Concentration of a solution represents the amount of solute dissolved in a given amount of solvent or solution. It is a macroscopic property and can be expressed in a variety of ways; qualitatively and quantitatively.

## Qualitative Expressions of Concentration

Quantitatively, a solution can be described as either

**dilute**: a solution that contains a small proportion of solute relative to solvent, or**concentrated**: a solution that contains a large proportion of solute relative to solvent.

## Semi-Quantitative Expressions of Concentration

Semiquantitatively, a solution can be described as

**unsaturated**: a solution in which more solute will dissolve, or**saturated**: a solution in which no more solute will dissolve.

In this case, the ** solubility** of a solute is the amount of solute that will dissolve in a given amount of solvent to produce a saturated solution.

## Qualitative Expressions of Concentration

There are a number of ways to express the relative amounts of solute and solvent in a solution; the prefered choice depends on convenience. For example, it is sometimes convinient to measure the volume of a solution rather than the mass of the solution.

It is good to note that some qualitative expressions for concentration are temperature-dependent (i.e., the concentration of the solution changes as the temperature changes), whereas others are not. This is particularly important for experiments in which the temperature does not remain constant.

Concentration expression | Measurements required | Temperature dependence |
---|---|---|

Percent composition | mass of solute mass of solution |
no(mass does not change with temperature) |

Percent composition (by volume) |
volume of solute volume of solution |
yes(volume changes with temperature) |

percent composition (mass/volume) |
mass of solute volume of solution |
yes(volume changes with temperature) |

molarity | moles of solute volume of solution |
yes(volume changes with temperature) |

molality | moles of solute mass of solvent |
no(neither mass nor moles changes with temperature) |

mole fraction | moles of solute moles of solvent |
no(moles do not change with temperature) |

- The
**mass/mass percent**of a solution is the percentage by mass of solute in the solution:

*Example: If 15 g of NaOH are dissolved in 285 g of water, what is the mass percent of NaOH in solution?*- First, the mass of the solution must be found by adding the solute and the solvent. In our case this value is 300 g.
- Second, we put the numbers into the equation and solve:

- The
**volume by volume percent**of a solution. This is convenient for measuring liquids and gases

*Example: If a 355 ml alcoholic beverage contains 43 ml of ethanol, what percent alcohol by volume is it?*- Again, the first step is finding the amount of solution that we have. This is given to us in the problem. It is 355 ml.
- Now, insert the numbers into the equation and solve:

- The
**mass/volume percent**of a solution is the percentage of a solute (mass) in a given volume of a solution:

*Example: How many grams of NaCl are needed to prepare 3.0 L of a 1% (m/v) NaCl solution?*- We know everything except the mass, or grams of solute, so we need to input our known quantities into the equation:

- We now divide both sides by 100

- Now, simply solve for grams of NaCl

- We know everything except the mass, or grams of solute, so we need to input our known quantities into the equation:

- In chemistry, the mole (from “molecule”), with the symbol=mol is used to measure the amount of a substance. Technically one mole is the amount of a substance which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12. This measure tells us the number of reacting particles.

The

**molarity (M)**of a solution is calculated by taking the moles of solute and dividing by the litres of solution; hence molarity = moles per litre.

*Example 1: What is the molarity of 2 moles of solute dissolved in 1 litre of solvent?*- Molarity = 2 mol /1 litre = 2 mol L -1 = 2 M

*Example 2: What is the molarity of 40 grams of NaOH dissolved in 2 litres of solvent?*

This calculation must be performed in two stages:- Convert grams to moles: The molecular weight of NaOH is 40 grams/mol (adding all atomic weights)

40 grams ÷ 40 grams/mol = 1 mol - Now calculating molarity: 1 mol ÷ 2 L = 0.5 M

- Convert grams to moles: The molecular weight of NaOH is 40 grams/mol (adding all atomic weights)
*Example 3: When 2 grams of NaCl (molecular weight 58.44 g mol -1) is dissolved in 100 mL of solute, what is the molarity of the solution?*

To do this calculation, we need to employ the relationship between molarity and moles, but in a rearranged format:- We know that

- which in other words is

- and therefore,

- since

- From equation (Y) above, ( x M) (0.1 L) = 2 g / 58.44
- Therefore: (x M) (0.1) = 0.034
**= 0.34 M**

- We know that
*Example 4: What is the molarity of a solution that contains 75 g of KNO*_{3}dissolved in 350 mL of solution?- We know that the KNO
_{3}is the solute so it is a simple matter of inputting numbers into the equation and solving. However, the solute is in grams and we want it in moles so we must use the molar mass of KNO_{3}as a conversion factor.

- We know that the KNO
*Example 5: How many grams of NaHCO*_{3}, are in 325 mL of a 4.50 M NaHCO_{3}solution?- We know the molarity and how much solution we have, so we use this as a conversion factor to determine the number of moles in 325 mL of solution.

- The above conversion gives us the number of moles of NaHCO
_{3}. It is easy to convert to the number of grams by simply using the molar mass of NaHCO_{3}.

- We know the molarity and how much solution we have, so we use this as a conversion factor to determine the number of moles in 325 mL of solution.

- The
**molality (m)**(yes this is the right spelling) of a solution is the concentration measured as moles of solute per kilogram of solvent (molality = moles per kilogram). Molalities are preferred in experiments that involve temperature changes of solutions, e.g. calorimetry and freezing point depression experiments.

*Example: A 1 m (not 1 M) NaCl solution contains 1 mole of NaCl per kilogram of water.*

**Remember!!!**- Measurement in moles is a measurement of the amount of a substance.
- Measurement in molarity is a measurement of the concentration of a substance – the amount (moles) per unit volume (litres).

- The
**Mole Fraction**,*X*, of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all components in the solution. To calculate mole fraction, we need to know the number of moles of each component present in the solution.- The mole fraction of A,
*X*A, in a solution consisting of A, B, C, … is calculated using the equation:

- To calculate the mole fraction of B,
*X*B, then

- The mole fraction of A,

## Dilutions of stock solutions

Dilution of stock solutions is frequently used to make solutions of any desired molarity. *What does it mean to dilute a solution?*

To dilute a solution means to add more solvent without adding more solute. (The resulting solution must be thoroughly mixed to ensure homogeneity.) The fact that the solute amount stays constant allows calculations to be made:

*moles before dilution = moles after dilution &&(X)*

From the definition of molarity (above):

*moles of solute = molarity * volume*

So we can substitute MV (molarity * volume) into equation (X) above:

*Molarity 1 * Volume 1 = Molarity 2 * Volume 2*

In short,

*M _{1}V_{1} = M_{2}V_{2}*

In other words

*Concentration 1 * Volume 1 = Concentration 2 * Volume 2*

“1” refers to the situation before dilution and “2” refers to after dilution. Any volume measurement can be used, so long as the same one is used on each side.

*Example 1: You have 60 ml of a 1.5 M solution of NaCl, but 0.6 M solution is needed. How many ml of 0.6 M can you make?*

We know that, M_{1}V_{1} = M_{2}V_{2}

1.5 M * 60 ml = 0.6 M * (V 2)

V_{2} = **150 ml**

*Example 2: You need 225 ml of 0.8 M NaOH solution and you have a 2.5 M stock solution. How would you make up the solution?*

2.5 M * (V_{1}) = 0.8 M * 225 ml

**V _{1} = 72 ml**

*Example 3: using M _{1}V_{1} = M_{2}V_{2} relationship, determine the following;*

*A) 60 ml of a solution are diluted to a volume of 120 ml. The concentration of the diluted solution is 2 M. What was the concentration of the original solution?*

**M _{1}** = (2 * 0.12)/0.06

**= 4 M**

*B) What volume of a 5 M solution is required to make 100 ml of a 2 M solution?*

V_{1} = (2 * 0.1)/5

= 0.04 L

**= 40 ml**

*C) 500 ml of a 5 M solution are diluted to 1 l. What is the concentration of the resulting solution?*

M_{2} = (5 * 0.5)/1

**= 2.5 M**

*D) 200 ml of a 1 M solution are diluted to make a 0.5 M solution. What is the volume of the resulting solution?*

V_{2} = (1 * 0.2)/0.5

= 0.4 L

= **400 ml**

Ratio method may also be used: The ratio of concentrations or volumes before or after dilution will depend on the dilution factor.

In the case of example 3(A) above, the volume is changing from 60 ml to 120 ml; which is basically a 2-fold dilution. We are required to determine the concentration of the original solution (C_{1}); which we know must be greater than the concentration of the dilute solution (C_{2}). The concentration of the diluted solution is 2 M; to get the concentration of the original solution, we simply multiply 2 M by the dilution factor.

**2M * 2 = 4 M**

In the case of example 3 (B) above, we can apply the same concept. The concentration is changing from 5 M in the original solution to 2 M in the diluted solution. The dilution is 2.5-fold dilution. We are required to calculate the volume of the original solution, which must be less than the volume of the diluted solution. Since the volume of the diluted solution is 100 ml, we need to divide this value (100) by the dilution factor;

**100 ÷ 2.5 = 40 ml**

You may practice with example 3 (C) and 3 (D). All the best!