Dilution PracticeRemember!

M1 V1 = M2 V2

M1 = concentration of stock solution

V1 = volume of stock solution

M2 = goal concentration

V2 = goal volume

#1 • With a 5.0M stock solution of NH3, you need to make

0.50L of 1.0 M solution. How much stock solution do you need?

• M1 = 5.0 M• V1 = ??• M2 = 1.0M• V2 = 0.50L• V1 = 0.1L = 100 mL• How much distilled water will you add to arrive at your

desired solution?• 500 mL – 100 mL stock solution = 400 mL water added

#2 • With a 6.0M stock solution of NaOH, you need to make

0.250L of 1.0 M solution. How much stock solution do you need?

• M1 = 6.0 M• V1 = ??• M2 = 1.0M• V2 = 0.250L• V1 = 0.0417L = 41.7 mL• How much distilled water will you add to arrive at your

desired solution?• 250 mL – 41.7 mL stock solution = 208.3 mL water added

#3

• With 100 mL of a 6.0M stock solution, how much 0.5M dilute solution can you make?

• M1 = 6.0 M

• V1 = 100mL = 0.1L

• M2 = 0.5M

• V2 = ??

• V2 = 1.2L

#4

• If you dilute 175 mL of a 1.6 M solution of LiCl to 1.0 L, determine the new concentration of the solution.

• M1 = 6.0 M

• V1 = 175mL = 0.175 L

• M2 = ?

• V2 = 1.0L

• M2 = 0.28 M

#5

• You need to make 10.0 L of 1.2 M KNO3. What molarity would the potassium nitrate solution need to be if you were to use only 2.5 L of it?

• M1 = ?

• V1 = 2.5 L

• M2 = 1.2 M

• V2 = 10.0 L

• M1 = 4.8 M

#6

• If I add 25 mL of water to 125 mL of a 0.15 M NaOH solution, what will the molarity of the diluted solution be?

• M1 = 0.15 M

• V1 = 125 mL = 0.125L

• M2 = ?

• V2 = 25 mL + 125 mL = 150 mL total = 0.150L

• M2 = 0.125 M

#7

• If I add water to 100 mL of a 0.15 M NaOH solution until the final volume is 150 mL, what will the molarity of the diluted solution be?

• M1 = 0.15 M

• V1 = 100 mL = 0.100L

• M2 = ?

• V2 = 150 mL = 0.150L

• M2 = 0.1 M

#8

• I have 345 mL of a 1.50 M NaCl solution. If I boil the water until the volume of the solution is 250. mL, what will the molarity of the solution be?

• M1 = 1.50 M

• V1 = 345 mL = 0.345L

• M2 = ?

• V2 = 250. mL = 0.250L

• M2 = 2.07 M

#9

• How much water would I need to add to 500. mL of a 2.4 M KCl solution to make a 1.0 M solution?

• M1 = 2.4 M

• V1 = 500 mL = 0.500L

• M2 = 1.0M

• V2 = ?

• V2 = 1.2 L total, this means you would need to add 0.7 L of pure water to your original solution

#10

• Draw what happens when calcium nitrate dissolves in water. What pieces does it separate into? In what ratio?

• Ca+2 (aq), 2 NO3- (aq)