Easy Dilution Practice Questions

Concept Explanation

Dilution is the process of reducing the concentration of a solute in a solution by adding more solvent, typically water, without adding more solute. This fundamental laboratory technique relies on the principle that the total number of moles of solute remains constant even as the volume of the liquid increases. In chemistry, we represent this relationship using the dilution equation: M1V1 = M2V2. In this formula, M1 and V1 represent the molarity and volume of the initial concentrated solution (often called the stock solution), while M2 and V2 represent the molarity and volume of the final diluted solution. Understanding this concept is a vital step before moving on to molarity practice questions or more complex laboratory procedures. According to Wikipedia's guide on dilution, the conservation of mass ensures that the product of concentration and volume stays stable throughout the process. This mathematical consistency allows scientists to precisely prepare reagents for experiments, ranging from simple classroom demonstrations to advanced pharmaceutical manufacturing.

Solved Examples

Review these step-by-step solutions to understand how to apply the dilution formula in various scenarios. If you are still mastering the basics of concentration, you might find our complete guide to molarity helpful for context.

1. Example 1: Finding Final Concentration
   If you have 50 mL of a 2.0 M NaCl solution and dilute it to a final volume of 200 mL, what is the new molarity?
   

   1. Identify the knowns: M1 = 2.0 M, V1 = 50 mL, V2 = 200 mL.

   2. Set up the equation: (2.0 M)(50 mL) = (M2)(200 mL).

   3. Solve for M2: M2 = (2.0 * 50) / 200.

   4. Calculate the result: M2 = 100 / 200 = 0.5 M.

2. Example 2: Finding Initial Volume Needed
   How many milliliters of a 12.0 M HCl stock solution are needed to prepare 500 mL of a 0.10 M HCl solution?
   

   1. Identify the knowns: M1 = 12.0 M, M2 = 0.10 M, V2 = 500 mL.

   2. Set up the equation: (12.0 M)(V1) = (0.10 M)(500 mL).

   3. Solve for V1: V1 = (0.10 * 500) / 12.0.

   4. Calculate the result: V1 = 50 / 12.0 = 4.17 mL.

3. Example 3: Finding Final Volume
   A chemist has 10 mL of a 5.0 M NaOH solution. To what volume must she dilute it to reach a concentration of 0.25 M?
   

   1. Identify the knowns: M1 = 5.0 M, V1 = 10 mL, M2 = 0.25 M.

   2. Set up the equation: (5.0 M)(10 mL) = (0.25 M)(V2).

   3. Solve for V2: V2 = (5.0 * 10) / 0.25.

   4. Calculate the result: V2 = 50 / 0.25 = 200 mL.

Practice Questions

Test your knowledge with these easy dilution practice questions. Ensure your units for volume are consistent on both sides of the equation.

1. If 25 mL of a 4.0 M solution is diluted to 100 mL, what is the final molarity?

2. A student needs to make 250 mL of a 0.5 M glucose solution from a 2.0 M stock solution. How much stock solution should they use?

3. You dilute 150 mL of a 1.2 M solution by adding enough water to reach a total volume of 600 mL. What is the new concentration?

4. How much water must be added to 50 mL of a 3.0 M KOH solution to prepare a 1.0 M solution? (Hint: Find V2 first, then subtract V1).

5. If a 10.0 M stock solution is used to make 2.0 L of 0.25 M solution, what volume of the stock solution was required?

6. A 0.75 M solution has a volume of 40 mL. If it is diluted to a final concentration of 0.15 M, what is the final volume?

7. To what volume should you dilute 100 mL of a 5.0 M CuS04 solution to obtain a 0.50 M solution?

8. If you take 5 mL of a concentrated 18.0 M H2SO4 and dilute it to 250 mL, what is the molarity of the resulting acid?

9. A technician dilutes 30 mL of an 8.0 M reagent to a new volume of 120 mL. What is the final concentration?

10. How many milliliters of a 6.0 M stock solution are needed to produce 300 mL of a 2.0 M solution?

Answers & Explanations

1. 1.0 M: Using M1V1 = M2V2, (4.0 M)(25 mL) = (M2)(100 mL). M2 = 100 / 100 = 1.0 M.

2. 62.5 mL: (2.0 M)(V1) = (0.5 M)(250 mL). V1 = 125 / 2.0 = 62.5 mL.

3. 0.3 M: (1.2 M)(150 mL) = (M2)(600 mL). M2 = 180 / 600 = 0.3 M.

4. 100 mL: First find V2: (3.0 M)(50 mL) = (1.0 M)(V2) -> V2 = 150 mL. Amount added = V2 - V1 = 150 mL - 50 mL = 100 mL.

5. 0.05 L (or 50 mL): (10.0 M)(V1) = (0.25 M)(2.0 L). V1 = 0.5 / 10.0 = 0.05 L.

6. 200 mL: (0.75 M)(40 mL) = (0.15 M)(V2). V2 = 30 / 0.15 = 200 mL.

7. 1000 mL (or 1.0 L): (5.0 M)(100 mL) = (0.50 M)(V2). V2 = 500 / 0.50 = 1000 mL.

8. 0.36 M: (18.0 M)(5 mL) = (M2)(250 mL). M2 = 90 / 250 = 0.36 M.

9. 2.0 M: (8.0 M)(30 mL) = (M2)(120 mL). M2 = 240 / 120 = 2.0 M.

10. 100 mL: (6.0 M)(V1) = (2.0 M)(300 mL). V1 = 600 / 6.0 = 100 mL.

Quick Quiz

Frequently Asked Questions

What is the difference between dilution and concentration?

Dilution is the process of adding solvent to decrease the concentration of a solute, while concentration involves removing solvent or adding more solute to increase the strength of the solution. Both processes change the ratio of solute to solvent but affect the molarity in opposite ways.

Does the dilution equation work with any volume units?

Yes, the dilution equation works with any volume units such as milliliters, liters, or even gallons, provided that the same unit is used for both V1 and V2. If you use liters for V1, you must use liters for V2 to ensure the mathematical relationship remains valid.

Why is it important to add acid to water during dilution?

Adding acid to water is a critical safety rule because the dissolution of many concentrated acids is highly exothermic, meaning it releases significant heat. Adding acid slowly to a large volume of water allows the water to absorb the heat, preventing the mixture from boiling and splashing acid on the user.

Can I use mass instead of molarity in the dilution formula?

While the standard formula uses molarity, a similar relationship (C1V1 = C2V2) can be used for other concentration units like mass percent or parts per million (ppm). However, for stoichiometry and general chemistry problems, molarity is the preferred unit because it relates directly to the number of particles in the solution.

What is a stock solution?

A stock solution is a highly concentrated reagent that is stored in the laboratory and used to prepare working solutions of lower concentrations through dilution. Using stock solutions saves storage space and ensures consistency across different experimental trials.

How do I calculate the volume of water added?

To find the volume of water added, you first calculate the final volume (V2) using the dilution equation and then subtract the initial volume (V1) of the stock solution. The formula for water added is: Volume of Water = V2 - V1.

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