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CHEMISTRY II

CHEMISTRY II. Solutions & Concentrations. Definitions. Solution: A homogeneous mixture of a solute and a solvent. Solute: The substance being dissolved. Solvent: The substance doing the dissolving.

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CHEMISTRY II

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  1. CHEMISTRY II Solutions & Concentrations

  2. Definitions • Solution: A homogeneous mixture of a solute and a solvent. • Solute: The substance being dissolved. • Solvent: The substance doing the dissolving. • Example: When you add salt (NaCl) to water, the salt is the solute and the water is the solvent.

  3. Concentrations • To prepare a solution with a specific concentration means to know the mass of the the solute and the volume or mass of the solvent. • Concentration Units: • Molarity ( M ) • Normality ( N ) • molality ( m ) • % Solution ( % )

  4. Molarity ( M ) • 1.00 M = 1.00 mole solute per 1.00 lsolvent. • 1 M = mol • l • Problem: Prepare 350.0 ml of .250M NaCl • ( the word prepare is asking the question “ how many grams of solute is needed.”) • Start with what is given and set up units to give you grams. NaCl 23.0 +35.5 = 58.5 g mol 0.250 mol l .350 l x 58.5 = 5.12 g x

  5. More MOLarity • If 23.5 g of CaBr2 is dissolved in 1.34 liters of water, what is the molar concentration? • Start with what is given and set up units to give you mol/liter. 23.5 g x mol g 1.00 l x = 0.0877 M 200.0 1.34 CaBr2 = 40.1 + 2(79.9) = 200.0

  6. Normal Concentration • If 34.5 g of AlCl3 were dissolved in 565 ml of water what is the normality and molarity of the solution? 34.5 g 1 mol g 3 ge mol x x 1 l x = 1.37 N 1 134 .565 Al = +3 1mol = 3 ge AlCl3 = 27.0 + 3(35.5) = 134

  7. Normal Volume • What volume of water would be required to make a 2.25 N solution of Fe(NO3)3 if 25.0 g were available. • (This spelling is for you Amanda,Claire and that other girl • whose name starts with an “A”) I’m so glad I took your class Mr.. Lamar I want Jared What is Normal volume? Alice or something Amanda Claire

  8. I Digress • What volume of water would be required to make a 2.25 N solution of Fe(NO3)3 if 25.0 g were available? 25.0 g 1 mol g 3 ge mol 1 l ge x x x = .138 l or 138 ml 242 1 2.25 Fe(NO3)3= 55.8 + 3[ (14.0 + 3(16.0)] = 242 Fe = +3, 1 mol = 3 ge

  9. Percent Solution • How many grams is necessary to make 300.0 ml of a 12.0% soln. Of Alcohol? 12.0g X , X = 36.0 g = 100 ml 300.0 ml What is the % Soln if 34.5 g of glucose is added to 250.0 ml of water? 34.5 g x 100 = 13.8 % 250.0

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