Molarity: Part Two


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The most typical molarity problem looks like this:

What is the molarity of "whatever" grams of "whatever" substance dissolved in "whatever" mL of solution.

To solve it, you convert grams to moles, then divide by the volume, like this:

The two steps just mentioned can be combined into one equation. Notice that moles is part of both equations, so one equation can be substituted into the other. Let's substitute the first into the second and rearrange just a bit to get this:

The M stands for molarity, the V for volume. In the second question, GMW has been substituted for molar mass. GMW stands for gram-molecular weight, which is a very common synonym for molar mass.


So let's try an example:

1) When 2.00 grams of KMnO4 (molec. wt = 158.0 g/mol) is dissolved into 100.0 mL of solution, what molarity results?

This next example is the most common type you'll see:

2) How many grams of KMnO4 are needed to make 500.0 mL of a 0.200 M solution?

Practice Problems

Now, it's time for you to try a couple without the answers visible!

1) 10.0 g of acetic acid (CH3COOH) is dissolved in 500.0 mL of solution. What molarity results?

2) How many mL of solution will result when 15.0 g of H2SO4 is dissolved to make a 0.200 M solution?

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