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30.00 mL of 6.00 M HCI is added to 570 mL of water to make a diluted solution. Calculate the new molarity

30.00 mL of 6.00 M HCI is added to 570 mL of water to make a diluted solution. Calculate the new molarity

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:** 30.00 mL of 6.00 M HCl is added to 570 mL of water to make a diluted solution. Calculate the new molarity. --- **Instructions for Solution:** To find the new molarity of the solution, use the dilution formula: \[ M_1 \times V_1 = M_2 \times V_2 \] where: - \( M_1 \) is the initial molarity (6.00 M) - \( V_1 \) is the initial volume (30.00 mL) - \( M_2 \) is the final molarity (unknown) - \( V_2 \) is the final volume (the sum of the initial solution volume and the water volume added, which is 600 mL) Step 1: Calculate the total volume \( V_2 \): \[ V_2 = 30.00 \, \text{mL} + 570 \, \text{mL} = 600.00 \, \text{mL} \] Step 2: Rearrange the dilution formula to solve for \( M_2 \): \[ M_2 = \frac{M_1 \times V_1}{V_2} \] Step 3: Substitute the known values into the equation: \[ M_2 = \frac{6.00 \, \text{M} \times 30.00 \, \text{mL}}{600.00 \, \text{mL}} \] Step 4: Calculate \( M_2 \): \[ M_2 = \frac{180.00 \, \text{M} \cdot \text{mL}}{600.00 \, \text{mL}} = 0.30 \, \text{M} \] Thus, the new molarity of the diluted HCl solution is 0.30 M.
Transcribed Image Text:**Problem Statement:** 30.00 mL of 6.00 M HCl is added to 570 mL of water to make a diluted solution. Calculate the new molarity. --- **Instructions for Solution:** To find the new molarity of the solution, use the dilution formula: \[ M_1 \times V_1 = M_2 \times V_2 \] where: - \( M_1 \) is the initial molarity (6.00 M) - \( V_1 \) is the initial volume (30.00 mL) - \( M_2 \) is the final molarity (unknown) - \( V_2 \) is the final volume (the sum of the initial solution volume and the water volume added, which is 600 mL) Step 1: Calculate the total volume \( V_2 \): \[ V_2 = 30.00 \, \text{mL} + 570 \, \text{mL} = 600.00 \, \text{mL} \] Step 2: Rearrange the dilution formula to solve for \( M_2 \): \[ M_2 = \frac{M_1 \times V_1}{V_2} \] Step 3: Substitute the known values into the equation: \[ M_2 = \frac{6.00 \, \text{M} \times 30.00 \, \text{mL}}{600.00 \, \text{mL}} \] Step 4: Calculate \( M_2 \): \[ M_2 = \frac{180.00 \, \text{M} \cdot \text{mL}}{600.00 \, \text{mL}} = 0.30 \, \text{M} \] Thus, the new molarity of the diluted HCl solution is 0.30 M.
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