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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.

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.

Chemistry
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.
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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