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12.3 Concentrations of Solutions – Sample Problem 1

12.3 Concentrations of Solutions – Sample Problem 1. You have 3.50 L of a solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of the solution? . Given Wanted 3.50 L of solution ? Molarity (M) 90.0 g NaCl

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12.3 Concentrations of Solutions – Sample Problem 1

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  1. 12.3 Concentrations of Solutions – Sample Problem 1 You have 3.50 L of a solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of the solution? GivenWanted 3.50 L of solution ? Molarity (M) 90.0 g NaCl Molarity = moles L Solution Step 1: Convert 90.0 g NaCl to mol NaCl 90.0 g NaCl x 1 mol = 1.54 mol NaCl 58.443 g **(1 x 22.990) + (1 x 35.453) = 58.443 g/mol • Step 2: Use molarity equation • Molarity = M = mol = 1.54 mol = 0.440 M • L 3.50 L

  2. 12.3 Concentrations of Solutions – Sample Problem 2 You have 0.8 L of a 0.5M HCl solution. How many moles of HCl does this solution contain? GivenWanted 0.8 L of HCl solution ? mol HCl Molarity = 0.5 M = 0.5 moles L Solution • Use molarity equation • Molarity = M = mol • L • 0.5 mol = ? mol HCl • 1 L 0.8 L • ? mol HCl = 0.8 L x 0.5 mol = 0.4 mol HCl • 1 L

  3. 12.3 Concentrations of Solutions – Sample Problem 3 To produce 40.0 g of silver chromate, you will need at least 23.4 g potassium chromate in solution as a reactant. All you have on hand is 5 L of a 6.0 M K2CrO4 solution. What volume of the solution is needed to give you the 23.4 g of K2CrO4 needed for the reaction? GivenWanted • 40.0 g silver chromate23.4 g K2CrO4 ? L solution 5 L of 6.0 M K2CrO4solution Solution Step 1: Convert 23.4 g K2CrO4 to mol K2CrO4 23.4 g K2CrO4 x 1 mol = 0.121 mol K2CrO4 194.1908 g * *(2 x 39.0983) + (1 x 51.9961) + (4 x 15.9994) = 194.1908 g/mol • Step 2: Use molarity equation • Molarity = M = 6 mol = 0.121 mol K2CrO4 • L ? L • ? L = 0.121 mol x 1 L = 0.020 L K2CrO4 • 6 mol

  4. 12.3 Concentrations of Solutions – Sample Problem 4 A solution was prepared by dissolving 17.1 g of sucrose, C12H22O11, in 125 g of water. Find the molal concentration of this solution? GivenWanted • 17.1 g C12H22O11 in 125 g water ? m Solution Step 1: Convert 125 g water to kg 125 g x 1 kg = 0.125 kg H2O 1000 g • Step 2: Convert 17.1 C12H22O11 to mol 17.1 g C12H22O11 x 1 mol = 0.050 mol C12H22O11 342.3014 g* *(12 x 12.011) + (22 x 1.008) + (11 x 15.9994) = 342.3014 g • Step 3: Use molality equation • Molality = m = mol= 0.050 mol = 0.400 m • kg 0.125 kg

  5. 12.3 Concentrations of Solutions – Sample Problem 5 A solution of iodine, I2, in carbon tetrachloride, CCl4, is used when iodine is needed for certain chemical tests. How much iodine must be added to prepare a 0.480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used? GivenWanted • 100.0 g CCl4? g I2 • 0.480 m solution of I2 in CCl4 Solution Step 1: Convert 100.0 g CCl4 to kg 100.0 g x 1 kg = 0.100 kg CCl4 1000 g • Step 2: Use molality equation • m = mol = ? mol I2 • kg 0.480 mol = ? mol I2= 0.480 mol x 0.100 kg = 0.048 mol I2 1 kg 0.100 kg 1 kg Step 3: Convert mol I2 to g I2 • 0.048 mol I2x 253.808 g* = 12.2 g I2 *(2 x 126.904) = 253.808 g/mol 1 mol

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