1. MOLARITY A measurement of the concentration of a solution Molarity (M) is equal to the moles of solute (n) per liter of solution M = n / V = mol / L Calculate the molarity of a solution prepared by mixing 1.5 g of NaCl in 500.0 mL of water. First calculate the moles of solute: 1.5 g NaCl ( 1 mole NaCl ) = 0.0257 moles of NaCl 58.45 g NaCl Next convert mL to L: 0.500 L of solution Last, plug the appropriate values into the correct variables in the equation: M = n / V = 0.0257 moles / 0.500 L = 0.051 mol/L

2. MOLARITY M = n / V = mol / L How many grams of LiOH is needed to prepare 250.0 mL of a 1.25 M solution? First calculate the moles of solute needed: M = n / V, now rearrange to solve for n: n = MV n = (1.25 mol / L ) (0.2500 L) = 0.3125 moles of solute needed Next calculate the molar mass of LiOH: 23.95 g/mol Last, use diminsional analysis to solve for mass: 0.3125 moles (23.95 g LiOH / 1 mol LiOH) = 7.48 g of LiOH

3. What is the molarity of hydroiodic acid if the solution is 47.0% HI by mass and has a density of 1.50 g/mL? First calculate the mass of solute in the 47.0% solution using the density. The 1.50 g/mL is the density of the solution but only 47.0% of the solution is the solute therefore: 47.0% of 1.50 g/mL = (0.470) (1.50 g/mL) = 0.705 g/mL density of solute Since molarity is given in moles per liter and not grams we must convert the g/mL to mol/mL using the molar mass. 0.705 g/mL (1 mole/ 128 g) = 0.00551 mol/mL Next convert mL to L: 0.00551 mol/mL (1000 mL/ 1L) = 5.51 mol/L = 5.51 M MOLARITY M = n / V = mol / L

4. MOLARITY & DILUTION M 1 V 1 = M 2 V 2 The act of diluting a solution is to simply add more water (the solvent) thus leaving the amount of solute unchanged. Since the amount or moles of solute before dilution (n b ) and the moles of solute after the dilution (n a ) are the same: n b = n a And the moles for any solution can be calculated by n=MV A relationship can be established such that M b V b = n b = n a = M a V a Or simply : M b V b = M a V a

5. Calculate the molarity of a solution prepared by diluting 25.0 mL of 0.05 M potassium iodide with 50.0 mL of water (the densities are similar). M 1 = 0.05 mol/LM 2 = ? V 1 = 25.0 mLV 2 = 50.0 + 25.0 = 75.0 mL M 1 V 1 = M 2 V 2 M 1 V 1 = M 2 = (0.05 mol/L) (25.0 mL) = 0.0167 M of KI V 2 75.0 mL MOLARITY & Dilution

6. Given a 6.00 M HCl solution, how would you prepare 250.0 mL of 0.150 M HCl? M 1 = 6.00 mol/LM 2 = 0.150 V 1 = ? mLV 2 = 250.0 mL M 1 V 1 = M 2 V 2 M 2 V 2 = V 1 = (0.150 mol/L) (250.0 mL) = 6.25 mL of 6 M HCl M 1 6.00 mol/L You would need 6.25 mL of the 6.00 M HCl reagent which would be added to about 100 mL of DI water in a 250.0 mL graduated cylinder then more water would be added to the mixture until the bottom of the menicus is at 250.0 mL. Mix well. MOLARITY & dilution

7. PRACTICE PROBLEMS _________1. What is the concentration of 250.0 mL of 0.60 moles of HCl? _________ 2. What is the concentration of 35.0 mL of 0. 0556 moles of KCl? _________ 3. How many grams of KCl is needed to prepare 50.0 mL of a 0.10 M solution? _________ 4. How many milliliters of water must be added to 30.0 mL of 9.0 M KCl to make a solution that is 0.50 M KCl? _________ 5. What volume of 0.7690 M LiOH will contain 55.3 g of LiOH? _________ 6. How many liters of water must be added to 100.0 mL of 4.50 M HBr to make a solution that is 0.250 M HCl? 2.4 M 3.00 L 1.80 L 0.38 g 1.59 M 510 mL