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2.2 Molarity

  • Page ID
    32262
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    There are several different ways to indicate the amount of solute dissolved in a given volume of solution. The most common method used in chemistry is as the number of moles of solute dissolved in one litre of solution. This is called molarity.

    Molarity =
    moles solute

    litre solution

    Symbol for Molarity = M

    As an example: The concentration of a solution might be expressed as a "0.50 M NaOH".

    You would read as "a 0.50 molar sodium hydroxide solution", meaning that there are 0.50 moles of NaOH dissolved per one litre of solution.

    CAUTION

    Be sure to note that molarity is calculated as the total volume of the entire solution, not just volume of solvent! The solute contributes to total volume.

    Other important things to note:

    • square brackets are often used to represent concentration.
      So we could also write our example as: [NaOH] = 0.50 M.
    • use the capital letter M for molarity, not a lower case m.

    It will be important for you to be able to do calculations involving solution concentrations. Generally you will have three types of problems:

    1. calculate concentration
    2. calculate the amount of solute (as the number of moles or mass) dissolved in a given volume of solution, and
    3. calculate the volume of a solution with a certain concentration that contains a certain amount of solute.

    In addition, you will be required to convert between molarity and another common unit of concentration, parts per million (ppm), and do calculations involving dilutions. We'll cover these in the next two sections.

    Molarity is expressed in terms of moles of solute, but the amount of solute is typically measured by its mass in grams. You'll need to be sure to can convert readily between moles and mass.

    An important note - there are many different ways you can set up and solve your chemistry equations. Some students prefer to answer multi-step calculations in one long step; others prefer to work out each step individually. Neither method is necessarily better or worse than the other method - whatever makes most sense to you is the one you should use. I will typically use unit analysis (also called dimension analysis or factor analysis).

    Here are some sample calculations.

    1. Antifreeze is a solution of ethylene glycol, C2H6O2 in water. If 4.50 L of antifreeze contains 27.5 g of ethylene glycol, what is the concentration of the solution?

    Solution:

    Determine what information the question asks for, and what information we know or can readily find:

    1. The question asks for concentration,
    which means finding molarity, or:
    moles solute

    litre solution
    2.

    The question gives the mass of ethylene glycol, but concentration is expressed in terms of moles. Thus we will need to calculate the molar mass of C2H6O2 in order to convert between mass and moles.

    Using a periodic table, we find the molar mass of C2H6O2 to be 62.1 g·mol-1

    3. The other part of the unit of concentration is litres of solution. The question tells us that we have 4.5 L of solution.

    Put this information together to solve the problem, arranging the information to end up with the desired unit:

    mol

    L
    =
    27.5 g
    ×
    1 mol

    62.1 g
    ×
    1

    4.5 L
    =
    0.098 mol

    L

    Our final answer: [C2H6O2 ] = 0.098

    2. What mass of sodium carbonate, Na2CO3 is present in 50.00 mL of a 0.750 M solution?

    Solution:

    1. The question asks for mass, meaning we are asked to find (calculate) grams (the unit for mass)
    2.
    We are given the concentration. I suggest you rewrite
    the concentration as shown to the right, to better see
    how the units will cancel out.
    0.75 mol

    L

    3.

    Because the question involves mass, we will need to know the molar mass of Na2CO3

    Using a periodic table we find the molar mass of Na2CO3 to be 106.0 g·mol-1

    4. The question gives us the volume in mL. Our unit of concentration uses L, so we will convert 50.00 mL into 0.050 L.

    Put this information together to solve the problem, setting up all units to cancel out except for grams:

    g
    =
    106.0 g

    mol
    ×
    0.75 mol

    L
    ×
    0.050 L
    =
    3.98 g
    answer


    3. What volume of 1.50 mol/L HCl solution contains 10.0 g of hydrogen chloride?

    Solution:

    1. The question asks for volume, so we want to calculate litres, L (the unit for volume).
    2.
    We are given the concentration. I suggest you rewrite
    the concentration as shown to the right, to better see
    how the units will cancel out.
    1.50 mol

    L

    3.

    Because the question involves mass, we will need to know the molar mass of HCl

    Using a periodic table we find the molar mass of HCl to be 36.5 g·mol-1

    Put this information together to solve the problem, setting up all units to cancel out except for L:

    L
    =
    1 L

    1.50 mol
    ×
    1 mol

    36.5 g
    ×
    10.0 g
    =
    0.183 L or 183 mL
    answer

    Note: other symbols and formulas are commonly used to represent solution concentration, number of moles, etc. You should be familiar with these:

    c - concentration
    n - number of moles
    V - volume

    c
    =
    n

    V

    2.2 Molarity is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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