SOLUTIONS
PRACTICE PROBLEMS

  1. The first type of solution problem concerns making DILUTIONS. Here is an example of a moderate dilution. Suppose that you have a 15% solution of glucose, and you want to add just enough to a petri plate to make a 0.3% solution in the agar. First, you estimate that a petri plate will contain 30 ml of final contents.

    C1V1 = C2V2 : 15% x V = 0.3% x 30; V =

  2. Here is an example of a very high dilution: A new antibiotic has been described. 100 µl of it is shipped to you as a 5% solution. From previous experience you know that you must dilute it to 0.000001% to run the 1 ml test you want to do. Were you to use the C1V1 = C2V2 formula, you would get:

    5% x V = 0.000001% x 1; and V = 0.0000002 ml = 0.0002 µl

    This is far below the precision of any micropipetter currently in existence. You have two choices: (A) add the 100 µl into some huge volume, or (B) use some of the 100 µl in some sort of serial dilution. Since the shipped sample is likely to be very expensive, you don't want to squander it all in one run. Suppose you used 10 µl (a volume you know you can rather reliably measure using "micropipetters"), and diluted this into 990 µl for a 100-fold dilution. You could do this several times in a series, and quickly get down to the range of concentration you need.

    Once: 0.05%; Twice: 0.0005%; Trice: 0.000005%; now what? (see #1)

    Why are µl so commonly used in laboratories today?

  3. Another type of solution problem concerns making up "Percent Solutions". If a liquid solute is involved, it is usually "v/v"; and if a solid solute is involved, it is usually "w/v". You want to make up a 70% ethanol solution for sterilizing gauze pads. If you have 100% ethanol ("absolute EtOH"), it comes out 70 ml EtOH + 30 ml water. What would it be if you had 95% ethanol? (Think: C1V1 = C2V2 )

  4. You now want to make up 0.7% NaCl: Add 0.7 gm NaCl per a total of 100 ml (weigh out 700 mg of NaCl, dissolve it in roughly 80 ml of water, and then make the volume up to 100 ml.

  5. But often solids have waters of hydration, and since these are somewhat variable, they cause confusion - UNLESS a refractometer is handy. Suppose you needed 100 ml of a 0.1% solution of CaCl2, a very hygroscopic compound. First make a small amount of a solution that is perhaps somewhere in the range of 20%: Weigh out 2 gms and add it to 10 ml of water: it dissolves quickly. Using the refractometer, you find that you have a 14% solution.

    C1V1 = C2V2 : 14% x V = 0.1% x 100; V = ?

  6. Oh, oh: and then come MOLAR solutions! Suppose that you need 100 ml of 0.1 M NaCl. Get the bottle of crystalline NaCl. It will probably have the 'formula weight" (FW) on the bottle: 58.43 (daltons).

    You would need 0.1 x 58.43 for 1000 mls, or 0.1 x 58.43 x 0.1 for 100 ml.
    = 584 mg.

  7. But suppose that you had a 3.31M NaCl solution sitting on the shelf, and that it had been precisely measured as such with a refractometer. Wouldn't it be easier to make a dilution rather than a weighing?

    C1V1 = C2V2 : 3.31M x V = 0.1M x 100 ml; V = ?

  8. Usually in biology, we don't often run into differences in molarity and normality.

  9. What about interconversion of M to %? 0.1 M NaCl = ?%
    0.1 x 58.43 = 5.843 g/liter = 0.5843 g/100 ml = 0.5843%

  10. And the other way? 3% glucose = ? M. Glucose is 180 daltons.
    3 gm/100 = 30 gm/liter; 30/180 -> 1/6 molar.

| Back to Lab Syllabus | Back to Lab 1 | Back to Solution Directions | On to Quiz #1 |