A student determined the average molarity of their sample of acetic acid to be 0.536 M. Calculate the (w/w)% of the solution. Assume the density of this solution is that of pure water, dH2O = 1000. g/L. Enter a number without the '%' symbol. (I'm struggling to find the Volume from the attached equation(s) that are "meant" to be used for this problem)
A student determined the average molarity of their sample of acetic acid to be 0.536 M. Calculate the (w/w)% of the solution. Assume the density of this solution is that of pure water, dH2O = 1000. g/L. Enter a number without the '%' symbol. (I'm struggling to find the Volume from the attached equation(s) that are "meant" to be used for this problem)
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
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A student determined the average molarity of their sample of acetic acid to be 0.536 M. Calculate the (w/w)% of the solution. Assume the density of this solution is that of pure water, dH2O = 1000. g/L. Enter a number without the '%' symbol.
(I'm struggling to find the Volume from the attached equation(s) that are "meant" to be used for this problem)
![**Molarity to (w/w)%**
To convert the molar concentration of acetic acid to a weight/weight percentage ((w/w)% concentration), follow these steps:
1. **Convert Molar Concentration:**
- Begin by converting the molarity (mol/L) of acetic acid to grams per milliliter (g/mL). This is achieved by multiplying the molarity by the molar mass of acetic acid and then adjusting from liters to milliliters.
- The formula used is:
\[
M_{HC_2H_3O_2} \cdot \frac{60.06 \, \text{g}}{1 \, \text{mol}} \cdot \frac{1 \, \text{L}}{10^3 \, \text{mL}} = \frac{m_{HC_2H_3O_2}}{V_{\text{solution}}}
\]
- \(M_{HC_2H_3O_2}\) is the molarity of acetic acid.
- 60.06 g/mol is the molar mass of acetic acid.
- \(V_{\text{solution}}\) is the volume of the solution.
2. **Calculate (w/w)%:**
- The (w/w)% is the ratio of the mass of acetic acid to the mass of the vinegar solution.
- To find the mass of the solution, use the relationship: density times volume equals mass.
- The formula for (w/w)% is:
\[
\left( \frac{w}{w} \right) \% = \frac{m_{HC_2H_3O_2}}{V_{\text{solution}}} \cdot \frac{1}{d_{\text{solution}}} \cdot 100
\]
- \(m_{HC_2H_3O_2}\) is the mass of acetic acid.
- \(d_{\text{solution}}\) is the density of the solution.
- This formula helps calculate the weight percentage by adjusting for the solution’s density and converting it into a percentage by multiplying by 100.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F36f9df77-a167-4a5a-b65f-f6706a95cade%2Fd93a7579-119f-4f50-a834-3b774e9a9fe5%2Fihgpomj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Molarity to (w/w)%**
To convert the molar concentration of acetic acid to a weight/weight percentage ((w/w)% concentration), follow these steps:
1. **Convert Molar Concentration:**
- Begin by converting the molarity (mol/L) of acetic acid to grams per milliliter (g/mL). This is achieved by multiplying the molarity by the molar mass of acetic acid and then adjusting from liters to milliliters.
- The formula used is:
\[
M_{HC_2H_3O_2} \cdot \frac{60.06 \, \text{g}}{1 \, \text{mol}} \cdot \frac{1 \, \text{L}}{10^3 \, \text{mL}} = \frac{m_{HC_2H_3O_2}}{V_{\text{solution}}}
\]
- \(M_{HC_2H_3O_2}\) is the molarity of acetic acid.
- 60.06 g/mol is the molar mass of acetic acid.
- \(V_{\text{solution}}\) is the volume of the solution.
2. **Calculate (w/w)%:**
- The (w/w)% is the ratio of the mass of acetic acid to the mass of the vinegar solution.
- To find the mass of the solution, use the relationship: density times volume equals mass.
- The formula for (w/w)% is:
\[
\left( \frac{w}{w} \right) \% = \frac{m_{HC_2H_3O_2}}{V_{\text{solution}}} \cdot \frac{1}{d_{\text{solution}}} \cdot 100
\]
- \(m_{HC_2H_3O_2}\) is the mass of acetic acid.
- \(d_{\text{solution}}\) is the density of the solution.
- This formula helps calculate the weight percentage by adjusting for the solution’s density and converting it into a percentage by multiplying by 100.
Transcribed Image Text:A student determined the average molarity of their sample of acetic acid to be 0.536 M. Calculate the (w/w)% of the solution. Assume the density of this solution is that of pure water, \(d_{\text{H}_2\text{O}} = 1000.\) g/L. Enter a number without the '%' symbol.
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