**Problem: Dilution of KBr Solution**You have 100.0 mL of a 0.545 M solution of KBr. This solution is diluted to a final volume of 500.0 mL. What is the new concentration of the solution?**Explanation:**To find the new concentration after dilution, use the dilution formula:\[ C_1V_1 = C_2V_2 \]Where:- \( C_1 \) is the initial concentration (0.545 M).- \( V_1 \) is the initial volume (100.0 mL).- \( C_2 \) is the final concentration.- \( V_2 \) is the final volume (500.0 mL).Rearranging the formula to solve for \( C_2 \):\[ C_2 = \frac{C_1V_1}{V_2} \]Plug in the values:\[ C_2 = \frac{0.545 \, \text{M} \times 100.0 \, \text{mL}}{500.0 \, \text{mL}} \]\[ C_2 = \frac{54.5 \, \text{M} \cdot \text{mL}}{500.0 \, \text{mL}} \]\[ C_2 = 0.109 \, \text{M} \]**Conclusion:**The new concentration of the KBr solution after dilution is 0.109 M.

Relation between molarity, volume for reaction between two solutions at - M1V1=M2V2 Where,…

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100.0 mL of a 0.545 M solution of KBr is diluted to 500.0 mL. What is the new concentration of the solution?

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: Dilution of KBr Solution**

You have 100.0 mL of a 0.545 M solution of KBr. This solution is diluted to a final volume of 500.0 mL. What is the new concentration of the solution?

**Explanation:**

To find the new concentration after dilution, use the dilution formula:

\[ C_1V_1 = C_2V_2 \]

Where:
- \( C_1 \) is the initial concentration (0.545 M).
- \( V_1 \) is the initial volume (100.0 mL).
- \( C_2 \) is the final concentration.
- \( V_2 \) is the final volume (500.0 mL).

Rearranging the formula to solve for \( C_2 \):

\[ C_2 = \frac{C_1V_1}{V_2} \]

Plug in the values:

\[ C_2 = \frac{0.545 \, \text{M} \times 100.0 \, \text{mL}}{500.0 \, \text{mL}} \]

\[ C_2 = \frac{54.5 \, \text{M} \cdot \text{mL}}{500.0 \, \text{mL}} \]

\[ C_2 = 0.109 \, \text{M} \]

**Conclusion:**

The new concentration of the KBr solution after dilution is 0.109 M.

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