You need 425 mL of a 0.300 M HCl solution. You have a bottle of 6.00 M HCl on hand. You measure out mL of M HCl and dilute it to a volume of mL.
You need 425 mL of a 0.300 M HCl solution. You have a bottle of 6.00 M HCl on hand. You measure out mL of M HCl and dilute it to a volume of mL.
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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![**Dilution Problem for Educational Use**
You need 425 mL of a 0.300 M HCl solution. You have a bottle of 6.00 M HCl on hand.
**Calculation Task:**
1. **Determine Volume to Measure:** You measure out ____ mL of 6.00 M HCl.
2. **Determine Dilution Volume:** Dilute it to a total volume of ____ mL.
**Explanation:**
This exercise involves preparing a diluted solution from a concentrated hydrochloric acid (HCl) stock solution. To solve this, apply the dilution formula:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) is the concentration of the stock solution (6.00 M).
- \( V_1 \) is the volume of the stock solution to be measured.
- \( C_2 \) is the desired concentration of the diluted solution (0.300 M).
- \( V_2 \) is the final volume of the diluted solution (425 mL).
By rearranging and solving the equation for \( V_1 \), you find the amount of the stock solution required to achieve the desired dilution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6bfb8691-d1db-4ddc-9d6d-d840f1ac88cb%2F82a1a969-1670-43c2-8301-a77441e2abc7%2F47b0box_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Dilution Problem for Educational Use**
You need 425 mL of a 0.300 M HCl solution. You have a bottle of 6.00 M HCl on hand.
**Calculation Task:**
1. **Determine Volume to Measure:** You measure out ____ mL of 6.00 M HCl.
2. **Determine Dilution Volume:** Dilute it to a total volume of ____ mL.
**Explanation:**
This exercise involves preparing a diluted solution from a concentrated hydrochloric acid (HCl) stock solution. To solve this, apply the dilution formula:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) is the concentration of the stock solution (6.00 M).
- \( V_1 \) is the volume of the stock solution to be measured.
- \( C_2 \) is the desired concentration of the diluted solution (0.300 M).
- \( V_2 \) is the final volume of the diluted solution (425 mL).
By rearranging and solving the equation for \( V_1 \), you find the amount of the stock solution required to achieve the desired dilution.
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