Beer’s Law Data
Practice Using Pipets
and Spectrophotometers
Please fill in the following data, then submit the form.
The error-checking on this Web page
includes checking to see if each absorbance reading you enter has the correct number
of decimal places. If you do not correctly enter a number, a small dialogue
box will pop up asking you to correctly enter the number. In that case,
enter the correct number in the space in the dialogue box, and it will be
transferred to the appropriate space on the Web form.
Copyright © 1998 by J. Stein Carter. All rights reserved.
Chickadee photograph Copyright © by David B. Fankhauser
This page has been accessed 
times since 14 Mar 2001.
Answers to Dilution Practice Problems
The answers to many of these problems can be derived through the
use of the formula:
V1 × C1 = V2 × C2
where V = volume and C = concentration.
1. Notice that here, you are not told an actual initial or
final concentration, but you are told that the final concentration is 1/10
of the initial concentration. The “10 mL” is the final volume. You are asked
to find the initial (“sample”) volume, and from that, to figure out how much
diluent to add.
| y mL × x conc = 10 mL × | x | conc |
| 10 |
| y = 1 mL aliquot |
| 10 – 1 = 9 mL diluent |
2. Notice that here, you are being asked to find the dilution
factor and the concentration factor. Thus, you must plug the numbers into
those two formulae, as given above. One thing that’s important to realize
is that the initial (aliquot) volume is 0.2 mL, but the final volume is the
TOTAL of the aliquot + diluent volumes, so 0.2 + 3.8 = 4.0 mL.
| DF = | Σ | = | 3.8 + 0.2 | = | 4 | = 20× as dilute |
| aliquot | 0.2 | 0.2 |
| CF = | aliquot | = | 1 | = 0.05 × as conc |
| Σ | 20 |
3. Here, again, you are not given an exact initial or final
concentration, but you are told that the final concentration is 1/100 of the
initial concentration. The “5 mL” is the final volume, and you are asked to
find the initial volume and the diluent volume.
| y mL × x conc = 5 mL × | x | conc |
| 100 |
| y = | 5 | = 0.05 mL aliquot |
| 100 |
| 5.00 – 0.05 = 4.95 mL diluent |
4. Here, you are given the initial (aliquot) volume (the
“0.6 ml”). Again, you are not given initial or final concentrations, but you
are told that the final concentration is 1/50 of the initial. The “trick”
to what it’s asking you to find is to realize that, even though it’s worded
such that it appears to be asking for the diluent volume first, in reality,
you must calculate the final, total volume, first, then from that, figure
out the diluent volume.
| 0.6 mL × x conc = y mL × | x | conc |
| 50 |
| 50 × 0.6 = y = 30 mL total |
| 30 – 0.6 = 29.4 mL diluent |
5. Again, you are not given initial or final concentrations,
but are told that the final is 1/400 of the initial. The wording that you
are to “prepare 20 mL” is a clue that the “20 mL” is the final volume. The
“How would you” is a clue that you are being asked to calculate the
aliquot and diluent volumes.
| y mL × x conc = 20 mL × | x | conc |
| 400 |
| y = | 20 | = 0.05 mL aliquot |
| 400 |
| 20.00 – 0.05 = 19.95 mL diluent |
6. Here, once again, the request is to calculate the dilution
factor, so that’s the formula you would need to use. Again, keep in mind
that the final volume is the TOTAL of the aliquot and diluent volumes.
| DF = | Σ | = | 2 + 8 | = | 10 | = 5× as dilute |
| aliquot | 2 | 2 |
7. Here, for the first time, you are given actual initial and
final concentrations (the “4 M” and “0.1 M”). The words, “You want” should
be a clue that the 1 L is the final volume. While you could do these
calculations all in liters, and end up with fractional liter final answers,
I think it’s a bit easier to convert the 1 L to 1000 mL, and do all the
calculations in milliliters.
| y mL × x 4 M = 1000 mL × 0.1 M |
| y = | 100 | = 25 mL NaCl soln |
| 4 |
| 1000 – 25 = 975 mL diluent |
8. The big thing to remember here is that the gas tank holds
12 gal, as stated, NOT 12 gal + 1 pt. When you add the STP, first, then fill
up (q. s.) the tank with gasoline, you are NOT adding 12 gal of gas,
but only 12 gal – 1 pt. Other than that, you need to know how to convert from
gallons to pints, so you know how much actual gasoline was added.
| 12 gal × | 4 qt | × | 2 pt | = 96 pt tank |
| gal | qt |
| DF = | Σ | = | 96 | = | 10 | = 96× as dilute |
| aliquot | 2 | 1 |
9. The way I solved this was to, first, use a proportion to
figure out, if there were 45 col in 0.2 mL, what’s that per milliliter? Then,
if the original culture was diluted by a factor of 106, the easiest
thing to do next is just multiply by the 106.
| 45 col | = | 222 col |
| 0.2 mL | mL |
| 225 col | × 106 = 2.25 × 108 col/mL in the original culture |
| mL |
10. You need to realize that the water is the diluent and the
glycerine is the aliquot. This question is asked “backwards” in that you are
given the diluent volume and asked to find both the initial and final volumes
of the glycerine solution from that. Thus, if the initial volume of
glycerine is y mL, the final volume of the solution will be that y mL added
to the 100 mL of water, so 100 + y mL, total.
The other piece of information you’re given is that the final concentration
is 2% (2/100) of the initial concentration. Thus, if the initial concentration
is “x,” then the final concentration is 2/100 × x.
| y mL × x conc = (100.00 + y) mL × | 2.00 | conc |
| 100.00 |
| y = (100.00 + y) × | 2.00 |
| 100.00 |
| 100.00 × y = 200.00 + 2.00 × y |
| 98.00 × y = 200.00 |
11. The big thing, here, is knowing how to convert from
gallons to teaspoons — sure shows how much easier metric system is to use
than English system! (The English people, of course, use the metric system,
and it’s only we here in the USA who don’t.)
| 1 gal × | 4 qt | × | 2 pt | × | 2 C | × | 16 T | × | 3 tsp | = 768 tsp |
| 1 gal | 1 qt | 1 pt | 1 C | 1 T |
| DF = | 768 tsp | = 768× as dilute |
| 1 tsp |