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As you come in,  Pass up the Calories in a Chip Lab.  Get paper, pencil, and calculator ready for notes on calculations.  NOTE: Solubility Graph Quiz.

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Presentation on theme: "As you come in,  Pass up the Calories in a Chip Lab.  Get paper, pencil, and calculator ready for notes on calculations.  NOTE: Solubility Graph Quiz."— Presentation transcript:

1 As you come in,  Pass up the Calories in a Chip Lab.  Get paper, pencil, and calculator ready for notes on calculations.  NOTE: Solubility Graph Quiz tomorrow!  Pass up the Calories in a Chip Lab.  Get paper, pencil, and calculator ready for notes on calculations.  NOTE: Solubility Graph Quiz tomorrow!

2 Concentration  Concentration - the amount of solute in a specific amount of solvent  Molarity (M)  Molality (m)  Molarity is the most common unit used to report the concentration of a solution in a laboratory setting.  Concentration - the amount of solute in a specific amount of solvent  Molarity (M)  Molality (m)  Molarity is the most common unit used to report the concentration of a solution in a laboratory setting.

3 Molarity  The number of moles of solute per liter of solution.  Write a mathematical equation to represent the definition of molarity.  Explain your equation. (A)  Agree or disagree. (B)  The number of moles of solute per liter of solution.  Write a mathematical equation to represent the definition of molarity.  Explain your equation. (A)  Agree or disagree. (B)

4 Molarity Equation moles of solute Molarity = Liters of solution Example: Calculate the molarity of 0.4 mol NaCl dissolved in 1.6 L solution. Answer: 0.25 M NaCl moles of solute Molarity = Liters of solution Example: Calculate the molarity of 0.4 mol NaCl dissolved in 1.6 L solution. Answer: 0.25 M NaCl

5 5 minute Independent Practice  You will have a 5 minute Partner time next. 8-) Practice Problems: 1.20.2 grams of potassium nitrate is dissolved in enough water to make 250.0 mL of solution. What is the molarity? 2.Calculate the number of grams of solute needed to prepare 2.0 L of 2.0M nitric acid. 3.Challenge: Give the concentrations of all the ions in each of the following solutions: 1.0.50M Co(NO 3 ) 2 2.1 M FeCl 3  You will have a 5 minute Partner time next. 8-) Practice Problems: 1.20.2 grams of potassium nitrate is dissolved in enough water to make 250.0 mL of solution. What is the molarity? 2.Calculate the number of grams of solute needed to prepare 2.0 L of 2.0M nitric acid. 3.Challenge: Give the concentrations of all the ions in each of the following solutions: 1.0.50M Co(NO 3 ) 2 2.1 M FeCl 3

6 5 Minute Partner Review  Check your Practice Problem answers with your partner.  Are your chemical formulas written correctly?  Did you make the appropriate conversions before plugging into the molarity equation?  What are your questions? (Write them down, and prepare to ask with our large group.)  Check your Practice Problem answers with your partner.  Are your chemical formulas written correctly?  Did you make the appropriate conversions before plugging into the molarity equation?  What are your questions? (Write them down, and prepare to ask with our large group.)

7 Molarity - Correct Answers 1.Potassium nitrate = KNO 3 20.2 g KNO 3 is converted to 0.20 mol KNO 3. The volume of 250.0mL is converted to.250L. The Molarity = 0.8 M KNO 3 2.Nitric Acid = HNO 3 4mol of HNO 3 is calculated by plugging into the Molarity equation. The mass of four moles of HNO 3 is 252 grams. 1.Potassium nitrate = KNO 3 20.2 g KNO 3 is converted to 0.20 mol KNO 3. The volume of 250.0mL is converted to.250L. The Molarity = 0.8 M KNO 3 2.Nitric Acid = HNO 3 4mol of HNO 3 is calculated by plugging into the Molarity equation. The mass of four moles of HNO 3 is 252 grams.

8 Molality  The number of moles of solute per kilogram of solvent.  Write a mathematical equation to represent the definition of molality.  Explain your equation. (B)  Agree or disagree. (A)  The number of moles of solute per kilogram of solvent.  Write a mathematical equation to represent the definition of molality.  Explain your equation. (B)  Agree or disagree. (A)

9 Molality Equation moles of solute Molality = kilograms of solvent Example: What is the molality of a solution that contains 2.3 moles of glucose dissolved in 500.0 g of water? Answer: 4.6 m moles of solute Molality = kilograms of solvent Example: What is the molality of a solution that contains 2.3 moles of glucose dissolved in 500.0 g of water? Answer: 4.6 m

10 5 Minute Independent Practice  Practice Problems: 1.Find the molality of 131 g Ba(NO 3 ) 2 dissolved in 750.0 g H 2 O. 2.How many grams of sodium chloride must dissolve in 750.0 grams of water to make a 0.50 molal solution?  Practice Problems: 1.Find the molality of 131 g Ba(NO 3 ) 2 dissolved in 750.0 g H 2 O. 2.How many grams of sodium chloride must dissolve in 750.0 grams of water to make a 0.50 molal solution?

11 Molality - Correct Answers 1.131 g Ba(NO 3 ) 2 is converted to 0.50 moles, and 750.0 g of H 2 O is converted to 0.750 kg H 2 O. The molality is 0.67 m Ba(NO 3 ) 2. 2.750.0 grams of H 2 O is converted to 0.750 kg of H 2 O. The equation is solved to give 0.375 moles NaCl, which converts to 21.9 grams. 1.131 g Ba(NO 3 ) 2 is converted to 0.50 moles, and 750.0 g of H 2 O is converted to 0.750 kg H 2 O. The molality is 0.67 m Ba(NO 3 ) 2. 2.750.0 grams of H 2 O is converted to 0.750 kg of H 2 O. The equation is solved to give 0.375 moles NaCl, which converts to 21.9 grams.

12 Preparing a Solution Lab  Today, you will be writing a lab procedure to prepare a solution of a specific Molarity.  Purpose  Materials  Procedure  Calculations  Each group will be preparing a different solution.  Today, you will be writing a lab procedure to prepare a solution of a specific Molarity.  Purpose  Materials  Procedure  Calculations  Each group will be preparing a different solution.

13 GroupMolarity (500 mL) Solute 10.05NaCl 20.005C 12 H 22 O 11 30.0075Na 2 SO 4 40.015Na 2 SO 4 50.05NH 4 Cl 60.025NH 4 Cl 70.009C 12 H 22 O 11


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