1. Moles and Molarity Part 2

2. Molarity Often, chemists use deal with chemicals in solution form. A SOLUTION is a type of MIXTURE in which one substance is completely dissolved in another. The substance that gets dissolved is known as a SOLUTE. Solutes are generally SOLIDS, but cant be LIQUIDS or GASES. The substance that does the dissolving is known as the SOLVENT. Solvents are generally LIQUIDS, but can be GASES too. WATER is called the UNIVERSAL SOLVENT because so many substance can be dissolved in it.

3. Concentration When dealing with solutions, the CONCENTRATION, or the amount of SOLUTE per unit of SOLVENT, is an important property of a solution. Concentration of solutions can be expressed in several ways. One of the most commonly used is MOLARITY. MOLARITY is defined as MOLES OF SOLUTE PER LITER OF SOLUTION. The unit for molarity is “M”.

4. Mathematically, M is calculated as: M = mol solute liters solution The mass of the solute must be in MOLES. The volume of the solvent must be in LITERS. If not, they must be converted to these units.

5. To convert grams to moles, use the mol = mass equation. FW To convert volume units to Liters, you must know some conversions: 1 liter = 1000 mL (divide volume in ml by 1000 to get liters) 1 gallon = 3.78 L (multiply gallons x 3.78 to get to liters) Others – ask!

6. Practice Problems 1. What is the molarity of a solution of 25.75 grams of Potassium Permanganate, KMnO 4, dissolved in enough water to make 675 ml of solution? a. Mol KMnO 4 = 25.75 g = 0.1629 mol 158.03 g/mol b. Liters solution = 675 mL = 0.675 L solution 1000 mL / L c. M = mol solute = 0.1629 mol = 0.241 M liters solution 0.675 L

7. 2. What is the molarity of a solution of 100.3 grams of Calcium Sulfate, CaSO 4, dissolved in enough water to make 1.25 L of solution? a. Mol CaSO 4 = 100.3 g = 0.7367 mol 136.14 g/mol b.Liters solution = 1.25 L (already in liters!) c. M = mol solute = 0.7367 mol = 0.589 M liters solution 1.25 L

8. 3. John needs to make one gallon (1.00 gallon) of salt water (NaCl) to supplement the water in his aquarium. The final solution needs to be 0.755 M. How should John make this solution? 1. Consider the molarity equation: M = mol solute liters solution 2. What do we already know from the problem? a. The final solution should be 0.755 M. b. He needs to make 1 gallon = 3.78 L

9. 3. John first needs to solve for mol solute (salt). mol solute = M x Liters Solution = 0.755 mol / L x 3.78 L = 2.854 mol 4. How many grams of salt in 2.854 mol? FW of NaCl = 58.44 g/mol so therefore: 2.854 mol x 58.44 g/mol = 166.7819g = 166.8 g 5. John should dissolve 166.8 grams of salt in enough water to make 1 gallon.

10. Practice Problems 1. If 35.6 g of Lead (II) Nitrate, Pb(NO 3 ) 2, are dissolved in enough water to make 775 mL of solution, what is the resulting molarity of the solution? 2. How many grams of Calcium Acetate, Ca(C 2 H 3 O 2 ) 2 does a lab worker need to prepare 500.0 mL of 0.675 M solution?