A 25 ml aliquot of a 2.00 M stock solution is placed in a 300. mL volumetric flask. When it is filled to line, what is the concentration of the new solution?
A 25 ml aliquot of a 2.00 M stock solution is placed in a 300. mL volumetric flask. When it is filled to line, what is the concentration of the new solution?
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:**
A 25 mL aliquot of a 2.00 M stock solution is placed in a 300 mL volumetric flask. When it is filled to line, what is the concentration of the new solution?
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**Explanation:**
To determine the concentration of the new solution, we can use the dilution formula:
\[ C_1 V_1 = C_2 V_2 \]
Where:
- \( C_1 \) is the initial concentration of the stock solution (2.00 M).
- \( V_1 \) is the volume of the stock solution used (25 mL).
- \( C_2 \) is the concentration of the new solution (what we're solving for).
- \( V_2 \) is the final volume of the new solution (300 mL).
Rearranging the formula to solve for \( C_2 \):
\[ C_2 = \frac{C_1 V_1}{V_2} \]
Substitute the known values:
\[ C_2 = \frac{(2.00 \, \text{M}) (25 \, \text{mL})}{300 \, \text{mL}} \]
Calculating this:
\[ C_2 = \frac{50 \, \text{M} \cdot \text{mL}}{300 \, \text{mL}} \]
\[ C_2 = 0.167 \, \text{M} \]
Therefore, the concentration of the new solution is 0.167 M.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff65fb7b2-ba85-4e67-8fe1-01f97a564d1e%2F38a17eae-62ef-41ce-8436-ca7e70908136%2Fyq6wfdf_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
A 25 mL aliquot of a 2.00 M stock solution is placed in a 300 mL volumetric flask. When it is filled to line, what is the concentration of the new solution?
---
**Explanation:**
To determine the concentration of the new solution, we can use the dilution formula:
\[ C_1 V_1 = C_2 V_2 \]
Where:
- \( C_1 \) is the initial concentration of the stock solution (2.00 M).
- \( V_1 \) is the volume of the stock solution used (25 mL).
- \( C_2 \) is the concentration of the new solution (what we're solving for).
- \( V_2 \) is the final volume of the new solution (300 mL).
Rearranging the formula to solve for \( C_2 \):
\[ C_2 = \frac{C_1 V_1}{V_2} \]
Substitute the known values:
\[ C_2 = \frac{(2.00 \, \text{M}) (25 \, \text{mL})}{300 \, \text{mL}} \]
Calculating this:
\[ C_2 = \frac{50 \, \text{M} \cdot \text{mL}}{300 \, \text{mL}} \]
\[ C_2 = 0.167 \, \text{M} \]
Therefore, the concentration of the new solution is 0.167 M.
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