Molarity

Molarity Virtual Lesson

Mrs. Nebzydoski

When calculating Molarity remember:

n of solute / L solution

If you did not get the correct answer on that problem, take a look at the solution here.

Do you see what you might have done wrong?

We are not always dealing with solids when we make solutions in the chemistry lab.

Sometime we have liquid reagents.

Liquid Reagents

When we are mixing chemicals that involve only liquids, we must use this equation to solve for molarity or volume.

Example: I have concentrated HCl in the storage room

• Concentrated HCl is 12 M

• I want to use 0.5 M HCl in a lab demo

• I must use the equation M1V1=M2V2 to mix the correct amount of HCl for lab

12M HCl is what I have, but I only need 250 mL of 0.5M HCl

I must first look at what I know....

I want 250 mL of 0.5 M HCl. (V1=250 mL and M1=0.5 M)

I have 12 M HCl (V2=12 M)

I need to solve for the V2

M1V1=M2V2

(.5M)(250mL) = (12M)(V2)

When I solve this equation, I get 10.4 mL for V2

This means that I need 10.4 mL of 12 M HCl.

I will take 10.4 mL of 12M HCl and dilute it to 250 mL.

This will make 250 mL of a 0.5M HCl solution.

M1V1=M2V2

Using this Equation, solve some of the following problems

Open Ended

If I have 18 M H₂SO₄ and I want to make 500 mL of a 1 M H₂SO₄ solution, what do I do?

M1V1=M2V2

M1=1 M, V1=500 mL

M2=18M, V2=?

In the previous question, did you get that you needed to take 27.8 mL of 18M sulfuric acid and dilute it to a volume of 500 mL? If you did, great job!

I would now like you to go to page 146 of your textbook and read about dilution.

• Try the sample problem on page 148 and the practice exercise on page 149

• Complete #4.73 and #4.74 on page 159 in your notebook and we will be going over it in class.