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If 0.25 L of a 9.1 M HNO3 solution is diluted to 2.0 L, what is the molarity of the new solution?

If 0.25 L of a 9.1 M HNO3 solution is diluted to 2.0 L, what is the molarity of the new 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:** If 0.25 L of a 9.1 M HNO₃ solution is diluted to 2.0 L, what is the molarity of the new solution? **Explanation:** This question involves a dilution calculation, which is based on the principle of conservation of moles. The formula used is: \[ M_1 \times V_1 = M_2 \times V_2 \] Where: - \( M_1 \) is the initial molarity, - \( V_1 \) is the initial volume, - \( M_2 \) is the final molarity, - \( V_2 \) is the final volume. Given: - \( M_1 = 9.1 \, \text{M} \) - \( V_1 = 0.25 \, \text{L} \) - \( V_2 = 2.0 \, \text{L} \) Substitute the known values into the equation to find \( M_2 \): \[ 9.1 \, \text{M} \times 0.25 \, \text{L} = M_2 \times 2.0 \, \text{L} \] Solve for \( M_2 \): \[ M_2 = \frac{9.1 \, \text{M} \times 0.25 \, \text{L}}{2.0 \, \text{L}} \] \[ M_2 = 1.1375 \, \text{M} \] Thus, the molarity of the new solution is approximately 1.14 M.
Transcribed Image Text:**Problem Statement:** If 0.25 L of a 9.1 M HNO₃ solution is diluted to 2.0 L, what is the molarity of the new solution? **Explanation:** This question involves a dilution calculation, which is based on the principle of conservation of moles. The formula used is: \[ M_1 \times V_1 = M_2 \times V_2 \] Where: - \( M_1 \) is the initial molarity, - \( V_1 \) is the initial volume, - \( M_2 \) is the final molarity, - \( V_2 \) is the final volume. Given: - \( M_1 = 9.1 \, \text{M} \) - \( V_1 = 0.25 \, \text{L} \) - \( V_2 = 2.0 \, \text{L} \) Substitute the known values into the equation to find \( M_2 \): \[ 9.1 \, \text{M} \times 0.25 \, \text{L} = M_2 \times 2.0 \, \text{L} \] Solve for \( M_2 \): \[ M_2 = \frac{9.1 \, \text{M} \times 0.25 \, \text{L}}{2.0 \, \text{L}} \] \[ M_2 = 1.1375 \, \text{M} \] Thus, the molarity of the new solution is approximately 1.14 M.
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