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
Chapter4: Types Of Chemical Reactions And Solution Stoichiometry
Section: Chapter Questions
Problem 47E
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![### Calculation of Moles of Ethanol (C₂H₅OH) in a Solution
**Question:**
What quantity in moles of \( \text{C}_2\text{H}_5\text{OH} \) are there in 63.4 mL of 0.250 M \( \text{C}_2\text{H}_5\text{OH} \)?
**Solution:**
1. **Understanding Molarity:**
- Molarity (M) is defined as the number of moles of solute (in this case, \( \text{C}_2\text{H}_5\text{OH} \)) per liter of solution.
- Given molarity (M) = 0.250 M, it means 0.250 moles of \( \text{C}_2\text{H}_5\text{OH} \) in 1 liter of the solution.
2. **Volume Conversion:**
- Volume given = 63.4 mL.
- Convert milliliters (mL) to liters (L):
\[ 63.4 \, \text{mL} = 63.4 \times 10^{-3} \, \text{L} \]
\[ 63.4 \, \text{mL} = 0.0634 \, \text{L} \]
3. **Calculate Moles:**
- Use the formula to calculate the moles of solute:
\[ \text{Moles of solute} = \text{Molarity} \times \text{Volume (in liters)} \]
- Substituting the values:
\[ \text{Moles of } \text{C}_2\text{H}_5\text{OH} = 0.250 \, \text{M} \times 0.0634 \, \text{L} \]
\[ \text{Moles of } \text{C}_2\text{H}_5\text{OH} = 0.01585 \, \text{moles} \]
**Answer:**
There are \(0.01585\) moles of \( \text{C}_2\text{H}_5\text{OH} \) in 63](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffbb15d99-77db-4308-835f-20d2902f98e0%2Ff3004213-36ea-45bc-bb28-342d5ecc390f%2Ffxz8p9_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculation of Moles of Ethanol (C₂H₅OH) in a Solution
**Question:**
What quantity in moles of \( \text{C}_2\text{H}_5\text{OH} \) are there in 63.4 mL of 0.250 M \( \text{C}_2\text{H}_5\text{OH} \)?
**Solution:**
1. **Understanding Molarity:**
- Molarity (M) is defined as the number of moles of solute (in this case, \( \text{C}_2\text{H}_5\text{OH} \)) per liter of solution.
- Given molarity (M) = 0.250 M, it means 0.250 moles of \( \text{C}_2\text{H}_5\text{OH} \) in 1 liter of the solution.
2. **Volume Conversion:**
- Volume given = 63.4 mL.
- Convert milliliters (mL) to liters (L):
\[ 63.4 \, \text{mL} = 63.4 \times 10^{-3} \, \text{L} \]
\[ 63.4 \, \text{mL} = 0.0634 \, \text{L} \]
3. **Calculate Moles:**
- Use the formula to calculate the moles of solute:
\[ \text{Moles of solute} = \text{Molarity} \times \text{Volume (in liters)} \]
- Substituting the values:
\[ \text{Moles of } \text{C}_2\text{H}_5\text{OH} = 0.250 \, \text{M} \times 0.0634 \, \text{L} \]
\[ \text{Moles of } \text{C}_2\text{H}_5\text{OH} = 0.01585 \, \text{moles} \]
**Answer:**
There are \(0.01585\) moles of \( \text{C}_2\text{H}_5\text{OH} \) in 63
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