A solution is a homogeneous mixture. This means that the composition of the mixture is uniform (= the same) throughout a sample of any size. The major component of the mixture is called the **solvent,** while the dissolved stuff is called the **solute**. Typically we are interested in the proportion, or concentration, of solute in the mixture There are several ways of expressing the concentration.

Molarity, abbreviated as __ M

Dilution Calculations

Osmolarity, abbreviated as __ osmol

In Latin "per cent" means "for every 100 hundred." When we express the concentration of a solute in % (w/v), we are reporting the mass of solute for every 100 mL of solution.

For example, a 4% (w/v) solution of NaCl in water is made by dissolving 4 g NaCl in enough water to give a total final volume of 100 mL.

Please note: the total final volume of solution is measured after completely mixing the water and salt. If we mix 4 g NaCl and 100 mL water, we can not be sure that the final volume will be 100 mL. When the components of a mixture are combined, sometimes the volume of the solution is less than that of the solvent alone; in other cases, the volume of the solution is greter than the volume of the solvent by itself; and in some cases the solution and solvent volume will be the same after mixing. Since it is impossible to predict the volume after mixing, enough solvent is added to dissolve the solute and bring it to the final desired total volume.

A solution need not be prepared to have exactly 100 mL final volume in order to calculate its concentration in % (w/v).

Example 1:

What is the concentration of a solution in % (w/v) made by dissolving 2.3 g NaCl in enough water to have 125 mL of solution?

If the concentration and volume of a solution are known, we can calculate the mass of the contained solute.

Example 2:

How many grams ethyl alcohol are contained in 1 fluid ounce (= 30 mL) bourbon if the concentration of ethyl alcohol is 35.6% (w/v)?

If we know the concentration of the solution, we can also calculate the volume of solution needed to provide a specified mass of solute.

Example 3:

What volume of a 25% (w/v) solution of HNO_{3} is needed to contain 12.6 g HNO_{3}?

Reactions are often performed in solutions. Since stoichiometric calculations to determine limiting reagents and theoretical yields must be performed in moles, it is very convenient to have a concentration unit that tells how many moles solute are present per unit of solution.

The **molarity** of a substance in solution is defined as the number of moles of solute per liter of solution.

Need to review how to calculate moles? click here.

Once again, please remember that the volume refers to the total mixture once all the solute has been dissolved. We can not predict the final volume of solution obtained by mixing one mole of a solute with one liter of solvent.

Example 4:

What is the molarity of ethyl alcohol in a solution made by diluting 0.25 moles ethyl alcohol to a final volume of 1.0 liter?

The solution need not be exactly 1.0 L in volume in order to calculate the molarity because the concentration ratio will be identical for any volume of the same solution.

Example 5:

What is the molarity of ethyl alcohol in bourbon whisky if 30 mL bourbon contains 10.68 g ethyl alcohol (C_{2}H_{6}O) ?

Since we know the contained mass of ethyl alcohol in grams, this must first be converted to moles of ethyl alcohol. Then, the volume of solution in mL must be expressed in units of liters.

If we know the molarity of a solution, the amount of contained solute in moles and grams can be calculated for a given volume of the solution.

Example 6:

How many grams of HNO_{3} are in 75 mL of a 8.0 M HNO_{3} solution?

It may seem that problems like Example 6 take many steps to work out, but in fact this is not so. The problem seems longer because each calculation was done separately as a ratio problem. Notice that in solving a ratio you must always cross-mutlipy and divide to find the answer. We can combine all the steps in one arithmetic calculation using the "factor-label" method. In this method of setting up the problem, we immediately go to the cross-multiply and divide step. Look back at steps 3 and 5 of Example 6. The unit in the numerator and denominator of the cross-multiplication always cancel, leaving the number on top in the units of the final answer. So we could write the arithmetic for Example 6 as:

Each ratio or fraction has the value of "1" because the numerator and denominator mean the same thing. That is, 63.02 g HNO_{3} = 1 mole HNO_{3}, so 63.02 g HNO_{3}/1 mole HNO_{3} = 1. When we multiply anything by 1, the answer is the same. (1 x 2 = 2). The result of doing a string of factor label calculations is to convert a number in one unit to an equivalent value in another unit. In example 6, 75 mL of 8.0 M HNO_{3} = 37.81 g HNO_{3}. In using this method to solve the problem, simply write the ratios being multiplied so that the units cancel to give the final answer in the units desired. With practice, you will find this method much faster than setting up each step as a separate ratio problem.

It is common laboratory practice to take a small volume of a concentrated solution and add more solvent to it to prepare a larger volume of a more dilute solution. This is analogous to taking one can of frozen orange juice concentrate and mixing it with "3 cans cold water" to prepare a quart of orange juice to drink. The key fact to appreciate is this: all the moles of solute in the concentrated solution end up in the more dilute solution. (Doesn't all the frozen orange pulp in the can end up in the quart of oj to drink?)

It is also helpful to note that when we multiply the Molarity of a solution by a particular volume in Liters, the result tells us the number of moles solute contained in that volume of solution:

Putting these two ideas together allows us to write the following useful equation:

Molarity_{dil} X Volume_{dil} = Molarity_{con} X Volume_{con},

where the subscript "dil" refers to the dilute solution and subscript "con" refers to the concentrated solution.

Example 7:

How many mL of 6.0 M HCl are needed to prepare 2.75 liters of 1.2 M HCl?

Most covalent compounds do not dissociate or ionize in water. However, an ionic compound separates into cations and anions on dissolving. In some chemical processes it is important to know the concentration of the total number of particles dissolved in solution. This is where the concentration unit osmolarity is useful. The __osmolarity__ of a solute in solution refers to the __Molarity of the compound x the number of particles produced per mole of compound when it dissolves in water.__

Example 8:

What is the osmolarity of a 1.0 M solution of ethyl alcohol (C_{2}H_{6}O) in water?

Ethyl alcohol is a covalent compound that does not ionize in water. Each mole of C_{2}H_{6}O which dissolves produces only one mole of particles. So,
osmolarity = 1.0 M x 1 mole of particles/mole compound = 1.0 osmol C_{2}H_{6}O.

Example 9:

What is the osmolarity of a 1.0 M solution of CaCl_{2} in water?

When 1 mole of CaCl_{2} dissolves in water, 3 moles of ions are produced;

CaCl_{2} --> Ca^{+2} + 2 Cl^{-}.

So, osmolarity = 1.0 M x 3 moles ions/mole compound = 3.0 osmol CaCl_{2}

copyright 1999, Larry McGahey, The College of St. Scholastica, Duluth MN.

All rights reserved.