Diluting Solutions to Prepare Workable Solutions

Today we've learned the dilution calculation.  For example, if I have a 2 L of 16 M solution, but all I need is a 0.8 L of 2 M solution, how can I diluted the more concentrated solution into the less concentrated one?

First of all
assume:-initial concentration of solution=M1
               -initial volume of solution=L1
               -diluted concentration=M2
               -diluted volume=L2

                     mol
Since M = ---------         
                      L

then  mol = M x L
which means   moles of chemical in concentrated solution = mole concentrated = M1 x L1
and                   moles of chemical in diluted solution = mole diluted = M2 x L2

But the amount of the chemical is not changed when the solution is diluted, only the concentration of the chemical is changed.
Therefore          moles of concentrated chemical = moles of diluted chemical
                                                            OR
                                               M1 x L1 = M2 x L2
So, 16 M x L1 = 2 M x 0.8 L
       16 M x L1 = 1.6 mol
        L1 = 1.6 mol / 16 M
        L1 = 0.1 L
If I need 0.1 L of the 16 M solution to make 0.8 L of 2 M solution......
I would take 0.1 L of the concentrated solution and add how much water?

     0.8 - 0.1 = 0.7 L