The extent to which a __solute__ dissolves in a solution is expressed by the solution's __concentration__.

**The Molarity Formula**

__Concentration__ is most often expressed as __molarity (M)__.

**Molarity (M) = number of moles of solute (mol) / liters of solution (L)**

-or-

The best way to understand how to use the molarity formula is to practice, practice, practice. So let's dive in to some molarity examples now.

Ready? Here we go...

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**How to Find Molarity**

*ex:* Calculate the molarity of a solution made by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.

_________

**answer:****molarity = moles NaOH / L solution**

➞ we already have the "L solution" ➞ 1.50 L

➞ but we need to calculate the "moles NaOH" before we can find the molarity...

Okay, so now we have __both__ our numerator, **0.288 mol NaOH**, and we always had our denominator (given), **1.50 L**. Let's divide the two and __solve for molarity__:

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*ex:* Give the concentration of each type of ion in a solution of 0.50 M Co(NO

_{3})

_{2}

_________

**answer:**Think mole ratios.

For each 1 mole of cobalt (II) nitrate dissolved, the solution contains 1 mole Co^{2+} and 2 moles NO_{3}^{-}

So a solution that contains 0.50 M Co(NO_{3})_{2} contains:

**0.50 M Co ^{2+}**

*and*

**1.0 M NO**

_{3}^{-}^{ }

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*ex:* Calculate the number of moles of chloride ions in 1.75 L of 1.0 M ZnCl

_{2}

_________

**answer:**NOTE - anytime you see [ X **"L"** of Y **"M" **], you often have to multiply because L** ^{.}**M = L

**(mol/L) =**

^{.}__mol__

(1.75 L / 1) (1.0 mol ZnCl_{2} / 1L) = 1.75 mol ZnCl_{2}

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*ex:* Blood serum is about 0.14 M NaCl. What volume of blood contains 1.0 mg NaCl?

_________

**answer:**If we convert 1.0mg NaCl to "moles NaCl," we can use the "0.14 moles / L" to get ourselves into units of __liters__ (volume):

➞ (1.0mg NaCl / 1) (1g NaCl / 1000mg NaCl) (1mol NaCl / 58.45g NaCl) = 1.7 x 10^{-5} moles NaCl

➞ (1.7 x 10^{-5} mol NaCl / 1) (1L solution / 0.14mol NaCl) = **1.2 x 10 ^{-4} L blood** (solution)

Here's a better look at it...

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*ex:* A chemist needs 1.00 L of an aqueous 0.200 M K

_{2}Cr

_{2}O

_{7}solution. How much solid K

_{2}Cr

_{2}O

_{7}does he need to weigh out?

_________

**answer:**We have [ X **"L"** of Y **"M" **] ➞ L** ^{.}**M = L

**(mol/L) =**

^{.}__mol__

➞ (1.00 L / 1) (0.200 mol K_{2}Cr_{2}O_{7} / 1L) = 0.200 mol K_{2}Cr_{2}O_{7}

➞ (0.200 mol K_{2}Cr_{2}O_{7} / 1) (294 g K_{2}Cr_{2}O_{7} / 1 mol K_{2}Cr_{2}O_{7}) = **58.8g K _{2}Cr_{2}O_{7}**

Here's an easy-to-follow version of the above two molarity calculations:

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**Dilutions**

People use the terms __dilute__ and __concentrated__ everyday to describe the contents of a liquid * qualitatively*.

__dilute__ = "watered-down"__concentrated__ = contains "a lot" of ingredients that can be tasted

**The Dilution Equation**

From a * quantitative* point of view, chemists can prepare "

__dilutions__" of solutions having an actual concentration.

Usually, highly concentrated "__stock solutions__" are purchased by a lab, and the chemists add water to achieve the desired * molarity* (concentration).

The Equation: **M _{1}V_{1} = M_{2}V_{2}**

-- M = molarity (mol / L)

-- V = volume (matching units, often mL or L)

As stated in the above image, the __moles of solute__ are * not* changed from situation 1 (M

_{1}V

_{1}) to situation 2 (M

_{2}V

_{2}).

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*ex:* What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H

_{2}SO

_{4}solution?

_________

**answer:**Well, we have:

M_{1} = 16 M M_{2} = 0.10 M

V_{1} = ?? V_{2} = 1.5 L

➞ M_{1}V_{1} = M_{2}V_{2}

➞ (16 M) (x L) = (0.10 M) (1.5 L)

➞ (16x) mol = 0.15 mol

➞ x = 0.0094 L or **9.4 mL**

Here's a better look at all that...

To prepare 1.5 L of a 0.10 M H_{2}SO_{4} solution, the chemist would __add__ 9.4 mL of 16 M H_{2}SO_{4} and then __add__ enough water until the total final volume is 1.5 L.

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Next up in **SECTION 4 - Types of Chemical Reactions and Solution Stoichiometry**, we'll talk about the different Types of Chemical Reactions.