A volume of 30.2 mL of a 2.16 M KMnO4 solution is mixed with 12.7 mL of a 0.492 M KMnO4 solution. Calculate the concentration of the final solution. Assume the volumes are additive. Round your answer to 3 significant digits. M x10 X
A volume of 30.2 mL of a 2.16 M KMnO4 solution is mixed with 12.7 mL of a 0.492 M KMnO4 solution. Calculate the concentration of the final solution. Assume the volumes are additive. Round your answer to 3 significant digits. M x10 X
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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![**Problem Statement:**
A volume of 30.2 mL of a 2.16 M KMnO₄ solution is mixed with 12.7 mL of a 0.492 M KMnO₄ solution. Calculate the concentration of the final solution.
Assume the volumes are additive. Round your answer to 3 significant digits.
**Answer Box:**
The final concentration is displayed in a box, with a placeholder for entering the concentration in molarity (M). There is functionality to multiply by 10 (×10), clear (×), or refresh (⟲) the input.
**Solution Explanation:**
To calculate the concentration of the final solution:
1. **Calculate the moles of KMnO₄ in each solution:**
- Moles from 30.2 mL of 2.16 M solution:
\[
\text{Moles} = 0.0302 \, \text{L} \times 2.16 \, \text{M} = 0.065232 \, \text{moles}
\]
- Moles from 12.7 mL of 0.492 M solution:
\[
\text{Moles} = 0.0127 \, \text{L} \times 0.492 \, \text{M} = 0.0062484 \, \text{moles}
\]
2. **Total moles of KMnO₄:**
\[
0.065232 + 0.0062484 = 0.0714804 \, \text{moles}
\]
3. **Total volume of the solution:**
\[
30.2 \, \text{mL} + 12.7 \, \text{mL} = 42.9 \, \text{mL} = 0.0429 \, \text{L}
\]
4. **Concentration of the final solution:**
\[
\text{Concentration} = \frac{0.0714804 \, \text{moles}}{0.0429 \, \text{L}} = 1.666732 \, \text{M}
\]
5. **Round to 3 significant digits:**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F84c202e2-51da-453e-8ae2-06a3b9c7648d%2Fedae8aeb-42ca-4f97-8565-d292fd0bc585%2Fqcpbr3n_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A volume of 30.2 mL of a 2.16 M KMnO₄ solution is mixed with 12.7 mL of a 0.492 M KMnO₄ solution. Calculate the concentration of the final solution.
Assume the volumes are additive. Round your answer to 3 significant digits.
**Answer Box:**
The final concentration is displayed in a box, with a placeholder for entering the concentration in molarity (M). There is functionality to multiply by 10 (×10), clear (×), or refresh (⟲) the input.
**Solution Explanation:**
To calculate the concentration of the final solution:
1. **Calculate the moles of KMnO₄ in each solution:**
- Moles from 30.2 mL of 2.16 M solution:
\[
\text{Moles} = 0.0302 \, \text{L} \times 2.16 \, \text{M} = 0.065232 \, \text{moles}
\]
- Moles from 12.7 mL of 0.492 M solution:
\[
\text{Moles} = 0.0127 \, \text{L} \times 0.492 \, \text{M} = 0.0062484 \, \text{moles}
\]
2. **Total moles of KMnO₄:**
\[
0.065232 + 0.0062484 = 0.0714804 \, \text{moles}
\]
3. **Total volume of the solution:**
\[
30.2 \, \text{mL} + 12.7 \, \text{mL} = 42.9 \, \text{mL} = 0.0429 \, \text{L}
\]
4. **Concentration of the final solution:**
\[
\text{Concentration} = \frac{0.0714804 \, \text{moles}}{0.0429 \, \text{L}} = 1.666732 \, \text{M}
\]
5. **Round to 3 significant digits:**
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