1. MOLARITY, MOLALITY, DILUTIONS & PERCENTS

2. Molarity mols M L Molarity involves a molar amount of solute, so if you are given something like grams, you must convert them into moles first. Molarity also involves the volume (in liters) of total solution, so if you are given something like mL, you must convert them into liters first.

3. 1.Set up a “column of information” that lists the info given in the problem, as well as which variable you are solving for (the ?) 2.Decide which rearrangement of the formula you need to solve the problem 3.Convert units given to those needed to match the units in the formula 4.Plug in the values and solve! Ex1 (Molarity): What is the molarity of a 2.3 L solution containing 54.3 g of H 3 PO 4 ? M = _________ mol = _________ L = _________ M = mol L M = mol L ? 54.3 g → mol 2.3 L 1 mol H 3 PO 4 = 97.9952 g H 3 PO 4 54.3 g H 3 PO 4.554 mol H 3 PO 4.554 M = mol L =.554 mol 2.3 L =.24 M

4. Molality m mols kg Molality involves a molar amount of solute, so if you are given something like grams, you must convert them into moles first. Molality also involves the mass (in kilograms) of solvent, so if you are given something like grams, you must convert them into kilograms first.

5. 1.Set up a “column of information” that lists the info given in the problem, as well as which variable you are solving for (the ?) 2.Decide which rearrangement of the formula you need to solve the problem 3.Convert units given to those needed to match the units in the formula 4.Plug in the values and solve! Ex2 (molality): How many grams of NaCl were added to 1.48 kg of ethanol if the resulting solution was 3.7 m? m = ________ mol = _________ kg = _________ m = mol kg m = mol kg 3.7 m ? (in g) 1.48 kg m = mol kg m ● kg = mol mol = m ● kg mol = (3.7 m)(1.48 kg) = 5.476 mol 5.476 mol NaCl 58.4428 g NaCl = I mol NaCl 320.02 g NaCl Grams to Kilograms  divide by 1000

6. Dilution  When chemists purchase solutions, they generally purchase “stock solutions” which are extremely concentrated solutions  This way a chemist can dilute the strong solution to any concentration that they wish. This stops the chemist from having to buy several concentrations There are four parts to the dilution equation : 1. M 1 = Molarity of the diluted (or desired) solution 2. M 2 = Molarity of the concentrated (or stock) solution 3. V 1 = volume of the diluted (or desired) solution 4. V 2 = volume of the concentrated (or stock) solution ***Units for volume can be mL or L, as long as they are the same. M 1 V 1 = M 2 V 2 M 1 V 1 = M 2 V 2

7. 1.Set up a “column of information” that lists the info given in the problem, as well as which variable you are solving for (the ?) 2.Decide which rearrangement you need to solve the problem 3.Convert units given to those needed to match the units in the formula 4.Plug in the values and solve! Ex3 (Dilution): How many mL of 12 M HCl is needed to produce 1.5 L of a solution that is 3.8 M? M 1 = _________ M 2 = _________ V 1 = _________ V 2 = _________ V 2 = M 1 V 1 M 2 V 2 = M 1 V 1 M 2 3.8 M 12 M 1.5 L ? V 2 = M 1 V 1 = M 2 V 2 = (3.8 M)(1.5 L) = 12 M.475 L 1000 mL = 1 L 475 mL HCl

8. Concentration  The amount of solute in a solution.  Describing Concentration  % by mass - medicated creams  % by volume- rubbing alcohol

9. Percent by Mass  Remember …  % = part x 100 whole  % by mass = mass solute x 100 mass solution

10. Example  What is the % by mass of a solution with 3.6 g of NaCl dissolved in 100.0 g of water?  % = (3.6 / 103.6) x 100 = 3.5% NaCl

11. Percent by Volume  Remember …  % = part x 100 whole  % by volume = volume solute x 100 volume solution

12. Example  What is the % by volume of 75.0 mL of ethanol dissolved in 200.0 mL of water?  % = (75.0 / 275.0) x 100 = 27.3%