Dilution Chem Worksheet 15-6

University:
Eastern Kentucky University
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CHE 101 | Introductory Chemistry
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2024
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Dilution Chem Worksheet 15-6 Conc. Solution M1 × V 1 = mol 0.5 L Water Concentrated solution is diluted with more solvent. A solution can be made less concentrated in a process called dilution. This is accomplished by adding more solvent. This process decreases the molarity of the solution - the moles of solute in a USEFUL EQUATIONS dilute solution remain constant while the volume of solvent is M1 × V1 = M2 × V2 molarity = increased. Let’s assume that you have 1 L = 1000 mL 0.500 L of a hydrochloric acid solution with a concentration of 12 M. This sample contains 6 moles of HCl. When this solution is placed in a larger flask and water is added until the volume reaches 2.00 L a more dilute solution is created. There are still 6 moles of HCl in the solution, but the new volume is 2.00 L. So, the concentration is now 6 moles/2.00 liters, or 3 M. A simple formula used when diluting solutions is molarity1 × volume1 = molarity2 × volume2. example Dilute Solution M1 × V 1 = mol Calculate the molarity of the solution that forms when 10 mL of a 6.0 M solution is diluted to a volume of 250 mL. - determine variables: M1 = 6.0 M V1 = 10.0 mL M2 = ? V2 = 250 mL - substitute and solve: 2L Solve the following dilution problems. 1. 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? 1.00x50/0.25 200 mL 2. 2.00 mL of a 0.75 M solution of potassium permanganate, K2MnO4 solution is used to make a 500.00 mL solution. What is the concentration of the new solution? 0.75x2.00/500.00 0.003M 3. 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.12.1x41.2/0.5 997.04 mL 4. 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.0.50x250/2500= 0.05M 5. 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? 0.25x80.0/0.10= 200mL 6. 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? 3.0x50.0/15.8= 94.94mL 7. 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? 0.50x250.0/7.0= 17.86m 8. 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? 0.01x400/0.5= 8mL © John Erickson, 2005

Dilution Chem Worksheet 15-6

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