Expanding on Molarity – Learning About Dilutions

1) 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?

M1V1 = M2V2

(0.15 M)(125 mL) = x (150 mL)

x = 0.13 M

2) 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?

M1V1 = M2V2

(0.15 M)(100 mL) = x (150 mL)

x = 0.1 M

3) How much 0.05 M HCl solution can be made by diluting 250 mL of 10 M HCl?

M1V1 = M2V2

(10 M)(250 mL) = (0.05 M) x

x = 50,000 mL

4) I have 345 mL of a 1.5 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?

M1V1 = M2V2

(1.5 M)(345 mL) = x (250 mL)

x = 2.1 M

5) 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?

M1V1 = M2V2

(2.4 M)(500 mL) = (1.0 M) x

x = 1200 mL

1200 mL will be the final volume of the solution. However, since there’s already 500 mL of solution present, you only need to add 700 mL of water to get 1200 mL as your final volume. The answer: 700 mL.

6) A stock solution of sodium sulfate, Na2SO4 has a concentration of 1.00 M. The volume of this solution is 50 mL. What volume of a 0.25 M solution could be made from the stock solution?

M1V1 = M2V2

(1.00 M)(50 mL) = (0.25 M) x

x = 200 mL

7) 2.00 mL of a 0.75 M solution of potassium permanganate, KMnO4 solution is used to make a 500.00 mL solution. What is the concentration of the new solution?

M1V1 = M2V2

(0.75 M)(2.00 mL) = x (500 mL)

x = 0.0030 M

8) A hydrochloric acid solution, HCl has a concentration of 12.1 M. A 41.2 mL sample is used to make a more dilute solution. If the new solution has a concentration of 0.5 M, determine the volume of the solution.

M1V1 = M2V2

(12.1 M)(41.2 mL) = (0.5 M) x

x = 997 à 1000 mL

9) A 0.50 M solution of sodium thiosulfate, Na2S2O3 is used to create a more dilute solution. If 250 mL of the concentrated solution is diluted to a volume of 2.5 L, determine the concentration of the new solution.

M1V1 = M2V2

(0.5 M)(0.25 L) = x (2.5 L)

x = 0.050 M

10) A stock solution of potassium nitrate, KNO3 has a concentration of 0.25 M. What volume of dilute potassium nitrate (0.10 M) can be formed with 80.0 mL of the concentrated solution?

M1V1 = M2V2

(0.25 M)(80.0 mL) = (0.10 M) x

x = 200 à 2.0 x 102 mL

11) What volume of concentrated nitric acid, HNO3 (15.8 M) should be added to water to form 500.0 mL of a 3.0 M nitric acid solution?

M1V1 = M2V2

(15.8 M) x = (3.0 M)(500.0 mL)

x = 95 mL

12) A sample of 7.0 mL of concentrated sulfuric acid, H2SO4 is used to make 250. mL of a 0.50 M sulfuric acid solution. What was the initial concentration of the sulfuric acid?

M1V1 = M2V2

x (7 mL) = (0.50 M) (250 mL)

x = 18 M

13) An instructor needs to make 400 mL of a silver nitrate solution that has a concentration of 0.01 M. How many milliliters of the 0.5 M solution should be used?

M1V1 = M2V2

(0.5M) x = (0.01 M) (400 mL)

x = 8 mL