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SOLUTIONS I

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SOLUTIONS I

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    1. SOLUTIONS I Concentrations of solutions Chapter 11.1-11.4 McM Chapter 13.5 Silberberg

    2. Goals & Objectives See the following Learning Objectives on pages 543-544. Understand these Concepts: 13.14 Master these Skills: 13.3-4.

    3. Concentrations of Solutions Solution a homogeneous mixture, consisting of a Solute the dissolved species usually present to the lesser extent and a Solvent the dissolving species usually present to the greater extent

    4. A Sodium Chloride Solution Sodium chloride is the solute water is the solvent (understood)

    6. Concentration A measure of the amount of solute dissolved per given amount of solvent or solution. Typical units Percent by mass (% w/w) Molarity (M, mol/L) Molality (m) Proof

    7. Proof Approximately twice the percent by volume of alcohol Proof %Alcohol by Volume 26 13% 80 40% 190 95%

    8. Percent by Mass of Solute %mass of solute = g solute x 100 g solution A 25.9% solution of glucose contains: 35.0g of glucose /35.g glucose + 100 g of water. The density of water is 1.00 g/ml

    10. Examples Determine the mass of sodium hydroxide, NaOH, required to prepare 250.0 grams of a solution that is 8.00% NaOH by mass. Calculate the mass of 8.00% NaOH solution that contains 32.0 grams of NaOH. Calculate the mass of NaOH in 300.0mL of 8.00% NaOH solution. The density of the solution is 1.09g/mL.

    14. Examples What volume of 12.9% KOH solution contains 40.0 grams of KOH? The density of the solution is 1.11g/mL.

    16. Molarity Moles of solute per liter of solution M = moles of solute liters of solution Abbreviate: mol/L A 2.0M (molar) solution of NaCl contains 2.0 moles of NaCl in 1.0 liter of solution. Or 4.0 moles of NaCl in 2.0 liters of solution etc.

    18. Examples Calculate the molarity of a solution prepared by dissolving 12.5 g of sulfuric acid (H2SO4) in sufficient water to give 1.75L of solution. Determine the mass of Ca(NO3)2 required to prepare 3.50 L of a solution that is 0.800M in Ca(NO3)2.

    21. Examples The density of commercial hydrochloric acid is 1.185g/mL. It is 36.31% HCl by mass. Determine its molarity.

    24. Using Solutions in Chemical Reactions Na2SO4 + BaCl2 ? BaSO4 + 2NaCl Determine the volume of 0.500M BaCl2 solution required for complete reaction with 4.32g of Na2SO4 according to the reaction

    26. Using Solutions in Chemical Reactions: Titration The process, operation, or method of determining the concentration of a substance in solution by adding to it a standard reagent of known concentration in carefully measured amounts until a reaction of definite and known proportion is completed, as shown by a color change or by electrical measurement, and then calculating the unknown concentration.

    27. Dilution of Solutions A more concentrated solution can be diluted with solvent to produce a solution that is more dilute. The moles of solute will be the same in both solutions. However, since the volumes will be different the molarity will change. M1V1 = M2V2

    28. Examples Determine the volume of 18.0M H2SO4 required to prepare 2.50 liters of 2.40M H2SO4 solution. If 50.0 ml of 12.0M HCl is diluted to a final total volume of 200.0 ml, determine the final concentration of the solution.

    31. Molality Moles of solute per kilogram of solvent m = moles of solute kg of solvent A 1.0m (molal) solution of NaCl contains 1.0 moles of NaCl in 1.0 kg of water.

    33. Examples Determine the molality of a solution prepared by dissolving 400 g of NaOH in 500mL of water. The density of water is 1.00g/mL. M.W. of NaOH is 40.0 amu. Calculate the molarity and molality of an aqueous solution that is 10.0% glucose. The density of the solution is 1.04g/mL. M.W. of glucose = 180.0amu.

    36. Examples Determine the molality of a solution that contains 7.25 g of benzoic acid, C6H5COOH, in 200 mL of benzene, C6H6. The density of benzene is 0.879g/mL and the molar mass of benzoic acid = 122 amu.

    38. Mole Fraction Moles of component A per total moles of all components in the solution. XA = molesA molesA + molesB + molesC XB = XA + XB + XC + ……… = 1

    39. Examples What are the mole fractions of glucose and water in 10.0% glucose solution.

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