If 21 mL of 1.0 M NH4Cl is used what volume does it need to be diluted to in order to make a 0.12 M solution? (give your answer in mL to zero decimal places)
If 21 mL of 1.0 M NH4Cl is used what volume does it need to be diluted to in order to make a 0.12 M solution? (give your answer in mL to zero decimal places)
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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![### How to Calculate Dilution of a Solution
**Question:**
If 21 mL of 1.0 M NH₄Cl is used, what volume does it need to be diluted to in order to make a 0.12 M solution? (Give your answer in mL to zero decimal places)
**Solution:**
To solve this problem, we will use the concept of dilution, which can be calculated using the formula:
\[ C_1 \times V_1 = C_2 \times V_2 \]
Where:
- \( C_1 \) is the concentration of the initial solution (1.0 M)
- \( V_1 \) is the volume of the initial solution (21 mL)
- \( C_2 \) is the desired concentration of the final solution (0.12 M)
- \( V_2 \) is the volume of the final solution we need to find
Rearranging the equation to solve for \( V_2 \):
\[ V_2 = \frac{C_1 \times V_1}{C_2} \]
Now plug in the values:
\[ V_2 = \frac{1.0 \text{ M} \times 21 \text{ mL}}{0.12 \text{ M}} \]
\[ V_2 = \frac{21}{0.12} \]
\[ V_2 = 175 \text{ mL} \]
Therefore, you need to dilute the 21 mL of 1.0 M NH₄Cl to a final volume of **175 mL** to obtain a 0.12 M solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0d0fca7b-87e2-42be-9458-260269753889%2F2514a7f9-90e5-4927-a698-9c69ff0cb6a7%2Fk05ebd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### How to Calculate Dilution of a Solution
**Question:**
If 21 mL of 1.0 M NH₄Cl is used, what volume does it need to be diluted to in order to make a 0.12 M solution? (Give your answer in mL to zero decimal places)
**Solution:**
To solve this problem, we will use the concept of dilution, which can be calculated using the formula:
\[ C_1 \times V_1 = C_2 \times V_2 \]
Where:
- \( C_1 \) is the concentration of the initial solution (1.0 M)
- \( V_1 \) is the volume of the initial solution (21 mL)
- \( C_2 \) is the desired concentration of the final solution (0.12 M)
- \( V_2 \) is the volume of the final solution we need to find
Rearranging the equation to solve for \( V_2 \):
\[ V_2 = \frac{C_1 \times V_1}{C_2} \]
Now plug in the values:
\[ V_2 = \frac{1.0 \text{ M} \times 21 \text{ mL}}{0.12 \text{ M}} \]
\[ V_2 = \frac{21}{0.12} \]
\[ V_2 = 175 \text{ mL} \]
Therefore, you need to dilute the 21 mL of 1.0 M NH₄Cl to a final volume of **175 mL** to obtain a 0.12 M solution.
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