Practice Problems: Dilutions With Varying Concentrations And Volumes

Understanding how to perform dilutions is essential in many scientific and medical fields. This practice problem set will help you master the concepts of diluting solutions with varying concentrations and volumes.

What Is a Dilution?

A dilution involves reducing the concentration of a solute in a solution by adding more solvent. The key formula used is:

C1V1 = C2V2

Where:

  • C1 = initial concentration
  • V1 = initial volume
  • C2 = final concentration
  • V2 = final volume

Practice Problems

Problem 1: Diluting a Concentrated Solution

You have 10 mL of a 5 M solution. How much water must you add to dilute it to a 1 M solution?

Solution:

Using the formula: C1V1 = C2V2

5 M × 10 mL = 1 M × V2

V2 = (5 M × 10 mL) / 1 M = 50 mL

Since the initial volume is 10 mL, the amount of water to add is:

50 mL – 10 mL = 40 mL

Problem 2: Preparing a Diluted Solution

You need 200 mL of a 0.5 M solution. How much of a 2 M stock solution do you need?

Solution:

Using the formula: C1V1 = C2V2

2 M × V1 = 0.5 M × 200 mL

V1 = (0.5 M × 200 mL) / 2 M = 50 mL

Therefore, you need 50 mL of the 2 M stock solution, and the remaining volume is water:

200 mL – 50 mL = 150 mL

Tips for Successful Dilutions

Always use accurate measuring tools and double-check your calculations. Remember, the total volume after dilution is the sum of the initial volume of concentrated solution and the added solvent.

Additional Practice

  • Calculate the volume of a 10 M solution needed to prepare 100 mL of a 1 M solution.
  • Determine how much water to add to dilute 25 mL of a 4 M solution to 0.5 M.
  • If you dilute 5 mL of a 3 M solution to 100 mL, what is the new concentration?

Practice these problems regularly to become confident in performing dilutions accurately.