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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![### Determining the Molality of an Antifreeze Solution
Antifreeze is often about 50.0% ethylene glycol (C₂H₆O₂) in water. To find the molality of this solution, we use the provided information and the following steps:
#### Information Given:
- Molecular Mass (Molar Mass) of Ethylene Glycol (C₂H₆O₂): 62.068 g/mol
- Composition of Antifreeze: 50.0% by weight of ethylene glycol in water.
#### What We Need to Find:
- Molality (m) of the solution.
### Step-by-Step Calculation:
1. **Understanding Molality:**
Molality (m) is defined as the number of moles of solute (in this case, ethylene glycol) per kilogram of solvent (water). The formula is:
\[
m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}
\]
2. **Calculating Moles of Ethylene Glycol:**
Let’s assume we have 100 g of antifreeze solution (since it's easy to work with percentage composition).
- Ethylene glycol is 50.0% by weight in the solution.
- Therefore, in 100 g of solution, we have 50 g of ethylene glycol.
Now, convert grams to moles using the given molar mass:
\[
\text{moles of } C_2H_6O_2 = \frac{50 \, \text{g}}{62.068 \, \text{g/mol}} = 0.805 \, \text{mol}
\]
3. **Calculating Mass of Water:**
- Since 50 g of the solution is ethylene glycol, the remaining 50 g is water.
- Convert grams to kilograms:
\[
\text{mass of water} = 50 \, \text{g} = 0.050 \, \text{kg}
\]
4. **Calculating Molality:**
\[
m = \frac{0.805 \, \text{mol}}{0.050 \, \text{kg}} = 16.1 \, \text{mol/kg}
\]
### Solution:
The molality of the ethylene glycol](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdfd84302-10e6-446d-ac4f-33b2622f5f16%2F9f2af93c-3ddf-4b47-9bcc-ad569c6a91d2%2Fau5iprp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Determining the Molality of an Antifreeze Solution
Antifreeze is often about 50.0% ethylene glycol (C₂H₆O₂) in water. To find the molality of this solution, we use the provided information and the following steps:
#### Information Given:
- Molecular Mass (Molar Mass) of Ethylene Glycol (C₂H₆O₂): 62.068 g/mol
- Composition of Antifreeze: 50.0% by weight of ethylene glycol in water.
#### What We Need to Find:
- Molality (m) of the solution.
### Step-by-Step Calculation:
1. **Understanding Molality:**
Molality (m) is defined as the number of moles of solute (in this case, ethylene glycol) per kilogram of solvent (water). The formula is:
\[
m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}
\]
2. **Calculating Moles of Ethylene Glycol:**
Let’s assume we have 100 g of antifreeze solution (since it's easy to work with percentage composition).
- Ethylene glycol is 50.0% by weight in the solution.
- Therefore, in 100 g of solution, we have 50 g of ethylene glycol.
Now, convert grams to moles using the given molar mass:
\[
\text{moles of } C_2H_6O_2 = \frac{50 \, \text{g}}{62.068 \, \text{g/mol}} = 0.805 \, \text{mol}
\]
3. **Calculating Mass of Water:**
- Since 50 g of the solution is ethylene glycol, the remaining 50 g is water.
- Convert grams to kilograms:
\[
\text{mass of water} = 50 \, \text{g} = 0.050 \, \text{kg}
\]
4. **Calculating Molality:**
\[
m = \frac{0.805 \, \text{mol}}{0.050 \, \text{kg}} = 16.1 \, \text{mol/kg}
\]
### Solution:
The molality of the ethylene glycol
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