Molarity By Dilution Worksheet Answers Explained
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Molarity, the measure of concentration of a solute in a solution, is a fundamental concept in chemistry that allows scientists to understand how substances interact in various chemical reactions. One way to adjust the concentration of a solution is through dilution, a process that involves adding more solvent to a solution to decrease the concentration of the solute. This article aims to explain the answers to a typical worksheet on molarity and dilution, helping to clarify the underlying principles through examples and explanations.
What is Molarity? 📚
Molarity (M) is defined as the number of moles of solute per liter of solution. It can be expressed with the following formula:
Molarity (M) = Moles of Solute / Liters of Solution
Understanding this formula is crucial for performing calculations regarding concentrations, especially when dealing with dilutions.
Understanding Dilution 🌊
Dilution refers to the process of reducing the concentration of a solution by adding more solvent. The relationship between the molarity of the concentrated solution (M1), the molarity of the diluted solution (M2), the volume of the concentrated solution (V1), and the volume of the diluted solution (V2) is described by the dilution equation:
M1 × V1 = M2 × V2
Where:
- M1 = Initial molarity
- V1 = Initial volume
- M2 = Final molarity
- V2 = Final volume
This equation helps to determine how much concentrated solution is needed to achieve a specific molarity after dilution.
Example Problem: Molarity by Dilution
Let's work through an example to illustrate the concept of molarity by dilution.
Problem: You have a 6 M solution of hydrochloric acid (HCl) and want to prepare 1 liter of a 2 M solution. How much of the 6 M solution do you need to use?
Step-by-Step Solution
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Identify the known values:
- M1 = 6 M (initial concentration)
- M2 = 2 M (final concentration)
- V2 = 1 L (final volume)
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Use the dilution equation: We need to find V1, the volume of the concentrated solution needed.
[ M1 × V1 = M2 × V2 ]
Substituting the known values:
[ 6 M × V1 = 2 M × 1 L ]
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Solve for V1:
[ V1 = \frac{2 M × 1 L}{6 M} = \frac{2}{6} L = \frac{1}{3} L = 0.333 L ]
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Convert to milliliters: To convert liters to milliliters:
[ 0.333 L × 1000 mL/L = 333.33 mL ]
Therefore, you would need to use 333.33 mL of the 6 M solution and dilute it with water to reach a total volume of 1 L. 🎉
Important Notes on Dilution Calculations 💡
- Always add the concentrated solution to water and not the other way around to ensure safety and prevent exothermic reactions.
- When working with concentrated acids or bases, wear appropriate safety equipment, including gloves and goggles.
- Ensure to mix thoroughly after dilution to achieve a uniform solution.
Common Mistakes When Calculating Molarity and Dilution 🚫
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Forgetting to Convert Units: It's easy to overlook unit conversions, especially between liters and milliliters. Always double-check your units.
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Incorrect Use of the Dilution Formula: Ensure that you are using the correct values for M1, V1, M2, and V2. Mislabeling can lead to incorrect results.
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Neglecting to Dilute Adequately: Not achieving the intended final volume can lead to incorrect concentrations, which can affect experimental results.
Practice Problems
To fully grasp the concept of molarity by dilution, consider solving the following practice problems:
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Problem 1: If you have 5 M sodium chloride (NaCl) solution, how many milliliters are needed to prepare 500 mL of a 1 M NaCl solution?
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Problem 2: A laboratory needs a 0.1 M solution of potassium nitrate (KNO3). If the stock solution is 2 M, what volume of the stock solution is necessary to make 250 mL of the diluted solution?
Here’s a quick table summarizing the answers:
<table> <tr> <th>Problem</th> <th>Initial Concentration (M)</th> <th>Final Concentration (M)</th> <th>Final Volume (L)</th> <th>Volume of Concentrated Solution Needed (mL)</th> </tr> <tr> <td>1</td> <td>5 M</td> <td>1 M</td> <td>0.5 L</td> <td>100 mL</td> </tr> <tr> <td>2</td> <td>2 M</td> <td>0.1 M</td> <td>0.25 L</td> <td>12.5 mL</td> </tr> </table>
Conclusion
Understanding molarity and the process of dilution is crucial for success in chemistry. By mastering the formulas and practicing with different problems, students can become proficient in these essential concepts. Whether you're preparing for an exam or conducting experiments in the lab, the ability to calculate and manipulate concentrations is a valuable skill. Keep practicing and remember to follow safety protocols when working with chemical solutions! 🚀