Dilution Practice Questions with Answers

Dilution is the process of decreasing the concentration of a solute in a solution, usually by adding more solvent like water. Whether you are working in a high school chemistry lab or a professional pharmaceutical setting, mastering Dilution Practice Questions with Answers is essential for accurately preparing chemical solutions. This guide will help you understand the mathematics behind changing concentrations and provide ample practice to sharpen your skills.

Concept Explanation

Dilution is the laboratory technique of adding additional solvent to a solution to reduce its molarity while keeping the total amount of solute constant. The fundamental principle of dilution is that the number of moles of solute before dilution is equal to the number of moles of solute after dilution. This relationship is mathematically expressed by the dilution equation: M1V1 = M2V2. In this formula, M1 and V1 represent the molarity and volume of the initial concentrated solution (stock solution), while M2 and V2 represent the molarity and volume of the final diluted solution.

According to Wikipedia's entry on dilution, the process does not change the amount of solute present, only how spread out that solute is within the liquid. For more background on how concentration is measured, you might find it helpful to review What Is Molarity? The Complete Guide. When performing these calculations, it is vital to ensure that your units for volume are consistent on both sides of the equation—if V1 is in milliliters, V2 must also be in milliliters.

Common applications of dilution include:

• Preparing working solutions from concentrated stock solutions.

• Creating a series of standard solutions for a calibration curve (serial dilution).

• Adjusting the strength of medications in clinical settings.

• Reducing the intensity of acids or bases for safe handling.

Solved Examples

Reviewing these step-by-step examples will help you understand how to apply the dilution formula in different scenarios.

Example 1: Finding Final Molarity
If you dilute 50.0 mL of a 2.0 M HCl solution to a final volume of 200.0 mL, what is the new concentration?

1. Identify the knowns: M1 = 2.0 M, V1 = 50.0 mL, V2 = 200.0 mL.

2. Set up the equation: (2.0 M)(50.0 mL) = (M2)(200.0 mL).

3. Solve for M2: M2 = (2.0 × 50.0) / 200.0.

4. Result: M2 = 100 / 200 = 0.50 M.

Example 2: Finding Initial Volume Needed
How many milliliters of 12.0 M H2SO4 are needed to prepare 500.0 mL of a 0.100 M solution?

1. Identify the knowns: M1 = 12.0 M, M2 = 0.100 M, V2 = 500.0 mL.

2. Set up the equation: (12.0 M)(V1) = (0.100 M)(500.0 mL).

3. Solve for V1: V1 = (0.100 × 500.0) / 12.0.

4. Result: V1 = 50.0 / 12.0 = 4.17 mL.

Example 3: Calculating Volume of Solvent Added
How much water must be added to 100 mL of 1.5 M NaOH to make it 0.50 M?

1. Identify the knowns: M1 = 1.5 M, V1 = 100 mL, M2 = 0.50 M.

2. Calculate V2: (1.5)(100) = (0.50)(V2) → V2 = 150 / 0.50 = 300 mL.

3. Subtract the initial volume to find added water: Vadded = V2 - V1 = 300 mL - 100 mL.

4. Result: 200 mL of water must be added.

Practice Questions

Test your knowledge with these dilution problems. If you find these challenging, you may want to practice how to solve molarity problems first.

1. A student takes 25.0 mL of a 4.0 M KNO3 solution and dilutes it to 500.0 mL. What is the concentration of the diluted solution?

2. To what volume should 10.0 mL of 15.0 M HNO3 be diluted to prepare a 0.30 M solution?

3. If you have 50.0 mL of a 10.0% (m/v) glucose solution and dilute it to 250.0 mL, what is the new percentage concentration?

4. How many milliliters of water must be added to 30.0 mL of a 5.0 M NaCl solution to produce a 1.5 M solution?

5. A laboratory technician needs 2.0 L of 0.50 M MgCl2. The available stock solution is 4.0 M. How much stock solution is required?

6. You dilute 5.0 mL of a 1.0 M solution to 50.0 mL. Then, you take 10.0 mL of that new solution and dilute it again to 100.0 mL. What is the final concentration?

7. A 0.75 M solution of CuSO4 is evaporated until the volume is reduced from 400 mL to 150 mL. What is the new molarity? (Hint: This is an "inverse" dilution or concentration problem).

8. What volume of 18.0 M H2SO4 is needed to make 250 mL of a 2.5 M solution?

9. If 45 mL of water is added to 15 mL of 6.0 M HCl, what is the resulting molarity?

10. A stock solution of 10.0 M NaOH is used to make 1.5 L of a 0.20 M solution. How many milliliters of the stock solution were used?

Answers & Explanations

1. 0.20 M: Use M1V1 = M2V2. (4.0 M)(25.0 mL) = (M2)(500.0 mL). M2 = 100 / 500 = 0.20 M.

2. 500 mL: (15.0 M)(10.0 mL) = (0.30 M)(V2). V2 = 150 / 0.30 = 500 mL.

3. 2.0%: The dilution formula works for percentages too. (10%)(50.0 mL) = (C2)(250.0 mL). C2 = 500 / 250 = 2.0%.

4. 70.0 mL: First find V2. (5.0 M)(30.0 mL) = (1.5 M)(V2). V2 = 150 / 1.5 = 100 mL. Water added = 100 mL - 30 mL = 70 mL.

5. 0.25 L (or 250 mL): (4.0 M)(V1) = (0.50 M)(2.0 L). V1 = 1.0 / 4.0 = 0.25 L.

6. 0.01 M: Step 1: (1.0)(5)/(50) = 0.1 M. Step 2: (0.1)(10)/(100) = 0.01 M.

7. 2.0 M: Use M1V1 = M2V2. (0.75 M)(400 mL) = (M2)(150 mL). M2 = 300 / 150 = 2.0 M.

8. 34.7 mL: (18.0 M)(V1) = (2.5 M)(250 mL). V1 = 625 / 18.0 = 34.72 mL.

9. 1.5 M: Total volume V2 = 15 mL + 45 mL = 60 mL. (6.0 M)(15 mL) = (M2)(60 mL). M2 = 90 / 60 = 1.5 M.

10. 30 mL: (10.0 M)(V1) = (0.20 M)(1500 mL). V1 = 300 / 10 = 30 mL.

Quick Quiz

Frequently Asked Questions

What is the dilution formula?

The dilution formula is M1V1 = M2V2, where M and V represent molarity and volume respectively. It is used to calculate how changing the volume of a solution affects its concentration.

Can I use milliliters instead of liters in the dilution equation?

Yes, you can use milliliters as long as you use the same unit for both V1 and V2. The units cancel out in the ratio, making the calculation valid for any volume unit.

Does adding more solute count as a dilution?

No, adding more solute increases the concentration, which is the opposite of dilution. Dilution specifically refers to adding more solvent to decrease the concentration.

How is serial dilution different from simple dilution?

Serial dilution involves a stepwise series of dilutions where the diluted solution from one step serves as the stock for the next. This is commonly used in microbiology to reduce cell concentrations.

What happens to the number of moles during dilution?

The total number of moles of solute remains exactly the same during a dilution. Only the volume of the liquid increases, which causes the molarity (moles per liter) to decrease.

Is the dilution equation useful for stoichiometry?

While the dilution equation helps prepare solutions, you often need to use the Stoichiometry Practice Questions methods to determine how those solutions react in chemical equations. Dilution is a precursor step to many stoichiometric experiments.

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