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15.3: Solution Concentration - Molarity

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  • Learning Objective

    • Learn to determine specific concentrations with several common units.

    Rather than qualitative terms (Section 11.2 - Definitions) we need quantitative ways to express the amount of solute in a solution; that is, we need specific units of concentration. In this section, we will introduce several common and useful units of concentration.

    Molarity (M) is defined as the number of moles of solute divided by the number of liters of solution:

    \[molarity \: =\: \frac{moles\: of\: solute}{liters\: of\: solution}\]

    which can be simplified as

    \[M\: =\: \frac{mol}{L},\; or\; mol/L\]

    As with any mathematical equation, if you know any two quantities, you can calculate the third, unknown, quantity.

    For example, suppose you have 0.500 L of solution that has 0.24 mol of NaOH dissolved in it. The concentration of the solution can be calculated as follows:

    \[molarity \: =\: \frac{0.24\: mol\: NaOH}{0.500\:L}=0.48\, M\; NaOH\]

    The concentration of the solution is 0.48 M, which is spoken as “zero point forty-eight molarity” or “zero point forty-eight molar.” If the quantity of the solute is given in mass units, you must convert mass units to mole units before using the definition of molarity to calculate concentration. For example, what is the molar concentration of a solution of 22.4 g of HCl dissolved in 1.56 L? First, convert the mass of solute to moles using the molar mass of HCl (36.5 g/mol):

    \[22.4\cancel{g\:HCl}\times \frac{1\: mol\: HCl}{36.5\cancel{g\:HCl}}=0.614\, M\; HCl\]

    Now we can use the definition of molarity to determine a concentration:

    \[M \: =\: \frac{0.614\: mol\: HCl}{1.56\:L}=0.394\, M\]

    Example \(\PageIndex{1}\):

    What is the molarity of a solution made when 32.7 g of NaOH are dissolved to make 445 mL of solution?


    To use the definition of molarity, both quantities must be converted to the proper units. First, convert the volume units from milliliters to liters:

    \[445\cancel{mL}\times \frac{1\: L}{1000\: \cancel{mL}}=0.445\, L\]

    Now we convert the amount of solute to moles, using the molar mass of NaOH, which is 40.0 g/mol:

    \[32.7\cancel{g\:NaOH}\times \frac{1\: mol\: NaOH}{40.0\cancel{g\:NaOH}}=0.818\, mol\: NaOH\]

    Now we can use the definition of molarity to determine the molar concentration:

    \[M \: =\: \frac{0.818\: mol\: NaOH}{0.445\:L}=1.84\, M\: NaOH\]

    Exercise \(\PageIndex{1}\)


    What is the molarity of a solution made when 66.2 g of C6H12O6 are dissolved to make 235 mL of solution?


    1.57 M

    The definition of molarity can be used to determine the amount of solute or the volume of solution, if the other information is given. Example 4 illustrates this situation.

    Example \(\PageIndex{1}\):

    How many moles of solute are present in 0.108 L of a 0.887 M NaCl solution?


    We know the volume and the molarity; we can use the definition of molarity to mathematically solve for the amount in moles. Substituting the quantities into the definition of molarity:

    \[0.887\, M \: =\: \frac{mol\: NaCl}{0.108\:L}\]

    We multiply the 0.108 L over to the other side of the equation and multiply the units together; “molarity × liters” equals moles, according to the definition of molarity. So

    mol NaCl = (0.887 M)(0.108 L) = 0.0958 mol

    Exercise \(\PageIndex{1}\)


    How many moles of solute are present in 225 mL of a 1.44 M CaCl2 solution?


    0.324 mol

    If you need to determine volume, remember the rule that the unknown quantity must be by itself and in the numerator to determine the correct answer. Thus rearrangement of the definition of molarity is required.

    Example \(\PageIndex{1}\):

    What volume of a 2.33 M NaNO3 solution is needed to obtain 0.222 mol of solute?


    Using the definition of molarity, we have

    \[2.33\, M \: =\: \frac{0.222\:mol}{L}\]

    To solve for the number of liters, we bring the 2.33 M over to the right into the denominator, and the number of liters over to the left in the numerator. We now have

    \[L \: =\: \frac{0.222\:mol}{2.33\, M}\]

    Dividing, the volume is 0.0953 L = 95.3 mL.

    Exercise \(\PageIndex{1}\)


    What volume of a 0.570 M K2SO4 solution is needed to obtain 0.872 mol of solute?


    1.53 L

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