**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.

We have 0.20 M of Ferric nitrate, which we want to dilute to 0.01 M of 50 mL. We have to calculate…

Answered: Determine the volume of 0.20 M iron…

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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.

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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Concept explainers

Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

Question

**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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