative to each other (and show your calculations) and draw an arrow indicating the direction of water movement. Dıici=2.07g/mL 15% v/v LİCI 4 M LİCI
Below is a diagram of two solutions separated by a water-permeable membrane. Label the tonicity of each solution relative to each other (and show your calculations) and draw an arrow indicating the direction of water movement. Dıici=2.07g/mL 15% v/v LİCI 4 M LİCI
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...
Related questions
Question
100%
![Below is a diagram of two solutions separated by a water-permeable membrane. Label the tonicity of each solution relative to each other (and show your calculations) and draw an arrow indicating the direction of water movement. D<sub>LiCl</sub> = 2.07 g/mL
**Diagram Explanation:**
The diagram depicts two compartments separated by a water-permeable membrane.
- The left compartment contains a solution of 15% v/v LiCl.
- The right compartment contains a 4 M LiCl solution.
**Explanation:**
1. **Calculate the concentration in grams/L for each solution:**
- **15% v/v LiCl:**
- Assume 100 mL of solution, which contains 15 mL of LiCl.
- Convert volume to mass using the density, D<sub>LiCl</sub> = 2.07 g/mL:
\[
\text{Mass of LiCl in 15% v/v solution} = 15 \, \text{mL} \times 2.07 \, \text{g/mL} = 31.05 \, \text{g}
\]
- Convert to g/L (assuming 100 mL solution is used as standard):
\[
\text{Concentration} = \frac{31.05 \, \text{g}}{0.1 \, \text{L}} = 310.5 \, \text{g/L}
\]
- **4 M LiCl:**
- Molar mass of LiCl = approximately 42.39 g/mol.
- Concentration in grams/L:
\[
\text{4 M LiCl} = 4 \times 42.39 \, \text{g/mol} = 169.56 \, \text{g/L}
\]
2. **Compare concentrations to determine tonicity:**
- 15% v/v LiCl = 310.5 g/L
- 4 M LiCl = 169.56 g/L
Since 310.5 g/L (15% v/v) > 169.56 g/L (4 M), the 15% v/v solution is hypertonic relative to the 4 M solution.
3. **Indicate direction of water movement:**
Water will move from the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F249affda-f293-4437-a106-927c7444442e%2F90e849ee-6dc9-4986-b83e-ef2129138b23%2Fjkx9o7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Below is a diagram of two solutions separated by a water-permeable membrane. Label the tonicity of each solution relative to each other (and show your calculations) and draw an arrow indicating the direction of water movement. D<sub>LiCl</sub> = 2.07 g/mL
**Diagram Explanation:**
The diagram depicts two compartments separated by a water-permeable membrane.
- The left compartment contains a solution of 15% v/v LiCl.
- The right compartment contains a 4 M LiCl solution.
**Explanation:**
1. **Calculate the concentration in grams/L for each solution:**
- **15% v/v LiCl:**
- Assume 100 mL of solution, which contains 15 mL of LiCl.
- Convert volume to mass using the density, D<sub>LiCl</sub> = 2.07 g/mL:
\[
\text{Mass of LiCl in 15% v/v solution} = 15 \, \text{mL} \times 2.07 \, \text{g/mL} = 31.05 \, \text{g}
\]
- Convert to g/L (assuming 100 mL solution is used as standard):
\[
\text{Concentration} = \frac{31.05 \, \text{g}}{0.1 \, \text{L}} = 310.5 \, \text{g/L}
\]
- **4 M LiCl:**
- Molar mass of LiCl = approximately 42.39 g/mol.
- Concentration in grams/L:
\[
\text{4 M LiCl} = 4 \times 42.39 \, \text{g/mol} = 169.56 \, \text{g/L}
\]
2. **Compare concentrations to determine tonicity:**
- 15% v/v LiCl = 310.5 g/L
- 4 M LiCl = 169.56 g/L
Since 310.5 g/L (15% v/v) > 169.56 g/L (4 M), the 15% v/v solution is hypertonic relative to the 4 M solution.
3. **Indicate direction of water movement:**
Water will move from the
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 1 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY