A particular beer is 6.50% ethanol by volume (C₂H₂O). If a single bottle of beer contains 750.0 mL of beer, what mass, in g of ethanol (C₂H₂O), is present in the bottle? The density of ethanol is 0.789 g ethanol/mL ethanol
A particular beer is 6.50% ethanol by volume (C₂H₂O). If a single bottle of beer contains 750.0 mL of beer, what mass, in g of ethanol (C₂H₂O), is present in the bottle? The density of ethanol is 0.789 g ethanol/mL ethanol
Chemistry
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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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Question
Please set up the answer the correct way. I am having a hard time with putting the units in the correct spots.
![### Problem Statement
A particular beer is **6.50 % ethanol by volume (C₂H₆O)**. If a single bottle of beer contains **750.0 mL of beer**, what mass, in **g of ethanol (C₂H₆O)**, is present in the bottle? The density of ethanol is **0.789 g ethanol/mL ethanol**.
### Diagram Explanation
#### Equation Setup:
The diagram helps set up the calculation as follows:
**Starting Amount:**
- Initial volume of beer available, given as 750.0 mL.
**Calculation Formula:**
The diagram provides a structure to solve the problem step-by-step visually, using multiplication and unit conversions.
**Given Variables:**
- Percentage of ethanol by volume in beer: 6.50%
- Volume of the beer: 750.0 mL
- Density of ethanol: 0.789 g ethanol/mL ethanol
#### Detailed Steps:
1. **Determine Volume of Ethanol:**
\[ \text{Volume of ethanol} = \text{Volume of beer} \times \left( \frac{\text{\% ethanol}}{100} \right) \]
Where:
- Volume of beer is 750.0 mL
- Percentage of ethanol by volume is 6.50%
- Hence, Volume of ethanol is \( 750.0 \, \text{mL} \times \left( \frac{6.50}{100} \right) = 48.75 \, \text{mL ethanol} \)
2. **Convert Volume of Ethanol to Mass of Ethanol:**
\[ \text{Mass of ethanol} = \text{Volume of ethanol} \times \text{Density of ethanol} \]
- Volume of ethanol is 48.75 mL
- Density of ethanol is 0.789 g/mL
- Hence, Mass of ethanol is \( 48.75 \, \text{mL ethanol} \times 0.789 \, \text{g/mL} \)
3. **Calculation:**
\[ \text{Mass of ethanol} = 48.75 \, \text{mL ethanol} \times 0.789 \, \text{g/mL} = 38.45 \, \text{g ethanol} \]
### Conclusion](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffbb15d99-77db-4308-835f-20d2902f98e0%2F37f09111-f233-4c27-9ed1-ca92ed560d77%2Fqjg8z0l_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A particular beer is **6.50 % ethanol by volume (C₂H₆O)**. If a single bottle of beer contains **750.0 mL of beer**, what mass, in **g of ethanol (C₂H₆O)**, is present in the bottle? The density of ethanol is **0.789 g ethanol/mL ethanol**.
### Diagram Explanation
#### Equation Setup:
The diagram helps set up the calculation as follows:
**Starting Amount:**
- Initial volume of beer available, given as 750.0 mL.
**Calculation Formula:**
The diagram provides a structure to solve the problem step-by-step visually, using multiplication and unit conversions.
**Given Variables:**
- Percentage of ethanol by volume in beer: 6.50%
- Volume of the beer: 750.0 mL
- Density of ethanol: 0.789 g ethanol/mL ethanol
#### Detailed Steps:
1. **Determine Volume of Ethanol:**
\[ \text{Volume of ethanol} = \text{Volume of beer} \times \left( \frac{\text{\% ethanol}}{100} \right) \]
Where:
- Volume of beer is 750.0 mL
- Percentage of ethanol by volume is 6.50%
- Hence, Volume of ethanol is \( 750.0 \, \text{mL} \times \left( \frac{6.50}{100} \right) = 48.75 \, \text{mL ethanol} \)
2. **Convert Volume of Ethanol to Mass of Ethanol:**
\[ \text{Mass of ethanol} = \text{Volume of ethanol} \times \text{Density of ethanol} \]
- Volume of ethanol is 48.75 mL
- Density of ethanol is 0.789 g/mL
- Hence, Mass of ethanol is \( 48.75 \, \text{mL ethanol} \times 0.789 \, \text{g/mL} \)
3. **Calculation:**
\[ \text{Mass of ethanol} = 48.75 \, \text{mL ethanol} \times 0.789 \, \text{g/mL} = 38.45 \, \text{g ethanol} \]
### Conclusion
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