Diluting Solution to Prepare Workable Solution

- chemicals are typically shipped around the world in their most concentrated form
- thus making it easier to transport and cost fairly less money
- but have to make concentrate form from any solution from an even more
  concentrated solution
Example 1: You have 2.00L of 16.0M HCl, but you need 0.800L of 2.00M HCl
How do you make it?
- numbers of moles are always constant=not changing
- only difference between the two would be there is more water in the less
 concentrated solution
Therefore, Equation would be= Moles solute before=Moles solute after
  M1L1=M2L2
                   The subscript 1 denotes “before” and subscript is “after”
         Solving for L1 wouls solve how much volume there is in the solution
Applying the formula, M1L1=M2L2
16.0M HCl×L1=2.00M HCl×0.800L
16.0M×L1=1.60moles HCl
L1=0.100L or 1.00×10ˆ2ml
0.800L-0.100L=0.700L→amount of water
Example 2: Concentrated HCl is 11.6 moles/L/ How would you make up 250ml of 0.500 moles/L HCl?
11.6M HCl×L1=0.500M HCl×250ml
L1=(M2×L2)/M1= (0.500 mole/L MCl×250ml)/(11.6 moles/L HCL)
  = 0.011L
  = 11ml 250-11ml= 239ml→amount of water

How to solve:
Preparing Solutions Sample Problem 1
Preparing Solutions Sample Problem 2
Preparing Solutions Sample Problem 3