Ten Examples

Go to Molarity Problems #11-25

Go to Molarity Problems #26-35

As should be clear from its name, molarity involves moles. Boy, does it!

The molarity of a solution is calculated by taking the moles of solute and dividing by the liters of solution.

moles of solute Molarity = –––––––––––––– liters of solution

This is probably easiest to explain with examples.

**Example #1:** Suppose we had 1.00 mole of sucrose (its mass is about 342.3 grams) and proceeded to mix it into some water. It would dissolve and make sugar water. We keep adding water, dissolving and stirring until all the solid was gone. We then made sure that when everything was well-mixed, there was exactly 1.00 liter of solution.

What would be the molarity of this solution?

**Solution:**

1.00 mol Molarity = ––––––– 1.00 L

The answer is 1.00 mol/L. Notice that both the units of mol and L remain. Neither cancels.

A symbol for mol/L is often used. It is a capital M. So, writing 1.00 M for the answer is the correct way to do it.

Some textbooks make the M using italics and some put in a dash, like this: 1.00-*M.* When you handwrite it; a block capital M is just fine.

When you say it out loud, say this: "one point oh oh molar." You don't have to say the dash (if it's there). By the way, you sometimes see 1.00 M like this: 1.00-molar. A dash is usually used when you write the word 'molar.'

And never forget this: replace the M with mol/L when you do calculations. The M is the symbol for molarity, the mol/L is the unit used in calculations.

**Example #2:** Suppose you had 2.00 moles of solute dissolved into 1.00 L of solution. What's the molarity?

**Solution:**

2.00 mol Molarity = ––––––– 1.00 L

The answer is 2.00 M.

Notice that no mention of a specific substance is mentioned at all. The molarity would be the same. It doesn't matter if it is sucrose, sodium chloride or any other substance. One mole of sucrose or sodium chloride or anything else contains the same number of chemical units. And that number is 6.022 x 10^{23} units, called Avogadro's Number.

**Example #3:** What is the molarity when 0.750 mol is dissolved in 2.50 L of solution?

**Solution:**

0.750 mol Molarity = ––––––––– = 0.300 M 2.50 L

Now, let's change from using moles to grams. This is much more common. After all, chemists use balances to weigh things and balances give grams, NOT moles.

By the way, here's another point: solutions are always considered to be fully-mixed. The solute has been completely dispersed throughout the entire extent of the solution. All samples from a well-mixed solution will show the same concentration when analyzed.

**Example #4:** Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 L of solution. What would be the molarity of the solution?

**Solution:**

There two steps to the solution of this problem. Eventually, the two steps will be merged into one equation.

Step One: convert grams to moles.Step Two: divide moles by liters to get molality.

In the above problem, 58.44 grams/mol is the molar mass of NaCl. (There is the term "formula weight" and the term "molecular weight." There is a technical difference between them that isn't important right now. The term "molar mass" is a moe generic term.) To solve the problem:

Step One: dividing 58.44 grams by 58.44 grams/mol gives 1.00 mol.Step Two: dividing 1.00 mol by 2.00 L gives 0.500 mol/L (or 0.500 M).

Comment: remember that sometimes, a book will write out the word "molar," as in 0.500-molar.

**Example #5:** Calculate the molarity of 25.0 grams of KBr dissolved in 750.0 mL.

**Solution:**

1) Convert grams to moles:

25.0 g –––––––––– = 0.210 mol 119.9 g/mol

2) Calculate the molarity:

0.210 mol –––––––– = 0.280 M 0.750 L

**Example #6:** 80.0 grams of glucose (C_{6}H_{12}O_{6}, mol. wt = 180. g/mol) is dissolved in enough water to make 1.00 L of solution. What is its molarity?

**Solution:**

1) Convert grams to moles:

80.0 g –––––––––– = 0.444 mol 180.0 g/mol

2) Calculate the molarity:

0.444 mol –––––––– = 0.444 M 1.00 L

Notice how the phrase "of solution" keeps showing up. The molarity definition is based on the volume of the solution, NOT the volume of pure water used. For example, to say this:

"A one molar solution is prepared by adding one mole of solute to one liter of water."

is totally incorrect. It should be "one liter of solution" not "one liter of water."

This is correct:

"A one molar solution is prepared by adding one mole of solute to sufficient water to make one liter of solution."

The most typical molarity problem looks like this:

What is the molarity of _____ grams of [a chemical] dissolved in _____ mL (or L) of solution.

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

mass ––––––––– = moles molar mass

moles –––––– = molarity volume

The two steps just mentioned can be combined into one equation. First, I'll rearrange the second equation (with M for molarity and V for volume):

moles = MV

Since the moles of the first equation equals the moles of the second equation, I can write this:

mass MV = ––––––––– <--- this is a very useful equation. Remember it!!! molar mass

**Example #7:** When 2.50 grams of KMnO_{4} (molar mass = 158.0 g/mol) is dissolved into 100. mL of solution, what molarity results?

2.50 g (x) (0.100 L) = –––––––––– 158.0 g/mol x = 0.158 M (to three sig figs)

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

**Example #8:** How many grams of KMnO_{4} are needed to make 500. mL of a 0.200 M solution?

x (0.200 mol/L) (0.500 L) = –––––––––– 158.0 g/mol x = 15.8 g (to three sig figs)

Note the use of mol/L. In the actual calculation, use mol/L rather than M. The M is the symbol for molarity and is used in discussing molarity. However, the unit g/mol is what is actually used in a calculation.

**Example #9:** 10.0 g of acetic acid (CH_{3}COOH) is dissolved in 500. mL of solution. What molarity results?

10.0 g (x) (0.500 L) = –––––––––– 60.05 g/mol x = 0.333 M (to three sig figs)

Make sure to always use liters with this equation. Never mL or cm

^{3}or anything else. Only liters. Or, if you prefer, dm^{3}.

**Example #10:** How many mL of solution will result when 15.0 g of H_{2}SO_{4} is dissolved to make a 0.200 M solution?

15.0 g (0.200 mol/L) (x) = –––––––––– 98.07 g/mol x = 0.765 L (to three sig figs)

In the requested volume, 76.5 mL.

**Bonus Example:** Carbon tetrachloride (CCl_{4}) has a density of 1.59 kg/L. What is the concentration (in mol/L) of pure CCl_{4}?

**Solution:**

Assume 1.00 L of CCl_{4}is present.(1.59 kg/L) (1.00 L) = 1.59 kg

(1.59 kg) (1000 g/kg) = 1590 g

1590 g / 153.823 g/mol = 10.336556 mol

10.336556 mol / 1.00 L = 10.3 M (to three sig figs)

Go to Molarity Problems #11-25