Molarity Dilution Worksheet Problems and Solutions

Molarity Dilution Worksheet: Understanding the Concept

Molarity dilution is a crucial concept in chemistry, particularly in preparing solutions for various experiments. It involves calculating the concentration of a solution after dilution, which is essential in ensuring the accuracy of experimental results. In this worksheet, we will explore molarity dilution problems and solutions to help you grasp this concept better.

What is Molarity?

Molarity (M) is defined as the number of moles of a substance dissolved in one liter of solution. It is a measure of the concentration of a solution. The formula for calculating molarity is:

M = moles of solute / liters of solution

What is Dilution?

Dilution is the process of adding a solvent to a solution to decrease its concentration. When a solution is diluted, the amount of solute remains the same, but the volume of the solution increases. This results in a decrease in the concentration of the solution.

Molarity Dilution Formula

The molarity dilution formula is used to calculate the new concentration of a solution after dilution. The formula is:

M1 × V1 = M2 × V2

Where: M1 = initial molarity V1 = initial volume (in liters) M2 = final molarity V2 = final volume (in liters)

Problems and Solutions

Let’s solve some molarity dilution problems to help you understand this concept better.

Problem 1: A 2.5 M solution of sodium chloride (NaCl) has a volume of 500 mL. If 200 mL of water is added to the solution, what is the new molarity of the solution?

Solution:

Initial molarity (M1) = 2.5 M Initial volume (V1) = 500 mL = 0.5 L Final volume (V2) = 500 mL + 200 mL = 700 mL = 0.7 L

Using the molarity dilution formula: M1 × V1 = M2 × V2 2.5 M × 0.5 L = M2 × 0.7 L

Solving for M2: M2 = (2.5 M × 0.5 L) / 0.7 L M2 = 1.79 M

Problem 2: A 0.5 M solution of glucose has a volume of 2 L. If the solution is diluted to a volume of 4 L, what is the new molarity of the solution?

Solution:

Initial molarity (M1) = 0.5 M Initial volume (V1) = 2 L Final volume (V2) = 4 L

Using the molarity dilution formula: M1 × V1 = M2 × V2 0.5 M × 2 L = M2 × 4 L

Solving for M2: M2 = (0.5 M × 2 L) / 4 L M2 = 0.25 M

Problem 3: A 1.2 M solution of sulfuric acid (H2SO4) has a volume of 1.5 L. If 500 mL of water is added to the solution, what is the new molarity of the solution?

Solution:

Initial molarity (M1) = 1.2 M Initial volume (V1) = 1.5 L Final volume (V2) = 1.5 L + 0.5 L = 2 L

Using the molarity dilution formula: M1 × V1 = M2 × V2 1.2 M × 1.5 L = M2 × 2 L

Solving for M2: M2 = (1.2 M × 1.5 L) / 2 L M2 = 0.9 M

💡 Note: When solving molarity dilution problems, it's essential to use the correct units for volume (liters) and molarity (moles per liter).

Conclusion

In this worksheet, we explored molarity dilution problems and solutions to help you understand this concept better. Remember to use the molarity dilution formula (M1 × V1 = M2 × V2) to calculate the new concentration of a solution after dilution. Practice solving problems to become more comfortable with this concept.

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