Dilution Problem Set Solutions

(posted 3/1/99)

  1. In order to make a solution to isolate DNA from bacteria, a cell biology student must first make up a 2 M solution of NaOH.  The student needs 5 ml and knows that NaOH MW = 40 g/mole.  How many grams of NaOH will the student need?

    2. 2 moles/L  x  5ml  x  1L/10^3 ml  x  40 g/mole = 0.4 g of NaOH
     

  2. For a later step in isolating DNA, another cell biology student has to make up a 5 mM solution of NaCl.  He weighs out the very last grains of NaCl and has 10 ug.  NaCl MW = 58 g/mole.  What volume of water, in ml, does he need to make his NaCl solution?

    3. 5 mMoles/L  x  1mole/10^3 mMoles = 0.005 moles/L

    1/0.005 moles/L  x  1/58 g/mole  x  0.00001 g =  3.45 x 10^-5 L = 3.45 x 10^-2 ml
     

  3. Design a serial dilution given that you start with 5 M NaCl and a final concentration of 250 uM.

    4.  
    1. To start with, figure out how many orders of magnitude (multiples of 10) you will need to dilute by.  Since you're going from M to uM, you will need an order of 10^-6.
    2. Next, what will be good multiples to do this in. You don't want too many tubes, but you also don't want a volume so small you can't pipet it.  So let's try a serial dilution in 3 tubes.
    3. Now, if you have three tubes with a TOTAL dilution of 10^-6, then each tube should be a dilution of 10^-2.
    4. Okay, now draw your diagram for clarification.  We're half way there.
      5. The way I have it now, our final concentration is 5 uM.  We want 250 uM, so this way is too much.
    5. So we diluted too much. Let's decrease our first two dilutions to 1/10 and see what happens.
    6. We are very close.  All we need now is a 1:2 dilution.  This is the final serial dilution.
      7. There are other ways to do this.  As long as you come up with 250 uM in the end and the volumes are something you can pipet in lab, your scheme is a success.
     
  4. What is the initial concentration in Mole/L?
    5.  First, determine the dilution in each tube:
      1/100
      2/10 = 1/5
      1/10
       
    Now determine the TOTAL dilution:
      1/100 x 1/5 x 1/10 = 1/5000
      The dilution factor (DF) = 5000
      so if Ci = DF x Cf
       Ci = 5000 x 5 uM
      Ci = 25000 uM = 0.025 M
       
     
  5. A biology grad student has a 1 M solution of a chemical called DTT.  She needs to do a dilution and comes to a cell biology student for help.  She wants you tell her what volume of 1 M DTT to use in order to have 100 ul of 100 mM DTT.  What do you tell her?  Feeling noble, you also tell her how much diluent (water) to use (the volume in ul).

    6. First, determine what order of magnitude the dilution will be.  Here we're going from 1 M to 100 mM = 1 x 10^-3 M.  If the difference is less than (or equal to) 2 orders of magnitude, then use the equation

    C1V1 = C2V2
    C1 = 1 M
    C2 = 100 mM = 0.1 M
    V1 = ?
    V2 = 100 ul
    Now, V1 = C2V2/C1
    V1= (0.1 x 100)/1
    V1 = 10 ul of 1 M DTT
    Which means, 100 - 10 = 90 ul of water
     
  6. What is the dilution if you put 250 ul of a solution into a total volume of 5 ml?
    7. First, get all the volumes in the same units.
    So, 5 ml = 5000 ul or 250 ul = 0.25 ml.
    Now:  250 ul/5000ul
    = 1/20
    That's the dilution. It's a straightforward answer, and you DO NOT need to know concentrations.