# Parts Per Million Example Chemistry Assignment

Skills to Develop

- To describe the concentration of a solution in the way that is most appropriate for a particular problem or application.
- To be familiar with the different units used to express the concentrations of a solution.

There are several different ways to quantitatively describe the concentration of a solution. For example, molarity is a useful way to describe solution concentrations for reactions that are carried out in solution. Mole fractions are used not only to describe gas concentrations but also to determine the vapor pressures of mixtures of similar liquids. Example \(\PageIndex{1}\) reviews the methods for calculating the molarity and mole fraction of a solution when the masses of its components are known.

The concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent:

\[ \text{molality (m)} =\dfrac{\text{moles solute}}{\text{kilogram solvent}} \label{Eq1}\]

Molality, therefore, has the same numerator as molarity (the number of moles of solute) but a different denominator (kilogram of solvent rather than liter of solution). For dilute aqueous solutions, the molality and molarity are nearly the same because dilute solutions are mostly solvent. Thus because the density of water under standard conditions is very close to 1.0 g/mL, the volume of 1.0 kg of \(H_2O\) under these conditions is very close to 1.0 L, and a 0.50 M solution of \(KBr\) in water, for example, has approximately the same concentration as a 0.50 m solution.

Another common way of describing concentration is as the ratio of the mass of the solute to the total mass of the solution. The result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb):

\[\text{mass percentage}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100 \label{Eq2}\]

\[\text{parts per million (ppm)}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6} \label{Eq3}\]

\[\text{parts per billion (ppb)}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{9} \label{Eq4}\]

In the health sciences, the concentration of a solution is often expressed as parts per thousand (ppt), indicated as a proportion. For example, adrenalin, the hormone produced in high-stress situations, is available in a 1:1000 solution, or one gram of adrenalin per 1000 g of solution.

The labels on bottles of commercial reagents often describe the contents in terms of mass percentage. Sulfuric acid, for example, is sold as a 95% aqueous solution, or 95 g of \(H_2SO_4\) per 100 g of solution. Parts per million and parts per billion are used to describe concentrations of highly dilute solutions. These measurements correspond to milligrams and micrograms of solute per kilogram of solution, respectively. For dilute aqueous solutions, this is equal to milligrams and micrograms of solute per liter of solution (assuming a density of 1.0 g/mL).

How do chemists decide which units of concentration to use for a particular application? Although molarity is commonly used to express concentrations for reactions in solution or for titrations, it does have one drawback—molarity is the number of moles of solute divided by the volume of the solution, and the volume of a solution depends on its density, which is a function of temperature. Because volumetric glassware is calibrated at a particular temperature, typically 20°C, the molarity may differ from the original value by several percent if a solution is prepared or used at a significantly different temperature, such as 40°C or 0°C. For many applications this may not be a problem, but for precise work these errors can become important. In contrast, mole fraction, molality, and mass percentage depend on only the masses of the solute and solvent, which are independent of temperature.

Mole fraction is not very useful for experiments that involve quantitative reactions, but it is convenient for calculating the partial pressure of gases in mixtures, as discussed previously. Mole fractions are also useful for calculating the vapor pressures of certain types of solutions. Molality is particularly useful for determining how properties such as the freezing or boiling point of a solution vary with solute concentration. Because mass percentage and parts per million or billion are simply different ways of expressing the ratio of the mass of a solute to the mass of the solution, they enable us to express the concentration of a substance even when the molecular mass of the substance is unknown. Units of ppb or ppm are also used to express very low concentrations, such as those of residual impurities in foods or of pollutants in environmental studies.

Table \(\PageIndex{1}\) summarizes the different units of concentration and typical applications for each. When the molar mass of the solute and the density of the solution are known, it becomes relatively easy with practice to convert among the units of concentration we have discussed, as illustrated in Example \(\PageIndex{3}\).

Unit | Definition | Application |
---|---|---|

molarity (M) | moles of solute/liter of solution (mol/L) | Used for quantitative reactions in solution and titrations; mass and molecular mass of solute and volume of solution are known. |

mole fraction (\(\chi\)) | moles of solute/total moles present (mol/mol) | Used for partial pressures of gases and vapor pressures of some solutions; mass and molecular mass of each component are known. |

molality (m) | moles of solute/kg of solvent (mol/kg) | Used in determining how colligative properties vary with solute concentration; masses and molecular mass of solute are known. |

mass percentage (%) | [mass of solute (g)/mass of solution (g)] × 100 | Useful when masses are known but molecular masses are unknown. |

parts per thousand (ppt) | [mass of solute/mass of solution] × 10^{3} (g solute/kg solution) | Used in the health sciences, ratio solutions are typically expressed as a proportion, such as 1:1000. |

parts per million (ppm) | [mass of solute/mass of solution] × 10^{6} (mg solute/kg solution) | Used for trace quantities; masses are known but molecular masses may be unknown. |

parts per billion (ppb) | [mass of solute/mass of solution] × 10^{9} (µg solute/kg solution) | Used for trace quantities; masses are known but molecular masses may be unknown. |

### Summary

Different units are used to express the concentrations of a solution depending on the application. The concentration of a solution is the quantity of solute in a given quantity of solution. It can be expressed in several ways: molarity (moles of solute per liter of solution); mole fraction, the ratio of the number of moles of solute to the total number of moles of substances present; mass percentage, the ratio of the mass of the solute to the mass of the solution times 100; parts per thousand (ppt), grams of solute per kilogram of solution; parts per million (ppm), milligrams of solute per kilogram of solution; parts per billion (ppb), micrograms of solute per kilogram of solution; and molality (m), the number of moles of solute per kilogram of solvent.

Example \(\PageIndex{1}\): Molarity and Mole Fraction

Commercial vinegar is essentially a solution of acetic acid in water. A bottle of vinegar has 3.78 g of acetic acid per 100.0 g of solution. Assume that the density of the solution is 1.00 g/mL.

- What is its molarity?
- What is its mole fraction?

**Given**: mass of substance and mass and density of solution

**Asked for**: molarity and mole fraction

**Strategy**:

- Calculate the number of moles of acetic acid in the sample. Then calculate the number of liters of solution from its mass and density. Use these results to determine the molarity of the solution.
- Determine the mass of the water in the sample and calculate the number of moles of water. Then determine the mole fraction of acetic acid by dividing the number of moles of acetic acid by the total number of moles of substances in the sample.

**Solution**:

A The molarity is the number of moles of acetic acid per liter of solution. We can calculate the number of moles of acetic acid as its mass divided by its molar mass. The volume of the solution equals its mass divided by its density. The calculations follow:

\[moles\; CH_3CO_2H=\dfrac{3.78\; \cancel{g}\; CH_3CO_2H}{60.05\; \cancel{g}/mol}=0.0629 \;mol \nonumber\]

\[volume=\dfrac{mass}{density}=\dfrac{100.0\; \cancel{g}\; solution}{1.00\; \cancel{g}/mL}=100\; mL\nonumber\]

\[ molarity\; of\; CH_3CO_2H=\dfrac{moles\; CH_3CO_2H }{\text{liter solution}}\nonumber\]

\[=\dfrac{0.0629\; mol\; CH_3CO_2H}{(100\; \cancel{mL})(1\; L/1000\; \cancel{mL})}=0.629\; M \;CH_3CO_2H \nonumber\]

This result makes intuitive sense. If 100.0 g of aqueous solution (equal to 100 mL) contains 3.78 g of acetic acid, then 1 L of solution will contain 37.8 g of acetic acid, which is a little more than \(\frac{1}{2}\) mole. Keep in mind, though, that the mass and volume of a solution are related by its density; concentrated aqueous solutions often have densities greater than 1.00 g/mL.

B To calculate the mole fraction of acetic acid in the solution, we need to know the number of moles of both acetic acid and water. The number of moles of acetic acid is 0.0629 mol, as calculated in part (a). We know that 100.0 g of vinegar contains 3.78 g of acetic acid; hence the solution also contains (100.0 g − 3.78 g) = 96.2 g of water. We have

\[moles\; H_2O=\dfrac{96.2\; \cancel{g}\; H_2O}{18.02\; \cancel{g}/mol}=5.34\; mol\; H_2O\nonumber\]

The mole fraction \(\chi\) of acetic acid is the ratio of the number of moles of acetic acid to the total number of moles of substances present:

\[X_{CH_3CO_2H}=\dfrac{moles\; CH_3CO_2H}{moles \;CH_3CO_2H + moles\; H_2O}=\dfrac{0.0629\; mol}{0.0629 \;mol + 5.34\; mol}\nonumber\]

\[=0.0116=1.16 \times 10^{−2}\nonumber\]

This answer makes sense, too. There are approximately 100 times as many moles of water as moles of acetic acid, so the ratio should be approximately 0.01.

Exercise \(\PageIndex{1}\): Molarity and Mole Fraction

A solution of \(HCl\) gas dissolved in water (sold commercially as “muriatic acid,” a solution used to clean masonry surfaces) has 20.22 g of \(HCl\) per 100.0 g of solution, and its density is 1.10 g/mL.

- What is its molarity?
- What is its mole fraction?

**Answer a**6.10 M HCl

**Answer b**\(\chi_{HCl} = 0.111\)

Example \(\PageIndex{2}\): Molarity and Mass

Several years ago, millions of bottles of mineral water were contaminated with benzene at ppm levels. This incident received a great deal of attention because the lethal concentration of benzene in rats is 3.8 ppm. A 250 mL sample of mineral water has 12.7 ppm of benzene. Because the contaminated mineral water is a very dilute aqueous solution, we can assume that its density is approximately 1.00 g/mL.

- What is the molarity of the solution?
- What is the mass of benzene in the sample?

**Given**: volume of sample, solute concentration, and density of solution

**Asked for**: molarity of solute and mass of solute in 250 mL

**Strategy**:

- Use the concentration of the solute in parts per million to calculate the molarity.
- Use the concentration of the solute in parts per million to calculate the mass of the solute in the specified volume of solution.

**Solution**:

a. A To calculate the molarity of benzene, we need to determine the number of moles of benzene in 1 L of solution. We know that the solution contains 12.7 ppm of benzene. Because 12.7 ppm is equivalent to 12.7 mg/1000 g of solution and the density of the solution is 1.00 g/mL, the solution contains 12.7 mg of benzene per liter (1000 mL). The molarity is therefore

\[ molarity=\dfrac{moles}{liter solution}=\dfrac{(12.7\; \cancel{mg}) \left(\frac{1\; \cancel{g}}{1000\; \cancel{mg}}\right)\left(\frac{1\; mol}{78.114\; \cancel{g}}\right)}{1.00\; L} \nonumber \]

\[=1.63 \times 10^{-4} M\nonumber\]

b. B We are given that there are 12.7 mg of benzene per 1000 g of solution, which is equal to 12.7 mg/L of solution. Hence the mass of benzene in 250 mL (250 g) of solution is

\[ mass of benzene=\dfrac{(12.7\; mg\; benzene)(250\; \cancel{mL})}{1000\; mL}\nonumber\]

\[=3.18\; mg =3.18 \times 10^{-3}\; g\; benzene\nonumber\]

Exercise \(\PageIndex{2}\): Molarity of Lead Solution

The maximum allowable concentration of lead in drinking water is 9.0 ppb.

- What is the molarity of \(Pb^{2+}\) in a 9.0 ppb aqueous solution?
- Use your calculated concentration to determine how many grams of \(Pb^{2+}\) are in an 8 oz glass of water.

**Answer a**4.3 × 10

^{−8}M**Answer b**2 × 10

^{−6}g

Example \(\PageIndex{3}\): Vodka

Vodka is essentially a solution of ethanol in water. Typical vodka is sold as “80 proof,” which means that it contains 40.0% ethanol by volume. The density of pure ethanol is 0.789 g/mL at 20°C. If we assume that the volume of the solution is the sum of the volumes of the components (which is not strictly correct), calculate the following for the ethanol in 80-proof vodka.

- the mass percentage
- the mole fraction
- the molarity
- the molality

**Given**: volume percent and density

**Asked for**: mass percentage, mole fraction, molarity, and molality

**Strategy**:

- Use the density of the solute to calculate the mass of the solute in 100.0 mL of solution. Calculate the mass of water in 100.0 mL of solution.
- Determine the mass percentage of solute by dividing the mass of ethanol by the mass of the solution and multiplying by 100.
- Convert grams of solute and solvent to moles of solute and solvent. Calculate the mole fraction of solute by dividing the moles of solute by the total number of moles of substances present in solution.
- Calculate the molarity of the solution: moles of solute per liter of solution. Determine the molality of the solution by dividing the number of moles of solute by the kilograms of solvent.

**Solution**:

The key to this problem is to use the density of pure ethanol to determine the mass of ethanol (\(CH_3CH_2OH\)), abbreviated as EtOH, in a given volume of solution. We can then calculate the number of moles of ethanol and the concentration of ethanol in any of the required units. A Because we are given a percentage by volume, we assume that we have 100.0 mL of solution. The volume of ethanol will thus be 40.0% of 100.0 mL, or 40.0 mL of ethanol, and the volume of water will be 60.0% of 100.0 mL, or 60.0 mL of water. The mass of ethanol is obtained from its density:

\[mass\; of\; EtOH=(40.0\; \cancel{mL})\left(\dfrac{0.789\; g}{\cancel{mL}}\right)=31.6\; g\; EtOH\nonumber\]

If we assume the density of water is 1.00 g/mL, the mass of water is 60.0 g. We now have all the information we need to calculate the concentration of ethanol in the solution.

B The mass percentage of ethanol is the ratio of the mass of ethanol to the total mass of the solution, expressed as a percentage:

\[\%EtOH=\left(\dfrac{mass\; of\; EtOH}{mass\; of\; solution}\right)(100)=\left(\dfrac{31.6\; \cancel{g}\; EtOH}{31.6\; \cancel{g} \;EtOH +60.0\; \cancel{g} \; H_2O} \right)(100) = 34.5\%\nonumber\]

C The mole fraction of ethanol is the ratio of the number of moles of ethanol to the total number of moles of substances in the solution. Because 40.0 mL of ethanol has a mass of 31.6 g, we can use the molar mass of ethanol (46.07 g/mol) to determine the number of moles of ethanol in 40.0 mL:

\[ moles\; EtOH=(31.6\; \cancel{g\; EtOH}) \left(\dfrac{1\; mol}{46.07\; \cancel{g\; EtOH}}\right)=0.686 \;mol\; CH_3CH_2OH\nonumber\]

Similarly, the number of moles of water is

\[ moles \;H_2O=(60.0\; \cancel{g \; H_2O}) \left(\dfrac{1 \;mol\; H_2O}{18.02\; \cancel{g\; H_2O}}\right)=3.33\; mol\; H_2O\nonumber\]

The mole fraction of ethanol is thus

\[ \chi_{EtOH}=\dfrac{0.686\; \cancel{mol}}{0.686\; \cancel{mol} + 3.33\;\cancel{ mol}}=0.171\nonumber\]

D The molarity of the solution is the number of moles of ethanol per liter of solution. We already know the number of moles of ethanol per 100.0 mL of solution, so the molarity is

\[ M_{EtOH}=\left(\dfrac{0.686\; mol}{100\; \cancel{mL}}\right)\left(\dfrac{1000\; \cancel{mL}}{L}\right)=6.86 \;M\nonumber\]

The molality of the solution is the number of moles of ethanol per kilogram of solvent. Because we know the number of moles of ethanol in 60.0 g of water, the calculation is again straightforward:

\[ m_{EtOH}=\left(\dfrac{0.686\; mol\; EtOH}{60.0\; \cancel{g}\; H_2O } \right) \left(\dfrac{1000\; \cancel{g}}{kg}\right)=\dfrac{11.4\; mol\; EtOH}{kg\; H_2O}=11.4 \;m \nonumber\]

Exercise \(\PageIndex{3}\): Toluene/Benzene Solution

A solution is prepared by mixing 100.0 mL of toluene with 300.0 mL of benzene. The densities of toluene and benzene are 0.867 g/mL and 0.874 g/mL, respectively. Assume that the volume of the solution is the sum of the volumes of the components. Calculate the following for toluene.

- mass percentage
- mole fraction
- molarity
- molality

**Answer a**mass percentage toluene = 24.8%

**Answer b**\(\chi_{toluene} = 0.219\)

**Answer c**2.35 M toluene

**Answer d**3.59 m toluene

In our daily routine we come across variety of substances which are not pure but instead a homogeneous mixture of two or more pure substances. Many of these homogeneous mixtures have great importance in our life. The utility of these homogeneous mixtures depend largely on their compositions. Sometime individual components are same but they exhibit different properties in mixtures of different compositions. For example, presence of fluoride ions (F) in water at a concentration of about 1 ppm prevents tooth decay but if its concentration in water becomes 1.5 ppm, it becomes harmful and cause tooth to become mottled. The concentration of fluoride ion beyond this limit becomes poisonous (generally used for rat poisoning).

**SOLUTE, SOLVENT AND SOLUTION**

A solution is a homogeneous mixture of two or more pure substances whose composition

may be altered within certain limits. Though the solution is homogeneous in nature, yet it retains the properties of its constituents. The substances which make up the solution are generally called its components. The solution of two components is referred to as binary solution. Similarly, the solutions of three and four components are called ternary and quaternary solutions respectively: In this unit, we shall limit our discussion to binary solutions only. The two components of the binary solutions are respectively called a solvent and a solute.

Solvent is that component in the solution whose physical state is the same as that of the resulting solution while the other component is called solute. For solutions in which both the components have the same physical state as that of solution, the component which is in excess is called solvent and the other one is called solute.

For example, in case of solution of sugar and water, sugar is the solute and water is the solvent. Each component of binary solution may be in solid, liquid or gaseous state. Consequently, there can be nine different types of solutions as shown in Table 11.1.

**Table 11.1. Different Types of Solutions**

**EXPRESSING CONCENTRATION OF SOLUTIONS OF SOLIDS IN LIQUIDS**

Concentration of solution means describing its composition. Qualitatively, it can be described by using the word dilute for solutions having very small quantity of solute and the word concentrated for solutions having large quantity of solute. However, this kind of description leads to confusion. Therefore. quantitative description is more appropriate. Quantitatively, concentration of a solution refers to the amount of solute present in the given quantity of solution or solvent. The concentration of the solution may be expressed in any of the following ways:

**1. Mass Percentage(% mass)**

Mass percentage may be defined as the number of parts by mass of solute per hundred parts by mass of solution. For example, a 5% (aqueous) solution of sugar by mass means that 100 g of solution contain 5 g of sugar.

If W B be the mass of solute (B) and W A be the mass of solvent (A), then

Mass percentage of B = W_{B} / W_{A} + W_{B} x 100

**2. Volume Percentage(% volume)**

This mode of concentration is used in case of solutions when solutes and solvents are both liquids. Volume percentage may be defined as the number of parts by volume of solute per hundred parts by volume of solution. For example, a 25% solution of ethyl alcohol (by volume) means that 100 cm3 of the solution contain 25 cm^{3 }of ethyl alcohol and 75 cm3 of water.

If VA and VB be the volumes of component A and B, then

Volume percentage of B = V_{B} + V_{A} + V_{B} x 100

**3. Normality (N)**

Normality of a solution is defined as the number of g-equivalents of the solute present in one litre (1 dm^{3}) of the solution or milliequivalents of solute present in one cm3 of solution. It is represented by N.

Mathematically,

Normality (N) = G- equivalents of solute (B) / volume of solution in litre

G-equivalents of solute (B) represents the ratio of its mass in gram (W B) to its gram-equivalent mass (GEM). Also, the ratio of the mass of solute in gram (W B) to the volume of the

solution in litre (V L) represents the strength of the solution.

Thus,

Normality (N) = WB (g) / GEM_{B} x V(L)

Strength of sol (gdm ^{-3}) / GEM_{B}

W_{B} (g) x 1000 / GEM_{B} x V_{(cm3)}

A solution having normality equal to unity is called a normal solution. Such a solution contains one gram equivalent of solute per litre of solution.

**4. Molarity (M)**

Molarity of a solution is defined as the number of gram mole of the solute present in one litre of the solution or millimoles of solute present in one cm3 of solution. It is represented by M.

Mathematically,

Molarity (M) = Moles of solute / volume of solution in litres

= Mass of solute in gram / (Molar mass of solute) x (Volume of soln. (dm^{3}))

Now, moles of the solute (n _{B}) represents the ratio of mass of solute in gram (W _{B}) to its molar mass (MMB). Thus, molarity can be represented by the expressions.

M = n_{B} / V_{(L)} or W_{B} (g) / (MM)_{B} x V_{(L)}

or strength (gdm ^{-3}) / (MM)_{B} OR W_{B} (g) x 1000 / (MM )_{B} x V _{(cm3) }

A solution having molarity equal to unity is called molar solution. Such a solution contains one mole of solute per litre of solution. The solutions having molarity equal to 0.5 M, 0.1 M and 0.01 M are called semimolar, decimolar and centimolar solutions respectively. Molarity is expressed in units of mol L^{-1} or mol dm^{-3} It may be noted that bothnormality as well as molarity of a solution change with change in temperature.

**5. Molality (m)**

Molality of a solution may be defined as the number of gram mole of the solute present in 1000 g ( 1 kg) of the solvent. It is represented by m.

Mathematically, Molality (m)

= mole of solvent / mass of solvent of solvent ( kg)

= Mass of solute (g) / (Molar mass of solute) x (Mass of solvent (kg))

or m = n_{B} / W_{A} (Kg) or E_{B} (g) / (MM)_{B} x W_{A} (kg)

or W_{B} (g) x 1000 / (MM) _{B} x W_{A} (g)

A solution containing one mole of solute per 1000 g of solvent has molality equal to one and is called a molal solution. Molality is expressed in units of moles per kilogram (mol kg-1 ). The molality of a solution does not change with temperature.

**6. Mole Fraction**

Mole fraction may be defined as the ratio of number of moles of one component to the total number of moles of all the components (solvent and solute) present in the solution. It is denoted by the letter x followed by the subscript representing the component. Let us suppose that a solution contains the components A and B and suppose that W A g of A and W B g of B are present in it.

Moles of (n_{A}) = W_{A} / (MM)_{A}

Moles of B (n_{B}) = W_{B }(MM)_{B}

(MM) _{A} and (MM)8 are molar masses of A and B respectively

Total number of moles of A and B = n A + n _{B}

Mole fraction of A, (x _{A}) = n _{A} / n _{A} + n _{B}

Now, the sum of the mole fractions of solute and solvent in binary solution is unity.

This generalisation can be extended to solutions having more than two components by saying that the sum of mol fractions of all the components in a solution is always unity. It may be noted that the mole fraction is independent of the temperature.

**7. Formality**

Formality of a solution may be defined as the number of gram formula masses (moles) of the ionic solute present in one litre of the solution. It is represented by F.

Mathematically,

Formality (F) = Number of moles of ions solute / Volume of solution in litres

= Mass of ionic solute (g) / (GFM) _{solute }x V _{sol }(dm ^{3})

Thus , f =

Commonly, the term formality is used to express the concentration of the ionic solids which do not exist as molecules but exist as network of ions. A solution containing one gram formula mass of solute per litre of the solution has formality equal to one and is called formal solution. It may be mentioned here that like normality and molarity. The formality of a solution changes with change in temperature.

**8. Parts Per Million (ppm)**

When a solute is present in very minute amounts, its concentration is expressed in parts per million. It may be defined as the number of parts by mass of solute per million parts by mass of the solution. It is abbreviated as ppm.

Parts per million (ppm) = Mass of solute / Mass of solution x 10^{6}

Ppm = W_{B }/ W_{A} + W_{B} x 10^{6}

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