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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![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
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