(3) - How many milliliters of 8.50 M stock HCI are needed to create 2.00 L of 4.00 M HCI solution?

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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**Problem 3:**

How many milliliters of 8.50 M stock HCl are needed to create 2.00 L of 4.00 M HCl solution?

**Explanation:**

This problem involves dilution, where a concentrated solution is used to prepare a diluted solution. The relationship for diluting solutions can be expressed by the formula:

\[ C_1V_1 = C_2V_2 \]

Where:
- \( C_1 \) is the concentration of the stock solution.
- \( V_1 \) is the volume of the stock solution needed.
- \( C_2 \) is the concentration of the final solution.
- \( V_2 \) is the volume of the final solution.

Here, \( C_1 = 8.50 \, M \), \( C_2 = 4.00 \, M \), and \( V_2 = 2.00 \, L \). By substituting these values into the formula, the required volume \( V_1 \) of the stock solution can be calculated.
Transcribed Image Text:**Problem 3:** How many milliliters of 8.50 M stock HCl are needed to create 2.00 L of 4.00 M HCl solution? **Explanation:** This problem involves dilution, where a concentrated solution is used to prepare a diluted solution. The relationship for diluting solutions can be expressed by the formula: \[ C_1V_1 = C_2V_2 \] Where: - \( C_1 \) is the concentration of the stock solution. - \( V_1 \) is the volume of the stock solution needed. - \( C_2 \) is the concentration of the final solution. - \( V_2 \) is the volume of the final solution. Here, \( C_1 = 8.50 \, M \), \( C_2 = 4.00 \, M \), and \( V_2 = 2.00 \, L \). By substituting these values into the formula, the required volume \( V_1 \) of the stock solution can be calculated.
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