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14.6: Molarity

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    ⚙️ Learning Objectives

    • Use molarity as a conversion factor in calculations.

    Of all the quantitative measures of solution concentration, molarity is the one used most frequently by chemists. Molarity is defined as the number of moles of solute per liter of solution.

    \[\mathrm{molarity=\dfrac{moles\: of\: solute}{Liters\: of\: solution}}\]

    This means that molarity may be expressed in units of mol/L, though the abbreviation M (must be capitalized) is used more often to abbreviate molarity. Equation \(\PageIndex{1}\) may be written more simply as:

    \[\mathrm M=\dfrac{\mathrm{mol}}{\mathrm L}\]

    Perhaps you have used a solution of 0.50 M HCl in lab. This concentration is read as "0.50 molar HCl" and means that there is 0.50 mol HCl dissolved for every one liter of solution. This concentration may also be written as 0.50 mol/L HCl. It is important to remember that “mol” in this expression refers to moles of solute and that “L” refers to liters of solution. In some instances, the molarity of a solution is instead called its molar concentration.

    Units of molarity are useful for discussing chemical reactions in which a solute is a product or reactant. Molar masses may then be used as conversion factors to convert amounts in moles to amounts in grams. Sometimes chemists use square brackets to indicate a reference to the molarity of a substance. For example, the expression \(\left[ \ce{Ag^+} \right]\) refers to the molarity of the silver ion in solution.

    ⛓ Kool-Aid

    It turns out that making a solution of known molarity is similar to the preparation of Kool-Aid (see Figure \(\PageIndex{1}\) below) in that the total amount of solution is specified rather than the amount of solvent. Note that the front side of the Kool-Aid package reads, "Makes 2 Quarts," while the back side instructs one to add all of the other ingredients first before adding water to make a total of two quarts of Kool-Aid. If instead, two quarts of water had been added to the other ingredients, more than two quarts of Kool-Aid would be prepared, and the drink would taste watered down. The procedure for making Kool-Aid and its resemblance to preparing solutions of known molarity lead many students to affectionately call it the "Kool-Aid Rule".

    Figure \(\PageIndex{1}\): The front and back sides of a package of Kool Aid, circa 1960s. (adapted from Roadsidepictures via Flickr)


    ⚡️ Important Notes About Molarity

    • The capital letter M is used for molarity, not a lower case m which is for a different concentration unit called molality.
    • Molarity is calculated from the volume of solution, not the volume of solvent! The solute contributes to the total volume of the solution.
    • The molarity of a solution is the same as its molar concentration.
    • Square brackets are often used to represent molar concentration. Example: [NH3] = 0.50 M.

    Before getting into more involved examples, let's look at the preparation of 1.00 L of a 1.00 M NaHCO3 solution. A 1.00 M NaHCO3 solution contains 1 mol NaHCO3, or 84.0 g NaHCO3 for every 1 L of solution (if necessary, review molar mass calculations). To make 1.00 L of this solution, a device that measures volume precisely will be needed. One option is to use a large graduated cylinder capable of measuring 1000 mL. However, there is a specific piece of equipment designed for this purpose that measures one volume very precisely called a volumetric flask. Volumetric flasks come in various sizes, so a 1000-mL volumetric flask would be selected to prepare 1.00 L of solution.

    Figure \(\PageIndex{2}\): An assortment of volumetric flasks. (Nuno Nogueira via Wikimedia Commons)

    Next, 84.0 g NaHCO3 would be weighed on a scale and added to the volumetric flask. Since solid residue could remain on the measuring device, it should be rinsed with water and added to the volumetric flask. Water is then added to the flask until it is ½ to ¾ full. The contents of the flask are then mixed thoroughly until all of the solid dissolves. Recall that volumes are not always additive due to particle size and intermolecular forces, so make sure that the solute completely dissolves before adding water to the 1000-mL etched mark on the neck of the flask. Finally, water would be added to the 1000-mL mark on the neck and the solution would be thoroughly mixed.

    Video \(\PageIndex{1}\) below shows how to make 1.00 L of a 1.00 M NaCl solution. Can you tell an important step that was skipped near the end of the video?

    Video \(\PageIndex{1}\): Making a 1.00 M NaCl solution. One important step was skipped at the very end. What is it? (CAYERCHEM via YouTube)

    You would be correct if you said they forgot to mix the solution thoroughly after the last portion of water was added to the flask. It is important that solutions are mixed thoroughly to obtain a uniform concentration of solute throughout the solution.

    ✅ Example \(\PageIndex{1}\)

    What is the molarity of 2.50 L of solution that contains 62.6 g Mg(NO3)2?


    Steps for Problem Solving  
    Identify the "given" information and what the problem is asking you to "find."
    Given: 62.6 g Mg(NO3)2; 2.50 L solution

    Find: Molarity = ? M
    List known relationship(s).
    1 mol Mg(NO3)2 = 148.33 g Mg(NO3)2

    \(\mathrm M=\dfrac{\mathrm{mol}}{\mathrm L}\)
    Prepare a concept map and use the proper conversion factor(s).
    Convert mass Mg(NO3)2 to moles:

      \({\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\mathrm g\;\mathrm{Mg}{({\mathrm{NO}}_3)}_2\;}}\;\xrightarrow[{148.33\;\mathrm g\;\mathrm{Mg}{({\mathrm{NO}}_3)}_2}]{1\;\mathrm{mol}\;\mathrm{Mg}{({\mathrm{NO}}_3)}_2}\;{\color[rgb]{0.0, 0.0, 1.0}\boxed{\;\mathrm{mol}\;\mathrm{Mg}{({\mathrm{NO}}_3)}_2\;}}\)

    Molarity is calculated by dividing moles by liters:

      \(\mathrm M=\dfrac{\mathrm{mol}\;\mathrm{Mg}{({\mathrm{NO}}_3)}_2}{2.50\;\mathrm L}\)
    Calculate the answer.
    Now substitute the known quantities and solve.

      \(62.6\;\cancel{\mathrm g\;\mathrm{Mg}({\mathrm{NO}}_3)_2}\times\dfrac{1\;\mathrm{mol}\;\mathrm{Mg}({\mathrm{NO}}_3)_2}{148.33\;\cancel{\mathrm g\;\mathrm{Mg}({\mathrm{NO}}_3)_2}}=0.42\underline20\;\mathrm g\;\mathrm{Mg}({\mathrm{NO}}_3)_2\)

      \(\mathrm M=\dfrac{0.42\underline20\;\mathrm{mol}\;\mathrm{Mg}{({\mathrm{NO}}_3)}_2}{2.50\;\mathrm L}=0.169\;{\textstyle\frac{\mathrm{mol}\;\mathrm{Mg}{({\mathrm{NO}}_3)}_2}{\mathrm L}}=\boxed{0.169\;\mathrm M\;\mathrm{Mg}{({\mathrm{NO}}_3)}_2}\)
    Think about your result.
    The molarity is 0.169 M, meaning that one liter of the solution would contain 0.169 mol Mg(NO3)2. Three significant figures are appropriate.


    ✏️ Exercise \(\PageIndex{1}\)

    1. Calculate the molar concentration of 0.500 L of solution in which 137 g NaCl are dissolved.
    2. What is the molarity of a solution that contains 66.2 g C6H12O6 dissolved in 235 mL of solution?
    Answer A

    4.69 M NaCl

    Answer B

    1.56 M C6H12O6

    Using Molarity as a Conversion Factor

    Molarity may be used as a conversion factor to convert between moles of solute and liters of solution. As such, molarity can be useful in a variety of stoichiometry problems.  For example, suppose we would like to know how many moles of NaCl are present in 108 mL of 0.887 M NaCl. Since molarity is defined as the moles of solute per liter of solution, this means that there are 0.887 mol NaCl for every 1 L of solution. However, the volume of solution is given in mL. Therefore, mL must first be converted to L or L must be converted to mL. As we recall, 1 L = 1000 mL, so 0.887 M NaCl may be expressed as a conversion factor in the form:

    \(\dfrac{0.887\;\mathrm{mol}\;\mathrm{NaCl}}{1\;\mathrm L\;\mathrm{solution}}=\dfrac{0.887\;\mathrm{mol}\;\mathrm{NaCl}}{1000\;\mathrm{mL}\;\mathrm{solution}}\)  or  \(\dfrac{1\;\mathrm L\;\mathrm{solution}}{0.887\;\mathrm{mol}\;\mathrm{NaCl}}=\dfrac{1000\;\mathrm{mL}\;\mathrm{solution}}{0.887\;\mathrm{mol}\;\mathrm{NaCl}}\)

    The conversion that enables cancellation of units would then be used in the solution map that takes us from mL of solution to mol NaCl. With the solution map in place, the moles of NaCl are easily calculated.

    \({\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\mathrm{mL}\;\mathrm{solution}\;}}\;\xrightarrow[{1000\;\mathrm{mL}\;\mathrm{solution}}]{0.887\;\mathrm{mol}\;\mathrm{NaCl}}\;{\color[rgb]{0.0, 0.0, 1.0}\boxed{\;\;\mathrm{mol}\;\mathrm{NaCl}\;\;}}\)



    Had the equation approach been used instead, the same answer would be obtained. Since:

    \(\mathrm M=\dfrac{\mathrm{mol}}{\mathrm L}\)

    this equation could be rearranged to solve for moles by multiplying each side by liters of solution. The result of this rearrangement is:

    \[\mathrm{mol}\;=\;\mathrm M\;\times\;\mathrm L\]

    Note that using the equations requires that volumes be expressed in liters, so one would first convert 108 mL to 0.108 L, then plug the known quantities into the equation:

    \(\mathrm{mol}\;=\;0.887\;\mathrm M\;\mathrm{NaCl}\;\times\;0.108\;\mathrm L\;=\;0.887\;\dfrac{\mathrm{mol}\;\mathrm{NaCl}}{\cancel{\mathrm L}}\;\times\;0.108\;{\cancel{\mathrm L}}\;=\;0.0958\;\mathrm{mol}\;\mathrm{NaCl}\)

    Whether using the conversion factor method or the equation method, it is still necessary to remember that M stands for mol/L, so the proper cancellation of units may be performed. However, once again, we can see that the same answer is obtained using either method.

    What if we were looking for volume and wanted to know how many liters of 2.35 M CuSO4 were needed to obtain 4.88 mol of CuSO4? A molarity of 2.35 M CuSO4 means that there are 2.35 mol CuSO4 for every 1 L of solution. As such, 2.35 M CuSO4 may be expressed as a conversion factor in the form:

    \(\dfrac{2.35\;\mathrm{mol}\;{\mathrm{CuSO}}_4}{1\;\mathrm L\;\mathrm{solution}}\) or \(\dfrac{1\;\mathrm L\;\mathrm{solution}}{2.35\;\mathrm{mol}\;{\mathrm{CuSO}}_4}\)

    The conversion that enables cancellation of units would then be used in the solution map that takes us from mol CuSO4 to L solution. With the solution map in place, the liters of solution are easily calculated.

    \({\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\mathrm{mol}\;{\mathrm{CuSO}}_4\;}}\;\xrightarrow[{2.35\;\mathrm{mol}\;{\mathrm{CuSO}}_4}]{1\;\mathrm L\;\mathrm{solution}}\;{\color[rgb]{0.0, 0.0, 1.0}\boxed{\;\;\mathrm L\;\mathrm{solution}\;\;}}\)

    \(4.88\;\cancel{\mathrm{mol}\,\;{\mathrm{CuSO}}_4}\times\dfrac{1\;\,\mathrm L\;\mathrm{solution}}{2.35\;\cancel{\mathrm{mol}\,\;{\mathrm{CuSO}}_4}}=2.08\;\,\mathrm L\;\mathrm{solution}\)

    ✅ Example \(\PageIndex{2}\)

    A chemist needs to prepare 250.0 mL of a 0.150 M solution of potassium permanganate, KMnO4.

    1. What mass of KMnO4 does she need to make the solution?
    2. Describe how this solution would be prepared.


    Steps for Problem Solving  
    Identify the "given" information and what the problem is asking you to "find."
    Given: 0.150 M KMnO4; 250.0 mL solution

    Find: Mass KMnO4 = ? g KMnO4

    List known relationship(s).

    1 mol KMnO4 = 158.04 g KMnO4

    0.150 mol KMnO4 for every 1 L (1000 mL) solution
    Prepare a concept map and use the proper conversion factor(s).
    \({\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\mathrm{mL}\;\mathrm{solution}\;}}\xrightarrow[{1000\;\mathrm{mL}\:\mathrm{solution}}]{0.150\;\mathrm{mol}\;{\mathrm{KMnO}}_4}{\color[rgb]{0.0, 0.0, 1.0}\boxed{\mathrm{mol}\;{\mathrm{KMnO}}_4}}\xrightarrow[{1\;\mathrm{mol}\;{\mathrm{KMnO}}_4}]{158.04\;\mathrm g\;{\mathrm{KMnO}}_4}{\color[rgb]{0.0, 0.5, 0.0}\boxed{\;\mathrm g\;{\mathrm{KMnO}}_4\;}}\)
    Calculate the answer.
    Now substitute the known quantities and solve.

    \(250.0\;\cancel{\mathrm{mL}\;\mathrm{solution}}\times\dfrac{0.150\;\cancel{\mathrm{mol}\;{\mathrm{KMnO}}_4}}{1000\;\cancel{\mathrm{mL}\;\mathrm{solution}}}\times\dfrac{158.04\;\mathrm g\;{\mathrm{KMnO}}_4}{1\;\cancel{\mathrm{mol}\;{\mathrm{KMnO}}_4}}=\boxed{5.93\;\mathrm g\;{\mathrm{KMnO}}_4}\)

    Think about your result.
    When 5.93 g KMnO4 are used to make 250.0 mL of solution, the molarity is 0.150 M.
    1. To prepare the solution, weigh out 5.93 g KMnO4 solid on a scale. Place the solid in a 250-mL volumetric flask. Add water to flask and mix to dissolve the solid. Add water until the bottom of the meniscus is on the line on the neck. Mix thoroughly.


    ✅ Example \(\PageIndex{3}\)

    What volume of a 0.200 M NaOH solution would contain 16.0 g NaOH?


    Steps for Problem Solving  
    Identify the "given" information and what the problem is asking you to "find."
    Given: 0.200 M NaOH; 16.0 g NaOH

    Find: Volume of Solution = ? L

    List known relationship(s).

    1 mol NaOH = 40.00 g NaOH

    0.200 mol NaOH for every 1 L solution
    Prepare a concept map and use the proper conversion factor(s).
    \({\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\;\mathrm g\;\mathrm{NaOH}\;\;}}\xrightarrow[{40.00\;\mathrm g\;\mathrm{NaOH}}]{1\;\mathrm{mol}\;\mathrm{NaOH}}{\color[rgb]{0.0, 0.0, 1.0}\boxed{\;\mathrm{mol}\;\mathrm{NaOH}\;}}\xrightarrow[{0.200\;\mathrm{mol}\;\mathrm{NaOH}}]{1\;\mathrm L\:\mathrm{solution}}{\color[rgb]{0.0, 0.5, 0.0}\boxed{\;\mathrm L\;\mathrm{solution}\;}}\)
    Calculate the answer.
    Now substitute the known quantities and solve.

    \(16.0\;\cancel{\mathrm g\;\mathrm{NaOH}}\times\dfrac{1\;\cancel{\mathrm{mol}\;\mathrm{NaOH}}}{40.00\;\cancel{\mathrm g\;\mathrm{NaOH}}}\times\dfrac{1\;\mathrm L\;\mathrm{solution}}{0.200\;\cancel{\mathrm{mol}\;\mathrm{NaOH}}}=\boxed{2.00\;\mathrm L\;\mathrm{solution}}\)

    Think about your result.
    When 16.0 g NaOH are used to make 2.00 L of solution, the molarity is 0.200 M.


    ✏️ Exercise \(\PageIndex{3}\)

    1. What mass of solute is present in 1.08 L of 0.0578 M H2SO4?
    2. What is the volume of 0.0444 M CH2O solution that contains 0.0773 mol CH2O?
    3. What volume of 1.50 M HCl solution, in mL, contains 10.0 g of hydrogen chloride?
    Answer A

    6.12 g H2SO4

    Answer B

    1.74 L

    Answer C

    183 mL or 0.183L



    This page is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Peggy Lawson (Oxbow Prairie Heights School), Funded by Saskatchewan Educational Technology Consortium; Lance S. Lund (Anoka-Ramsey Community College), Marisa Alviar-Agnew, and Henry Agnew.

    14.6: Molarity is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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