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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Question
![**Question:**
To what volume should 16 mL of a 6.7 M HCl solution be diluted to prepare a 0.30 M HCl solution?
**Explanation:**
This exercise asks for the final volume required to dilute a concentrated hydrochloric acid (HCl) solution. It involves using the concept of dilution in chemistry.
**Formula for Dilution:**
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) is the initial concentration (6.7 M HCl),
- \( V_1 \) is the initial volume (16 mL),
- \( C_2 \) is the final concentration (0.30 M HCl),
- \( V_2 \) is the final volume (unknown).
Using this formula, you can solve for \( V_2 \):
\[ (6.7\, \text{M}) \times (16\, \text{mL}) = (0.30\, \text{M}) \times V_2 \]
From this equation, rearrange to solve for \( V_2 \):
\[ V_2 = \frac{6.7\, \text{M} \times 16\, \text{mL}}{0.30\, \text{M}} \]
Calculate the result:
\[ V_2 = \frac{107.2\, \text{mL} \cdot \text{M}}{0.30\, \text{M}} \]
\[ V_2 = 357.33\, \text{mL} \]
Therefore, the 16 mL of a 6.7 M HCl solution should be diluted to 357.33 mL to prepare a 0.30 M HCl solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf4fd61a-46d9-45b2-8502-88fe4eac45cd%2Fedf1e3c5-1f92-424c-b1a5-d453a25c69f8%2F88vsqs_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
To what volume should 16 mL of a 6.7 M HCl solution be diluted to prepare a 0.30 M HCl solution?
**Explanation:**
This exercise asks for the final volume required to dilute a concentrated hydrochloric acid (HCl) solution. It involves using the concept of dilution in chemistry.
**Formula for Dilution:**
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) is the initial concentration (6.7 M HCl),
- \( V_1 \) is the initial volume (16 mL),
- \( C_2 \) is the final concentration (0.30 M HCl),
- \( V_2 \) is the final volume (unknown).
Using this formula, you can solve for \( V_2 \):
\[ (6.7\, \text{M}) \times (16\, \text{mL}) = (0.30\, \text{M}) \times V_2 \]
From this equation, rearrange to solve for \( V_2 \):
\[ V_2 = \frac{6.7\, \text{M} \times 16\, \text{mL}}{0.30\, \text{M}} \]
Calculate the result:
\[ V_2 = \frac{107.2\, \text{mL} \cdot \text{M}}{0.30\, \text{M}} \]
\[ V_2 = 357.33\, \text{mL} \]
Therefore, the 16 mL of a 6.7 M HCl solution should be diluted to 357.33 mL to prepare a 0.30 M HCl solution.
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