- the process of decreasing the concentration by adding a solvent(usually water)
- the amount of solute does not change, only the volume
- because concentration is mol/L we use the following equations:
C=n/v and n=Cv so that C1V1=C2V2
*always remember to convert mL into L
EX.
How much water must be added to 10.0mL of 10.0M Na2SO4 to give a solution with a concentration of 0.50M?
V1=100mL-->.10L C1V1=C2V2
C1=0.10M
C2=400mL C1V1 = V2 ----> (10.0)(10) = 200mL changeV=200-10=190mL
C2 (0.50)
EX.
If a 75.0mL sample of 0.500M KCl is added to 100mL of 0.250M KCl what will the resulting solutions concentrations be?
0.075L x 0.500M = 0. 0375mol
L
0.100L x 0.250M = 0.025mol
L
Then add both mols together,
0.0375+0.025= 0.0625mol
After divide the mols by adding both amounts of litres together,
0.0625M = 0.357M
0.175L
Janine R
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