MOLARITY A Measurement Of The Concentration Of A

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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 Na. Cl in 500. 0 m. L of water. First calculate the moles of solute: 1. 5 g Na. Cl (1 mole Na. Cl) = 0. 0257 moles of Na. Cl 58. 45 g Na. Cl Next convert m. L 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

MOLARITY M=n/ V = mol / L How many grams of Li. OH is needed to prepare 250. 0 m. L 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 Li. OH: 23. 95 g/mol Last, use diminsional analysis to solve for mass: 0. 3125 moles (23. 95 g Li. OH / 1 mol Li. OH) = 7. 48 g of Li. OH

MOLARITY M=n/ V = mol / L What is the molarity of hydroiodic acid if the solution is 47. 0% HI by mass and has a density of 1. 50 g/m. L? First calculate the mass of solute in the 47. 0% solution using the density. The 1. 50 g/m. L is the density of the solution but only 47. 0% of the solution is the solute therefore: 47. 0% of 1. 50 g/m. L = (0. 470) (1. 50 g/m. L) = 0. 705 g/m. L density of solute Since molarity is given in moles per liter and not grams we must convert the g/m. L to mol/m. L using the molar mass. 0. 705 g/m. L (1 mole/ 128 g) = 0. 00551 mol/m. L Next convert m. L to L: 0. 00551 mol/m. L (1000 m. L/ 1 L) = 5. 51 mol/L = 5. 51 M

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 (nb) and the moles of solute after the dilution (na) are the same: nb = na And the moles for any solution can be calculated by n=MV A relationship can be established such that M b V b = n a = M a. V a Or simply : Mb. Vb = Ma. Va

MOLARITY & DILUTION Calculate the molarity of a solution prepared by diluting 25. 0 m. L of 0. 05 M potassium iodide with 50. 0 m. L of water (the densities are similar). M 1 = 0. 05 mol/L M 2 = ? V 1 = 25. 0 m. L V 2 = 50. 0 + 25. 0 = 75. 0 m. L M 1 V 1 = M 2 V 2 M 1 V 1 = M 2 = (0. 05 mol/L) (25. 0 m. L) = 0. 0167 M of KI V 2 75. 0 m. L

MOLARITY & DILUTION Given a 6. 00 M HCl solution, how would you prepare 250. 0 m. L of 0. 150 M HCl? M 1 = 6. 00 mol/L M 2 = 0. 150 V 1 = ? m. L V 2 = 250. 0 m. L M 1 V 1 = M 2 V 2 = V 1 = (0. 150 mol/L) (250. 0 m. L) = 6. 25 m. L of 6 M HCl M 1 6. 00 mol/L You would need 6. 25 m. L of the 6. 00 M HCl reagent which would be added to about 100 m. L of DI water in a 250. 0 m. L graduated cylinder then more water would be added to the mixture until the bottom of the menicus is at 250. 0 m. L. Mix well.

MOLARITY & Stoichiometry In the past chapters, you measured out a solid reactant and needed to predict the mass of the product. You learned to use the molar mass and mole ratio to achieve that goal. The same reasoning applies with solution chemistry but instead to using the molar mass you will use the molarity to convert to moles. Stoichiometry Flow chart

MOLARITY & Stoichiometry How much calcium carbonate will be precipitated by adding 25. 0 m. L calcium chloride to 25. 0 m. L of 0. 56 M potassium carbonate? Ca. Cl 2 + K 2 CO 3 Ca. CO 3 + 2 KCl V = 25. 0 m. L M = 0. 56 mol/L m=? V = 25. 0 m. L First convert volume of A to moles of A: 0. 025 L K 2 CO 3 (0. 56 mol/L) = 0. 014 moles of K 2 CO 3 Now convert moles of A to moles of B: 0. 014 mol K 2 CO 3(1 mol Ca. CO 3/1 mol K 2 CO 3) =0. 014 mol Ca. CO 3 Next convert moles of B to grams of B: 0. 014 mol Ca. CO 3 (100 g/mol) = 1. 40 g of Ca. CO 3

MOLARITY & Stoichiometry What would be the molarity of the potassium chloride solution from the last problem? Ca. Cl 2 + K 2 CO 3 Ca. CO 3 + 2 KCl V = 25. 0 m. L M = 0. 56 mol/L M=? V = 25. 0 m. L First convert volume of A to moles of A: 0. 025 L K 2 CO 3 (0. 56 mol/L) = 0. 014 moles of K 2 CO 3 Now convert moles of A to moles of B: 0. 014 mol K 2 CO 3(2 mol KCl /1 mol K 2 CO 3) =0. 028 mol KCl Next convert moles of B to molarity of B: 0. 028 mol KCl / 0. 050 L = 0. 56 M of KCl

PRACTICE PROBLEMS #24 a 2. 4 M _____1. What is the concentration of 250. 0 m. L of 0. 60 moles of HCl? 3. 00 L _____ 2. What volume of 0. 7690 M Li. OH will contain 55. 3 g of Li. OH? 1. 80 L 3. How many liters of water must be added to _____ 100. 0 m. L of 4. 50 M HBr to make a solution that is 0. 250 M HCl? 39. 6 g _____ 4. How many grams of barium sulfate that will precipitate when 500. 0 m. L of 0. 340 M Ba. Cl 2 and 300. 0 m. L of 1. 70 M Na 2 SO 4 are mixed? 5. How would you prepare 850. 0 m. L of a 0. 020 M ferric chloride solution if Weigh g of of the. Fe. Cl hydrated salt to a graduated cylinder with. you start out with 4. 59 crystals 3 6 H 2 O? 800. 0 m. L then add enough DI water to make 850 m. L exactly.

PRACTICE PROBLEM #24 b 1. What is the concentration of 35. 0 m. L of 0. 0556 moles of KCl? 1. 59 M 2. How many grams of KCl is needed to prepare 50. 0 m. L of a 0. 10 M solution? 0. 38 g 3. How many milliliters of water must be added to 30. 0 m. L of 9. 0 M KCl to make a solution that is 0. 50 M KCl? 510 m. L 4. How many grams of calcium carbonate will precipitate when 500. 0 m. L of 0. 340 M Ca. Cl 2 and 300. 0 m. L of 1. 70 M Na 2 CO 3 are mixed? 17. 0 g 5. What is the concentration of the product solution (assuming the volumes are additive) when 500. 0 m. L of 0. 340 M Ca. Cl 2 and 300. 0 m. L of 1. 70 M Na 2 CO 3 are mixed? 0. 425 M

GROUP STUDY PROBLEM #24 a ______1. How many grams of K 3 PO 4 is needed to prepare 500. 0 m. L of a 0. 00567 M solution? ______2. What volume of 0. 7690 M Li. OH will contain 55. 3 g of Li. OH? ______3. How many milliliters of water must be added to 600. 0 m. L of 3. 40 M HCl to make a solution that is 0. 500 M HCl? ______ 4. What is the concentration of the product solution (assuming the volumes are additive) when 24. 9 m. L of 0. 70 M Ba. Cl 2 and 45. 0 m. L of 0. 15 M Na 3 PO 4 are mixed?

GROUP STUDY PROBLEM #24 b ______1. What is the concentration of 250. 0 m. L of 0. 60 moles of Na. OH? ______2. What mass of Mg. Cl 2 will be required to prepare 500 m. L of 0. 150 M solution? ______3. How many milliliters of water must be added to 50. 0 m. L of 1. 97 M Na. OH to make a solution that is 0. 025 M Na. OH? ______ 4. How many grams of barium sulfate that will precipitate when 500. 0 m. L of 0. 340 M Ba. Cl 2 and 300. 0 m. L of 1. 70 M Na 2 SO 4 are mixed? ______ 5. What is the concentration of the product solution (assuming the volumes are additive) when 500. 0 m. L of 0. 340 M Ba. Cl 2 and 300. 0 m. L of 1. 70 M Na 2 SO 4 are mixed?