Making Solutions
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LIQUID SOLUTIONS

Throughout your scientific endeavors you will need to make a number of solutions of solids dissolved in liquids, and of liquids dissolved in liquids. It is obviously necessary to know how to do this reliably. (You'll look like an ignorant klutz in a lab job if you don't know how to do this.) We shall discuss three types of solutions: %(w/v), %(v/v), and molar.

  1. %(w/v): Read as "percent weight per volume"

    This type of solution is extremely common in biological sciences - including medicine. In order to understand the basic principle, you must realize that the word 'percent' is derived from the Latin words 'per' (for, to, in) and 'centum' (one hundred). Examples:

    1. Make 100 mL of 1% NaCl (mol.wt.=58). Ans.: percent solutions don't use the molecular weight, so disregard that number. You need only gms of NaCl per the number of hundreds of milliliters needed. Thus a 1% aqueous solution of NaCl in water is made by taking 1 gram of NaCl and dissolving it in water up to 100 ml.
    2. Make 300 mL of 3% sucrose (table sugar) (mol.wt. = 342). You need 3 grams of sucrose for each hundred mL, and you need three of those hundred mL. Thus you need 9 grams of sucrose and dissolve it in about 200 mL water and then add more water up to 300 mL.

    3. Harder: 'Normal Saline' is 0.9%(w/v) in water. How would you make a liter of NS? Ans: since a liter = ten hundreds, you will need 10 x 0.9 grams = 9.0 grams of NaCl. Thus dissolve 9 grams of NaCl in about 950 mL water and then add more water up to 1 liter (1,000 mL).
    4. More complicated: You want to make 300 mL of L-broth, knowing that Bacteriological L-broth is (0.3% glucose, 0.7% tryptone). Since both glucose and tryptone are solids, you should assume 'w/v'. Ans.: your are making three hundreds, and thus 3 x 0.3 gms of glucose and 3 x 0.7 gms of tryptone (0.9 gms of glucose + 2.1 gms of tryptone); make up to 300 mL with water.

  2. %(v/v): Read as "percent volume per volume"
    1. You need 100 mL of 40% ethanol. Ans.:

      A 40% aqueous solution of ethanol is made by taking 40 ml of EtOH and adding enough water to make 100 ml. Really simple do-it-in-your-head stuff!

    2. Make 200 mL of 5% ethanol in water. Ans.: you have two hundreds; thus 2 x 5 mL = 10 mL of ethanol; add enough water to bring the volume up to 200 mL.

  3. Molar*** Solutions require more sophisticated knowledge: you will need to know the molecular weight of the compound first. A 1M aqueous solution of NaCl will require about 58 grams of NaCl to be dissolved in water up to 1000 mls. Where did the 58 come from? Look at your local reagent bottle of solid NaCl. It will tell you the formula weight ('F.W.) is 58. (More esoterically, FW (or Molec. wt.) is the number of grams that one Avagadro's number of molecules of that substance weighs.)

    1. Make a liter of 1 M NaCl (mol.wt.=58). Ans.: simply take 58 grams of NaCl and dissolve it in water and make up to 1,000 mL with water.
    2. How would you make 500 mL of aqueous 0.1 M glucose (mol.wt. 180)?
      (The following steps are devised to help you see how this can be done in your head without any writing at all.)
      1. IF you took 180 gms of glucose and made it up to a liter, you would have 1M.
      2. BUT you need 0.1M. So you take a tenth of 180 = 18 grams.
      3. But you don't need a liter: you only need 500 mL, which is a half liter. Thus half of 18 = 9 grams of glucose.
      4. Thus dissolve 9 gms of glucose and make it up to 500 mL


  1. Now comes a different twist: suppose that you have a solution of something that is too concentrated. How do you dilute it to the concentration that you need?

    1. In head: Suppose you have some 20% ethanol in a bottle, and you need 100 mL of 10% ethanol. Ans.: You need to cut the ethanol's concentration in half, and you do that by adding an equal amount of water. Thus were you to take 50mL of 20% ethanol, and add to it 50mL of water, you should get 100 mL of 10% ethanol.

    2. Need paper: Suppose you have 100 ml of a 5% solution of glucose, and you need 25 ml of a 2% solution. Use an equation that your instructor finds difficult to forget:

      C1V1 = C2V2
      (concentrations times volumes)
      2% x 25 ml = 5% x (Volume of the 5%)
      50 = 5 x (Volume of the 5%) = 10 ml of the 5% solution is needed

      So, take that 10 of 5%, and add 15 ml of water to give 25 ml of 2%.


*** What is a "mole?" You are familiar with other terms for grouping things - pairs (2), dozens (12), and gross (144) are examples of clusters of a rather small number of items. For larger numbers, we have Astronomical Units (distance between the sun and earth), speed of light, parsecs in astronomy, and in chemistry the "mole" which is 602,000,000,000,000,000,000,000 (six hundred-two sextillion) which is defined as the number of gas molecules in 22,414 mL at standard temperature and pressure (STP) - regardless of the type of gas. What is neat about this 6.02x1023, is that if you take the molecular weight of something, and weigh out that many grams of it, you have 1 mole of it - that number of molecules. Thus if water has a molecular weight of 18, then eighteen grams of water has that many molecules of H2O, which is 1 mole of water. "Molar" is one mole of molecules diluted into a liter by some sort of solvent - usually water. Thus 1 M NaCl is one mole of NaCl (mol.wt. = 57) or 57 grams of NaCl made up to 1 liter of water (not added to a liter, but made up to a total of a liter).


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