There are three parts to this question, which require increasing amounts of thought. However, all are based on the standard mathematics of dilutions.

You are provided with a large quantity of a solution of a chemical at a standard concentration and a large supply of water with which to dilute the solution. You have three 100ml beakers in which to mix dilutions and a 10ml syringe to transfer fluids from the large beakers to the small beakers or between the small beakers. The 10ml syringe can only measure exactly 10mls.

Part 1:

For certain applications you require non-standard dilutions. Try to use the three beakers and syringe to create a quantity of dilutions in each of these cases (in increasing order of complexity)

- 1/3
- 1/7
- 1/18
- 5/12
- 29/35
- 71/882

Part 2:

Once you have a feel for the way in which to create dilutions, try to create dilutions 1/2, 1/3, 1/4, 1/5, 1/6, 1/7 ... and so on up to 1/19, 1/20. Which can you make and which can you not make? Why? Do you think that an exact dilution 1/362880 will be possible?

Part 3:

I mix a dilution with a concentration $c_A$ in the first beaker and a smaller concentration $c_B$ in the second beaker. I pour some liquid (unmeasured amounts) from beakers A and B into beaker C to make a new dilution with concentration $c_C$. Use common sense to understand and algebra to prove that $$c_A> c_C> c_B$$

On a scrap of paper I found this method for making a 1/13 dilution using the above configuration of beakers:

- Fill beaker A with 100ml of water.
- Put 10ml of solution and 20ml of water into beaker B.
- Roughly fill beaker B with fluid from beaker A.
- Roughly fill beaker A with fluid from beaker B.
- Repeat steps 3. and 4. 10 times.

How accurately will this make the dilution? You might like to use a spreadsheet to work this out. Can you prove how repeating this mixing process will eventually provide the dilution to any required level of accuracy?

Can you extend this method to produce a way of mixing a 1/23 dilution using the above configuration of beakers?