17.3: Measuring Concentration
Lesson Objectives
The student will:
 define the terms concentrated and dilute.
 define concentration and list the common units used to express the concentration of solutions.
 define molarity, mass percent, ppm, and molality.
 write the formula for molarity and use the formula to perform calculations involving the molarity, moles of solute, and volume of a solution.
 calculate mass percent, ppm, molality, and mole fraction.
Vocabulary
 concentrated
 concentration
 dilute
 mass percent
 molality
 molarity
 parts per million
Introduction
Concentration is the measure of how much a given substance is mixed with another substance. Solutions can be said to be dilute or concentrated. A concentrated solution is one in which there is a large amount of solute in a given amount of solvent. A dilute solution is one in which there is a small amount of solute in a given amount of solvent. A dilute solution is a concentrated solution that has been, in essence, watered down. Think of the frozen juice containers you buy in the grocery store. In order to make juice, you mix the frozen juice from inside these containers with about 3 or 4 times the amount of water. Therefore, you are diluting the concentrated juice. The terms “concentrated” and “dilute,” however, only provide a qualitative way of describing concentration. In this lesson, we will explore some quantitative methods of expressing solution concentration.
Molarity
Of all the quantitative measures of concentration, molarity is the one used most frequently by chemists. Molarity is defined as the number of moles of solute per liter of solution. The symbol given for molarity is \begin{align*}\mathrm{M}\end{align*}



 molarity = mol/L = \begin{align*} \frac {\mathrm{moles} \ \mathrm{of} \ \mathrm{solute}} {\mathrm{liters} \ \mathrm{of} \ \mathrm{solution}}\end{align*}


Example:
What is the concentration, in mol/L, when \begin{align*}2.34 \ \mathrm{mol}\end{align*} of \begin{align*}\mathrm{NaCl}\end{align*} has been dissolved in \begin{align*}500. \ \mathrm{mL}\end{align*} of \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*}?
Solution:


 \begin{align*}[\mathrm{NaCl}] = \frac {2.34 \ \mathrm{mol}} {0.500 \ \mathrm{liter}} = 4.68 \ \mathrm{M}\end{align*}

The concentration of the \begin{align*}\mathrm{NaCl}\end{align*} solution is \begin{align*}4.68 \ \mathrm{mol/L}\end{align*}.
Example:
What would be the mass of potassium sulfate in \begin{align*}500. \ \mathrm{mL}\end{align*} of a \begin{align*}1.25 \ \mathrm{mol/L}\end{align*} potassium sulfate solution?
Solution:


 \begin{align*}\mathrm{M} = \frac {\mathrm{mol}} {\mathrm{L}}\end{align*}, so \begin{align*}\mathrm{mol} = \mathrm{M} \times \mathrm{L}\end{align*}



 \begin{align*}(1.25 \ \mathrm{mol/L}) (0.500 \ \mathrm{L}) = 0.625 \ \mathrm{mol}\end{align*}



 \begin{align*}\mathrm{mol} = \frac {\mathrm{mass}} {\mathrm{molar \ mass}}\end{align*}



 so \begin{align*}\mathrm{mass} = (\mathrm{mol}) \cdot (\mathrm{molar \ mass})\end{align*}



 \begin{align*}\mathrm{mass} = (0.625 \ \mathrm{mol})(174.3 \ \mathrm{g/mol}) = 109 \ \mathrm{g}\end{align*}

Therefore, the mass of the \begin{align*}\mathrm{K}_2\mathrm{SO}_4\end{align*} that dissolves in \begin{align*}500. \ \mathrm{mL}\end{align*} of \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*} to make this solution is \begin{align*}109 \ \mathrm{g}\end{align*}.
Mass Percent
Mass percent is the number of grams of the solute in the number of grams of solution. Mass percent is a term frequently used when referring to solid solutions. It has the formula:


 percent by mass = \begin{align*} \frac {\mathrm{mass \ of \ solute}} {\mathrm{solute \ mass} + \mathrm{solvent \ mass}} \times 100\%\end{align*}

or


 percent by mass = \begin{align*} \frac {\mathrm{mass \ solute}} {\mathrm{mass \ solution}} \times 100\%\end{align*}

Example:
An alloy is prepared by adding \begin{align*}15 \ \mathrm{g}\end{align*} of zinc to \begin{align*}65 \ \mathrm{g}\end{align*} of copper. What is the mass percent of zinc?
Solution:


 percent by mass = \begin{align*}\frac {\mathrm{mass \ solute}} {\mathrm{mass \ solution}} \times 100\% = \frac {15 \ \mathrm{g}} {80. \ \mathrm{g}} \times 100\%\end{align*} = 19%

Parts Per Million
Parts per million is another unit for concentration. Parts per million denotes that there is 1 milligram of solute for every kilogram of solvent. It is used most frequently when dealing with environmental issues. You may have heard about parts per million when scientists are referring to drinking water or poisons in fish and other food products. To calculate parts per million, the following formula is used.


 \begin{align*}\mathrm{ppm} = \frac {\mathrm{mass \ of \ solute}} {\mathrm{mass \ of \ solution}} \times 10^6\end{align*}

Example:
Mercury levels in fish have often been at the forefront of the news for people who love to eat fresh fish. Salmon, for instance, contains \begin{align*}0.01 \ \mathrm{ppm}\end{align*} compared to shark which contains \begin{align*}0.99 \ \mathrm{ppm}\end{align*}. In the United States, canned tuna is the most popular selling fish and has a mercury level of \begin{align*}0.12 \ \mathrm{ppm}\end{align*}, according to the FDA statistics. If one were to consume \begin{align*}1.00 \ \mathrm{kg}\end{align*} of canned tuna over a certain time period, how much mercury would be consumed?
Solution:


 \begin{align*}\mathrm{ppm} = \frac {\mathrm{mass \ of \ solute}} {\mathrm{mass \ of \ solution}} \times 10^6\end{align*}



 \begin{align*}\text{mass of solute} = \frac {(\mathrm{mass \ of \ solution})(\mathrm{ppm})} {1 \times 10^6}\end{align*}



 \begin{align*}\text{mass of solute} = \frac {(1000. \ \mathrm{g})(0.12)} {1 \times 10^6} = 1.2 \times 10^{4} \ \mathrm{g}\end{align*}

Molality
Molality is another way to measure concentration of a solution. It is calculated by dividing the number of moles of solute by the number of kilograms of solvent. Molality has the symbol \begin{align*}\mathrm{m}\end{align*}.


 \begin{align*} \text{molality (m)} = \frac {\mathrm{mol \ of \ solute}} {\mathrm{kg \ of \ solvent}}\end{align*}

Molarity, if you recall, is the number of moles of solute per volume of solution. Volume is temperature dependent. As the temperature rises, the molarity of the solution will actually decrease slightly because the volume will increase slightly. Molality does not involve volume, and mass is not temperature dependent. Thus, there is a slight advantage to using molality over molarity when temperatures move away from standard conditions.
Example:
Calculate the molality of a solution of hydrochloric acid where \begin{align*}12.5 \ \mathrm{g}\end{align*} of hydrochloric acid has been dissolved in \begin{align*}115 \ \mathrm{g}\end{align*} of water.
Solution:


 \begin{align*}\mathrm{mol \ HCl} = \frac {12.5 \ \mathrm{g}} {36.46 \mathrm{g/mol}} = 0.343 \ \mathrm{mol}\end{align*}



 \begin{align*}\text{molality} \ \mathrm{HCl} = \frac {\mathrm{mol \ solute}} {\mathrm{kg \ solvent}} = \frac {0.343 \ \mathrm{mol}} {0.115 \ \mathrm{kg}} = 2.98 \ \mathrm{m}\end{align*}

Here is a video of a teacher writing on an electronic blackboard. It shows how to calculate molarity, molality, and mole fraction (6d): http://www.youtube.com/watch?v=9br3XBjFszs (7:36).
Lesson Summary
 Concentration is the measure of how much of a given substance is mixed with another substance.
 Molarity is the number of moles of solute per liter of solution.
 Mass percent is the number of grams of the solute in the number of grams of solution, multiplied by 100%.
 Parts per million means that there is 1 milligram of solute for every kilogram of solvent. Therefore, it is the mass of solute per mass of solution multiplied by 1 million.
 Molality is calculated by dividing the number of moles of solute by the kilograms of solvent. It is less common than molarity but more accurate because of its lack of dependence on temperature.
Review Questions
 Calculate the mass percent of silver when a silver/nickel solution is made with \begin{align*}34.5 \ \mathrm{g}\end{align*} of silver and \begin{align*}72.3 \ \mathrm{g}\end{align*} of nickel.
 What would be the ppm of silver for the data presented in question 1?
 Why is it a good idea to learn mass percent when molarity and molality are the most commonly used concentration measures?
 Most times when news reports indicate the amount of lead or mercury found in foods, they use the concentration measures of \begin{align*}\mathrm{ppb}\end{align*} (parts per billion) or ppm (parts per million). Why use these over the others we have learned?
 What is the molarity of a solution prepared by dissolving \begin{align*}2.5 \ \mathrm{g}\end{align*} of \begin{align*}\mathrm{LiNO}_3\end{align*} in sufficient water to make \begin{align*}60. \ \mathrm{mL}\end{align*} of solution?
 \begin{align*}0.036 \ \mathrm{mol/L}\end{align*}
 \begin{align*}0.041 \ \mathrm{mol/L}\end{align*}
 \begin{align*}0.60 \ \mathrm{mol/L}\end{align*}
 \begin{align*}0.060 \ \mathrm{mol/L}\end{align*}
 A solution is known to have a concentration of \begin{align*}325 \ \mathrm{ppm}\end{align*}. What is the mass of the solute dissolved in \begin{align*}1.50 \ \mathrm{kg}\end{align*} of solvent?
 \begin{align*}0.32 \ \mathrm{mg}\end{align*}
 \begin{align*}0.49 \ \mathrm{mg}\end{align*}
 \begin{align*}325 \ \mathrm{mg}\end{align*}
 \begin{align*}488 \ \mathrm{mg}\end{align*}
 Calculate the molality of a solution of copper(II) sulfate where \begin{align*}11.25 \ \mathrm{g}\end{align*} of the crystals has been dissolved in \begin{align*}325 \ \mathrm{g}\end{align*} of water.
 \begin{align*}0.0346 \ \mathrm{m}\end{align*}
 \begin{align*}0.0705 \ \mathrm{m}\end{align*}
 \begin{align*}0.216 \ \mathrm{m}\end{align*}
 None of the above.
 What is the mass of magnesium chloride present in a \begin{align*}250 \ \mathrm{g}\end{align*} solution found to be \begin{align*}21.4\%\end{align*} \begin{align*}\mathrm{MgCl}_2\end{align*}?
 \begin{align*}21.4 \ \mathrm{g}\end{align*}
 \begin{align*}53.5 \ \mathrm{g}\end{align*}
 \begin{align*}196.5 \ \mathrm{g}\end{align*}
 \begin{align*}250 \ \mathrm{g}\end{align*}
 What is the concentration of each of the following solutions in \begin{align*}\mathrm{mol/L}\end{align*}?
 \begin{align*}3.50 \ \mathrm{g}\end{align*} of potassium chromate dissolved in \begin{align*}100 \ \mathrm{mL}\end{align*} of water
 \begin{align*}50.0 \ \mathrm{g}\end{align*} of magnesium nitrate dissolved in \begin{align*}250 \ \mathrm{mL}\end{align*} of water.
 Find the mass of aluminum nitrate required to produce \begin{align*}750 \ \mathrm{g}\end{align*} of a \begin{align*}1.5\end{align*} molal solution.
 The Dead Sea contains approximately \begin{align*}332\ \mathrm{grams}\end{align*} of salt per kilogram of seawater. Assume this salt is all \begin{align*}\mathrm{NaCl}\end{align*}. Given that the density of the Dead Sea water is approximately \begin{align*}1.20 \ \mathrm{g/mL}\end{align*}, calculate:
 the mass percent of \begin{align*}\mathrm{NaCl}\end{align*}.
 the mole fraction of \begin{align*}\mathrm{NaCl}\end{align*}.
 the molarity of \begin{align*}\mathrm{NaCl}\end{align*}.
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