3. The manufacturer of the vinegar used in this experiment claims that the vinegar contains 5% acetic acid by weight. Using your results, molarity of vinegar, and 1.0 g/ml for a density of vinegar, calculate the % acetic acid by weight. Can you tell if the manufacturer's claims is reasonable?
3. The manufacturer of the vinegar used in this experiment claims that the vinegar contains 5% acetic acid by weight. Using your results, molarity of vinegar, and 1.0 g/ml for a density of vinegar, calculate the % acetic acid by weight. Can you tell if the manufacturer's claims is reasonable?
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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Can you help me solve question 3? “The manufacturer of vinegar…”

Transcribed Image Text:3. The manufacturer of the vinegar used in this experiment claims that the vinegar contains 5% acetic acid by weight. Using your results, molarity of vinegar, and 1.0 g/ml for a density of vinegar, calculate the % acetic acid by weight. Can you tell if the manufacturer's claims are reasonable?
![**Data**
- **Molarity of the NaOH solution, M:** 0.250 M
- **Initial buret reading, mL:**
- 2.10 mL
- 0.00 mL
- 6.0 mL
- **Final buret reading, mL:**
- 38.60 mL
- 34.30 mL
- 40.4 mL
- **Volume of NaOH solution, mL:**
- 36.50 mL
- 34.30 mL
- 34.4 mL
- **Volume of CH₃COOH, mL:** 10.00 mL
- **Concentration of CH₃COOH, M:**
- 0.9135
- 0.8575
- 0.8600
- **Mean concentration of CH₃COOH, M:** 0.8777
**Calculation**
\[
\frac{m_1V_1}{n_1} = \frac{m_2V_2}{n_2}
\]
- **Molarity (mol/L):** Number of moles (mol) = Molarity × Volume
- 0.009125 = 0.25 × 0.0365
- Conversion factor used: \( \frac{x 10}{1} \)
Example calculations:
- \( 0.085 = 0.25 \times 0.0344 \)
- \( 0.86 = \frac{0.0086}{0.010} \)
The calculations were done to find the molarity of the acetic acid solution in different trials, and the mean concentration was calculated as 0.8777 M.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F21a85cdb-354b-48e5-a602-13d990cff496%2F93a02d9a-c07b-4638-a1b3-6c6122ddef14%2F7drp5kt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Data**
- **Molarity of the NaOH solution, M:** 0.250 M
- **Initial buret reading, mL:**
- 2.10 mL
- 0.00 mL
- 6.0 mL
- **Final buret reading, mL:**
- 38.60 mL
- 34.30 mL
- 40.4 mL
- **Volume of NaOH solution, mL:**
- 36.50 mL
- 34.30 mL
- 34.4 mL
- **Volume of CH₃COOH, mL:** 10.00 mL
- **Concentration of CH₃COOH, M:**
- 0.9135
- 0.8575
- 0.8600
- **Mean concentration of CH₃COOH, M:** 0.8777
**Calculation**
\[
\frac{m_1V_1}{n_1} = \frac{m_2V_2}{n_2}
\]
- **Molarity (mol/L):** Number of moles (mol) = Molarity × Volume
- 0.009125 = 0.25 × 0.0365
- Conversion factor used: \( \frac{x 10}{1} \)
Example calculations:
- \( 0.085 = 0.25 \times 0.0344 \)
- \( 0.86 = \frac{0.0086}{0.010} \)
The calculations were done to find the molarity of the acetic acid solution in different trials, and the mean concentration was calculated as 0.8777 M.
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