Lect+9+08 - Solutions Chapter 11 Solution Concentrations...

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Solutions Chapter 11 Solution Concentrations Making solutions An understanding of Strong Electrolytes Plain language understanding of Colligative properties Vapor Pressure Lowering (Raoult’s law, ideal solutions) Boiling point elevation Freezing point depression Henry’s law
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Using Molar Concentration (Need to know molar mass of solute) Molarity is the conversion factor between the volume of a solution and the number of moles of solute it contains moles solute = M * V moles solute = (moles solute/L soln) * (L soln) What volume of 0.153 M CaSO 4 contains 0.100 mole of solute? 0.100moles solute = 0.153 (moles-L -1 ) * V (L) V = 0.654 L 50.0 mL of 0.153 M CaSO 4 is mixed with 75.0 mL of 0.120 M CaSO 4 ? What is the concentration of the final solution? total moles CaSO 4 = .0500 * 0.153 + 0.0750 * .120 = 0.0167 moles total volume = 50.0 + 75.0 mL (assumes volumes are additive Final concentration = 0.0167 moles / 0.1250 L = 0.134 M
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Making a solution with a definite Molarity Weigh Solute Transfer Solute Add Solvent until desired Solution Volume reached
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Using Wt% Concentrated phosphoric acid is 90.0% H 3 PO 4 . What is the molarity of H 3 PO 4 in this solution? mole H 3 PO 4 1 L soln = 0.900 kg H 3 PO 4 1.00 kg soln 1 L soln kg soln ρ need density mole H 3 PO 4 kg H 3 PO 4 MW 1 ρ = 1.328 kg soln 1 L soln gives: Molarity of H 3 PO 4 = 12.1 M
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Ions dissolved in water are
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This note was uploaded on 04/01/2008 for the course CHEM 230 taught by Professor Sharp during the Winter '08 term at University of Michigan.

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Lect+9+08 - Solutions Chapter 11 Solution Concentrations...

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