**aqueous** (water-as-solvent) solutions, but the same measures of concentration can apply to any other solvent. When the solven is not given, or if we recognize the solutes as water-soluble (solutes most frequently dissoved in water), we assume that the solution is aqueous.

It makes sense that some solutions are stronger than others: adding a tiny bit of acid to a gallon of water, for example, might not even be detectable. But add a *lot* of acid and you have a dangerous solution that needs to be handled carefully.

We need a way to represent the relative strength of a solution. We'll call it **concentration**.

There are several methods of concentration measurement, each used in different kinds of situations.

If we dissolved just a couple of crystals of table salt (NaCl) in water, we might not even be able to taste it. But if we dissolved a bunch of NaCl in the *same* amount of water (and you might be surprised how much will dissolve), it would taste very salty.

On a relative scale, one solution is **dilute** and the other is **concentrated**. We'd like to be able to put this on some sort of numerical scale so that we could say just exactly how dilute or concentrated a solution is.

Remember that the **solute** (usually a solid) is what's being dissolved in a **solvent** (usually a liquid)

All of the concentration measurement methods covered below consist of some measure of the amount of solute (in grams, moles or atoms/molecules) divided by the amount of solvent (in units of mass, volume, moles or number of atoms/molecules).

**Molarity**, abbreviated** (M) **is probably the most commonly used measure of solution concentration. Molarity is the number of moles of solvent divided by the number of liters of solution. We refer to a solution, for example, as a "1.5 molar solution" or "1.5 M".

Note that we divide by the total number of liters of **solution**, *including the solute*, not the number of liters of solvent *to which the solute was added* — it's an important (if often small) distinction.

This figure ( → ) shows how to make an X-molar (X M) solution, where X is the desired molar concentration. We have a target volume (1 liter in this case, but it could be anything). The solute is dissolved in a smaller volume of solvent, then the total volume is adjusted to the final desired amount.

**Molarity** (M) is the number of moles of solute divided by the total volume of the solution in liters. A 1 M solution has 1 mole of solute for every 1 L of solution.

Calculate the molarity of a NaCl solution formed by dissolving 62 g of NaCl in water and adjusting the total volume to 0.50 liters.

**Solution**

Then divide it by the total number of liters of solution, 0.5 L in this case:

How many grams of (NH_{4})_{2}SO_{4} (ammonium sulfate) need to be added to water to make 200 ml of a 1.6 M solution?

In this problem we know the concentration, we just need to know the amount of solute needed to achieve it. Fist find that number of moles:

Then convert it to grams.

Note that in *making* this solution, we'd want to dissolve the solute in less than 200 ml of water, then bring the total volume to 200 ml afterward. Adding solutes to solvent can cause *either* expansion or contraction (weird, right?) of the solution.

**Molality***, abbreviated by lower-case **m**, is the number of moles of solute divided by the number of **Kilograms** of solvent. We say a solution is, for example, a "3.0 **molal** solution."

This figure ( → ) shows how to make an X-molal (X m) solution, where **X** is the number of moles of solute and **n** is the number of Kilograms of solvent.

The volume of the solvent can be used instead of the mass if its density is known. For example, at 4˚C, 1 liter of H_{2}O has a mass of 1 Kg. This measure of concentration has a couple of advantages: (1) only a balance is required to prepare a solution of known concentration and (2) variations of the density of the solvent with temperature (some can be significant) are irrelevant. Nevertheless, molality isn't used that often.

*Note: In modern chemistry the term molality has fallen out of use in favor of just using the units: mol/Kg

**Molality** (m) is the number of moles of solute divided by the number of Kilograms of solvent. A 1 m solution contains 1 mole of solute for every 1 Kg of solvent. Use of the unit mol/Kg is now preferred over molality.

Calculate the molality of a solution of MgCl_{2} that is formed by adding 998.2 g of water to 10.6 g of solid MgCl_{2}.

**Solution**

$$ \require{cancel} \begin{align} 10.6 \cancel{g \, MgCl_2} &\left( \frac{1 \, mol \, MgCl_2}{95.2 \cancel{g \, MgCl_2}} \right) \\[5pt] &= 0.1113 \, mol \, MgCl_2 \end{align}$$

Now convert the mass of the solvent to kilograms ("*kilo*" means 1000; there are 1000 *grams* in a *kilogram*)

$$ \require{cancel} \begin{align} 998.2 \cancel{g \, H_2O} &\left( \frac{1 \, Kg}{1000 \cancel{g}} \right) \\[5pt] &= 0.9982 \, Kg \, H_2O \end{align}$$

Finally, the concentration is just the ratio of those amounts:

$$\frac{0.1113 \, mol \, MgCl_2}{0.9982 \, Kg \, H_2O} = 0.11 \, m \, MgCl_2$$

While molality can be quite useful as a measurement of concentration, it isn't too convenient for converting to **molarity**. The reason is that we usually don't know how much volume the solute is going to occupy in the solution. Some solutes even cause **contraction of the solvent**. For example, when 900 ml of distilled **H _{2}O** is mixed with 100 ml of ethanol (

The **mole fraction** is used in some calculations because it is massless. The mole fraction of one constituent of a solution is the number of moles of that constituent divided by the total number of moles of all components of the solution. The sum of the mole fractions of all components is one.

Mole fraction calculations work with any property that is proportional to the number of molecules (therefore moles) present, including partial pressure for gases.

Mole fraction is particularly useful when we learn the fine details of the liquid state and in **statistical mechanics**, the microscopic approach to deriving thermodynamic characteristics of materials.

The **mole fraction** ( **χ _{i}** ) of the i

Calculate the mole fraction of both components of a mixture of water (H_{2}O) and ethanol (C_{2}H_{5}OH) that is 50% by mass in each component.

**Solution**

Now it's a simple matter to find the two mole fractions. Mole fraction is usually given the symbol χ, the Greek letter "chi."

X
#### The Greek alphabet

alpha | Α | α |

beta | Β | β |

gamma | Γ | γ |

delta | Δ | δ |

epsilon | Ε | ε |

zeta | Ζ | ζ |

eta | Η | η |

theta | Θ | θ |

iota | Ι | ι |

kappa | Κ | κ |

lambda | Λ | λ |

mu | Μ | μ |

nu | Ν | ν |

xi | Ξ | ξ |

omicron | Ο | ο |

pi | Π | π |

rho | Ρ | ρ |

sigma | Σ | σ |

tau | Τ | τ |

upsilon | Υ | υ |

phi | Φ | φ |

chi | Χ | χ |

psi | Ψ | ψ |

omega | Ω | ω |

$$\chi_{ethanol} = \frac{17.13}{17.13 + 55.55} = 0.236$$

$$\chi_{H_2O} = \frac{55.55}{17.13 + 55.55} = 0.764$$

We frequently use units like parts per million (ppm) or parts per trillion (ppt). The most commonly used are

parts per million (ppm)

parts per billion (ppb)

parts per trillion (ppt)

These concentration measurements are used frequently to describe low concentrations where low concentrations are significant, like toxins. For example, the U.S. Food And Drug Administration (USFDA) sets a limit on the allowable concentration of mercury (Hg) in food at 1 ppm because it is so toxic in very small amounts.

Calculate the molarity of mercury (Hg) at a concentration of 1 ppm in water.M

Wide equations, scroll L ↔ R

**Solution**

$$1 \cancel{atom \, Hg} \left( \frac{1 \, mol \, Hg}{6.02 \times 10^{23} \cancel{atoms}} \right) = 1.66 \times 10^{-24} \, moles$$

Now for every million molecules of water, what is the volume of that water?

$$ \require{cancel} 10^6 \cancel{molecules \, H_2O} \left( \frac{18 \cancel{g \, H_2O}}{6.02 \times 10^{23} \cancel{molecules}} \right) \left( \frac{1 \, L \, H_2O}{1000 \cancel{g \, H_2O}} \right) = 3 \times 10^{-20} \, L$$

The molarity is the number of moles of Hg divided by the number of liters of solution. Notice that we're actually making an approximation here, namely that the addition of a small amount of mercury to water doesn't significantly change its volume.

$$molarity = \frac{1.66 \times 10^{-24} \, mol}{3 \times 10^{-20} \, L} = 5.55 \times 10^{-5} \, M$$

You can see that 1 ppm is a very small concentration. Some substances are toxic at much smaller concentrations.

1. |
Calculate the molar concentration of a 415 ml solution containing 0.745 moles of HCl. ## Solution$$ \begin{align} Molarity &= \frac{\text{moles of solute}}{\text{liters of solution}}\\[5pt] &= \frac{0.745 \, mol \, HCl}{0.415 \, L} \\[5pt] &= 1.79 \, M \end{align}$$ |

2. |
Calculate the molar concentration of an acetic acid (CH ## Solution$$ \begin{align} Molarity &= \frac{\text{moles of solute}}{\text{liters of solution}}\\[5pt] &= \frac{3.21 \, mol \, HOAc}{4.50 \, L} \\[5pt] &= 0.71 \, M \end{align}$$ |

3. |
How many moles of KI (potassium iodide) are present in 125 ml of 0.5 M KI? ## Solution$$0.5 \, M \: \text{means} \: \frac{0.5 \, mol \, KI}{1 \, L}$$ Now set up a proportion to find the number of moles of KI: $$ \begin{align} \frac{0.5 \text{mol KI}}{1 \,L} &= \frac{x \text{mol KI}}{0.125 \, L} \\[5pt] x(1) &= 0.5(0.125) \\[5pt] &= 0.062 \; \text{moles of KI} \end{align}$$ |

4. |
How many liters of water are required to prepare a solution of 7.25 M MgCl ## Solution$$7.25 \, M \: \text{means} \: \frac{7.25 \, mol \, MgCl_2}{1 \, L}$$ Now set up a proportion to find the number of liters of solution: $$ \begin{align} \frac{7.25 \text{mol MgCl_2}}{1 \,L} &= \frac{4.89 \text{mol KI}}{x \, L} \\[5pt] 7.25 x &= 4.89(1) \\[5pt] &= 0.674 \; \text{liters of solution} \\[5pt] &= 674 \; \text{mL} \end{align}$$ |

5. |
Calculate the molar concentration of a solution prepared by adding 34 g of NaCl (table salt) to 230 ml of H ## Solution$$ \require{cancel} \begin{align} \text{Molarity} &= \frac{\text{moles of solute}}{\text{liters of solution}}\\[5pt] &= \frac{34 \cancel{g \, NaCl} \left( \frac{1 \, mol \, NaCl}{54.45 \cancel{g \, NaCl}} \right)}{0.230 \, L} \\[5pt] &= 2.53 \, M \end{align}$$ Be careful here. Molarity is moles of solute divided by liters of |

6. |
Calculate the concentration of a solution prepared by dissolving 5.68 g of NaOH in enough water to make 400 ml of solution. ## Solution$$ \require{cancel} \begin{align} \text{Molarity} &= \frac{\text{moles of solute}}{\text{liters of solution}}\\[5pt] &= \frac{5.68 \cancel{g \, NaOH} \left( \frac{1 \, mol \, NaOH}{40 \cancel{g \, NaOH}} \right)}{0.400 \, L} \\[5pt] &= 0.35 \, M \end{align}$$ Be careful here. Molarity is moles of solute divided by liters of |

7. |
If a 2.34 g sample of dry ice (CO ## Solution$$ \require{cancel} \begin{align} \text{Molarity} &= \frac{\text{moles of solute}}{\text{liters of solution}}\\[5pt] &= \frac{2.34 \cancel{g \, CO_2} \left( \frac{1 \, mol \, CO_2}{44 \cancel{g \, CO_2}} \right)}{0.500 \, L} \\[5pt] &= 0.11 \, M \end{align}$$ This assumes that the CO |

8. |
How many grams of beryllium chloride (BeCl ## Solution$$ \begin{align} \text{0.05 m means} \: &\frac{0.05 \, mol \, BeCl_2}{1 \, L \, solvent} \\[5pt] \frac{0.05 \, mol \, BeCl_2}{1 \, L} &= \frac{x \, mol \, BeCl_2}{0.125 \, L} \\[5pt] x &= 0.05(0.125) \\[5pt] &= 0.006 \, mol \, BeCl_2 \end{align}$$ Then calculate the number of grams of BeCl $$ \require{cancel} 0.006 \cancel{mol \, BeCl_2} \left( \frac{80 \, g \, BeCl_2}{1 \cancel{mol \, BeCl_2}} \right) = 0.05 \, g \, BeCl_2$$ |

9. |
How many grams of BeCl ## Solution$$\text{molality} = \frac{\text{mol solute}}{\text{Kg solvent}}$$ The density of water is 1 g/mL or 1 Kg/L, so the mass of our solvent is 0.125 Kg. Now solve for the number of moles: $$ \begin{align} 0.05 \, m &= \frac{x}{0.125} \\[5pt] x &= 0.05(0.125) \\[5pt] x &= 0.00625 \, mol \, BeCl_2 \end{align}$$ Now calculate the mass of 0.00625 moles of BeCl $$ \require{cancel} \begin{align} 0.00625 \cancel{mol \, BeCl_2} &\left( \frac{80 \, g \, BeCl_2}{1 \cancel{mol \, BeCl_2}} \right) \\[5pt] &= 0.5 \, g \, BeCl_2 \end{align}$$ |

10. |
The density of ethanol is 0.789 g/ml. How many grams of ethanol should be mixed with 225 ml of water to make a 4.5% (volume/volume) mixture? ## SolutionRemember that 4.5% is 4.5/100 or 0.045. $$0.045 = \frac{V_{ethanol}}{225 + V_{ethanol}}$$ Now rearrange and solve for V $$ \begin{align} 0.045 (225) + 0.045 \, V_{ethanol} &= V_{ethanol} \\[5pt] 0.955 \, V_{ethanol} &= 10.125 \\[5pt] V_{ethanol} &= 10.60 \, ml \end{align}$$ Now use the density of ethanol to find the mass: $$ \require{cancel} 10.60 \cancel{ml} \left( \frac{0.789 \, g}{1 \cancel{ml}} \right) = 8.36 \text{g ethanol}$$ |

11. |
Calculate the volume of a 0.50 M solution of calcium hydroxide (Ca(OH) ## SolutionCalculate the number of moles of Ca(OH) $$ \require{cancel} \begin{align} 25 \cancel{g \, Ca(OH)_2} &\left( \frac{1 \, mol \, Ca(OH)_2}{74.1 \cancel{g \, Ca(OH)_2}} \right) \\[5pt] &= 0.337 \, mol \, Ca(OH)_2 \end{align}$$ Now set up a proportion to find the volume in liters: $$ \begin{align} \frac{0.5 \, mol \, Ca(OH)_2}{1 \, L} &= \frac{0.337 \, mol \, Ca(OH)_2}{\bf{x} \, L} \\[5pt] &= 0.674 \text{ L of solution} \end{align}$$ |

12. |
Calculate the mole fraction of sulfuric acid in a solution made by adding 3.4 g of sulfuric acid (H ## Solution$$ \require{cancel} \begin{align} 3.4 \cancel{g \, H_2SO_4} &\left( \frac{1 \, mol \, H_2SO_4}{98 \cancel{g \, H_2SO_4}} \right) \\[5pt] &= 0.0347 mol \, H_2SO_4 \end{align}$$ $$3,500 \, ml \, H_2O \approx 3,500 \, g \, H_2O$$ $$ \require{cancel} 3500 \cancel{g \, H_2O} \left( \frac{1 \, mol \, H_2O}{18 \cancel{g \, H_2O}} \right) = 194.4 \, mol \, H_2O$$ $$\chi_{H_2SO_4} = \frac{0.037}{194.4 + 0.037} = 1.9 \times 10^{-4}$$ |

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