Dilutions and Percent Solutions Unit 8 – Solutions Chapter 15

Dilutions and Molarity Use this formula to make a more dilute solution from a concentrated solution Molarity1Volume1 = Molarity2Volume2 (Concentrated) (Dilute) (Moles before) = (Moles after!) M1V1 = M2V2

Sample Problem #1 How many milliliters of 1.0 Molar HCl are required to make 100 mL of 0.025 M HCl? M1V1 = M2V2 M1 = 1.0M V1 = ? M2 = 0.025M V2 = 100mls (1.0M)V1 = (0.025M)(100mls) V1 = (0.025M)(100mls) (1.0M) = 2.5 mls of 1.0M HCl required, then dilute to 100mls

Sample Problem #2 How much water should you add to the volume of 1.0M HCl you calculated above to make the solution? We need 2.5 mls of concentrated 1.0M HCl. Then we dilute to a final total volume of 100mls. Therefore, 100mls – 2.5mls = 97.5 mls of water should be added. **Water added = V2 – V1

Percent Solutions Solutions can also be represented as percent of solute in a specific mass or volume of solvent. For a solid dissolved in water, you use percent by mass. % by mass = mass solute x 100 mass of solution *Mass of solution = solute mass + solvent mass

For a liquid mixed with another liquid % by volume = volume solute x 100 Total volume of solution

Example 1 What percent solution do you have if you dissolve 80 grams of NaCl in 1 liter of water? * 1 Liter = 1000mls = 1000grams for water % mass = 80 grams NaCl X 100 (80g + 1000g H2O) = 7.4 %

Example #2 What percent solution will you have if you mix 40mls of ethanol with 200 mls of water? %volume = 40mls x 100 (200mls + 40mls) = 16.7%

What if the previous problem was worded as follows? What percent solution will you have if you dilute 40mls of ethanol to a final volume of 200mls? % volume = 40mls x 100 200mls* = 20% *Denominator is always total volume. You must pay attention the wording in the problem!