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100.0 mL of a 0.755 M solution of KBr is diluted to 500.0 mL. What is the new concentration of the solution?

100.0 mL of a 0.755 M solution of KBr is diluted to 500.0 mL. What is the new concentration of the solution?

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:** 100.0 mL of a 0.755 M solution of KBr is diluted to 500.0 mL. What is the new concentration of the solution? --- To find the new concentration after dilution, use the formula: \[ C_1V_1 = C_2V_2 \] Where: - \( C_1 \) is the initial concentration (0.755 M) - \( V_1 \) is the initial volume (100.0 mL) - \( C_2 \) is the final concentration (unknown) - \( V_2 \) is the final volume (500.0 mL) Rearrange the formula to solve for \( C_2 \): \[ C_2 = \frac{C_1V_1}{V_2} \] Substitute the known values: \[ C_2 = \frac{0.755 \, \text{M} \times 100.0 \, \text{mL}}{500.0 \, \text{mL}} \] \[ C_2 = \frac{75.5}{500.0} \] \[ C_2 = 0.151 \, \text{M} \] Therefore, the new concentration of the KBr solution is 0.151 M.
Transcribed Image Text:**Problem Statement:** 100.0 mL of a 0.755 M solution of KBr is diluted to 500.0 mL. What is the new concentration of the solution? --- To find the new concentration after dilution, use the formula: \[ C_1V_1 = C_2V_2 \] Where: - \( C_1 \) is the initial concentration (0.755 M) - \( V_1 \) is the initial volume (100.0 mL) - \( C_2 \) is the final concentration (unknown) - \( V_2 \) is the final volume (500.0 mL) Rearrange the formula to solve for \( C_2 \): \[ C_2 = \frac{C_1V_1}{V_2} \] Substitute the known values: \[ C_2 = \frac{0.755 \, \text{M} \times 100.0 \, \text{mL}}{500.0 \, \text{mL}} \] \[ C_2 = \frac{75.5}{500.0} \] \[ C_2 = 0.151 \, \text{M} \] Therefore, the new concentration of the KBr solution is 0.151 M.
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