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.

%(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). 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.

'Normal Saline' is 0.9%(w/v) in water. How would you make a liter of NS? Bacteriological L-broth is (0.3% glucose, 0.7% tryptone). Since both glucose and tryptone are solids, you should assume 'w/v'.

%(v/v): Read as "percent volume per volume"

A 40% aqueous solution of ethanol is made by taking 40 ml of EtOH and adding enough water to make 100 ml. Real simple!

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.)

How would you make aqueous 0.1 M glucose? The molecular weight of glucose = 180. Therefore dissolve 18 gms of glucose up to a liter with water.

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? 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%.

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