1. Aim # 19: How can we prepare a more dilute solution from a more concentrated one?
H.W. # 19
Study pp (sec. 15.6)
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Do Now: 1. How many moles of calcium chloride, CaCl2, are needed to prepare 500. mL of a .250 M solution?
2. How many grams of CaCl2 are needed?

2. II Because only solvent is added to the stock solution,
I Stock Solution- a solution of known concentration from which other, more dilute solutions may be prepared by adding more solvent to a measured portion (volume) of it.
II Because only solvent is added to the stock solution,
moles of solute moles of solute before dilution = after dilution
If M = moles of solute , then volume of solution(L)
moles of solute = M x V
where M1 = molarity before dilution
and M1V1 = M2V V1 = volume before dilution M2 = molarity after dilution V2 = volume after dilution

3. M2 = M1V1 = (2.0 mol/L)(25.0 mL) V2 (40.0 mL)
Problem: If 25.0 mL of a 2.0 M solution of KCl is diluted to mL , what will be the concentration of the final solution?
M1V1 = M2V2
M2 = M1V1 = (2.0 mol/L)(25.0 mL) V (40.0 mL)
M2 = 1.25 mol/L = 1.3 mol/L

4. Problem: To what volume should 15. 0 mL of a 10
Problem: To what volume should 15.0 mL of a 10.0 M solution of sulfuric acid be diluted to prepare a M solution?
Problem: How would you prepare 150 mL of 0.10 M NaNO from a 0.25 M stock solution?