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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![**Problem Statement:**
Determine the volume of 0.20 M iron (III) nitrate needed to prepare 50.00 mL of a 0.0100 M solution of Fe(NO₃)₃.
**Explanation:**
This question requires the use of the dilution formula:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) is the concentration of the initial solution (0.20 M in this case).
- \( V_1 \) is the volume of the initial solution (what we need to find out).
- \( C_2 \) is the concentration of the final solution (0.0100 M in this case).
- \( V_2 \) is the volume of the final solution (50.00 mL in this case).
By solving for \( V_1 \), we get:
\[ V_1 = \frac{C_2V_2}{C_1} \]
Substitute the given values into the equation:
\[ V_1 = \frac{(0.0100 \, \text{M}) (50.00 \, \text{mL})}{0.20 \, \text{M}} \]
Finally, calculating the volume:
\[ V_1 = \frac{0.50 \, \text{M·mL}}{0.20 \, \text{M}} = 2.50 \, \text{mL} \]
So, the volume of 0.20 M iron (III) nitrate needed is 2.50 mL.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe0ae19d5-3750-41df-884f-8ca3aa31e002%2Fad71b092-0a05-4b95-81f3-fa54acf2e0c1%2Fwvc0hu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Determine the volume of 0.20 M iron (III) nitrate needed to prepare 50.00 mL of a 0.0100 M solution of Fe(NO₃)₃.
**Explanation:**
This question requires the use of the dilution formula:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) is the concentration of the initial solution (0.20 M in this case).
- \( V_1 \) is the volume of the initial solution (what we need to find out).
- \( C_2 \) is the concentration of the final solution (0.0100 M in this case).
- \( V_2 \) is the volume of the final solution (50.00 mL in this case).
By solving for \( V_1 \), we get:
\[ V_1 = \frac{C_2V_2}{C_1} \]
Substitute the given values into the equation:
\[ V_1 = \frac{(0.0100 \, \text{M}) (50.00 \, \text{mL})}{0.20 \, \text{M}} \]
Finally, calculating the volume:
\[ V_1 = \frac{0.50 \, \text{M·mL}}{0.20 \, \text{M}} = 2.50 \, \text{mL} \]
So, the volume of 0.20 M iron (III) nitrate needed is 2.50 mL.
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