CH101 Lec10

CH101 Lec10 - CHAPTER 10: Solutions Heterogeneous- a...

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CHAPTER 10: Solutions Heterogeneous - a mixture that is not uniform throughout. Homogeneous - a mixture that is uniform throughout. -The solute is dissolved in the solvent . -There can be more than one solute in a solution -If only one liquid is present, it is the solvent. -If more than one liquid is present, the liquid there is more of is the solvent. -There can only be one solvent in a solution Solution = Solute(s) + Solvent Solutions are homogeneous.
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Solute(s) Solute(s) Solvent Solvent Soft drink Substance Gas (CO 2 ) Solid (C 12 H 22 O 11 ) Liquid (H 2 O) Air Gas (O 2 ) Gas (N 2 ) Sterling silver Solid (Cu) Solid (Ag) Vinegar Examples of Solutions Sea Water Solid (NaCl)
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Section 10.1 Concentration Measure of the amount of solute in the solvent. Consider three kinds: Molarity (mol/L or M)- moles of solute/L of solution Percent by mass (% m/m)- g solute/100 g solution (Commonly used for solid solutions) Percent by volume (% v/v)- mL solute/100 mL of solution (Commonly used for liquid/liquid solutions) “Proof” = twice the % v/v
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Some practice with molarity: Calculate the molarity of a solution made with 0.24 mol MgCl 2 in 200 mL of water. What are [Mg 2+ ] and [Cl - ] in this solution? How many grams of KCl would it take to make 25 mL of a 0.1 M solution? M = mol/L = (0.24 mol)/(0.200 L) = 1.2 M = [MgCl 2 ] MgCl 2 Mg 2+ + 2Cl - Each mole of MgCl 2 dissolves to give 1 mol Mg 2+ and 2 mol Cl 1- [Mg 2+ ] = [MgCl 2 ] = 1.2 M [Cl - ] = 2[MgCl 2 ] = 2.4 M (0.1 mol KCl )(0.025 L)(74.5 g KCl ) = 0.186 g KCl ( L solution ) ( mol KCl )
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Density Density (d) - a measure of the mass per unit volume of a substance. d water = 1 g/mL 1 mL of water weighs 1 g! d lead = 11.3 g/mL 1 mL of lead weighs 11.3 g! So, most dense to least dense: lead >> carbon >> water Dense substances sink to the bottom of a mixture: increasing density
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Solvation For the reaction : A(s) A(aq) H solution Enthalpy of solution A substance is soluble if solution or solution is negative.
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Solvation Think about the process of dissolving a solid: breaking IM forces or ionic bonds in solute H solute >0 + Making new IM interactions between solvent and solute H mixing <0 Breaking IM forces in the solvent ( H solvent >0) H solution = H solute + H solvent + H mixing
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Solvation H solution = H solute + H solvent + H mixing endothermic exothermic Soluble if
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This note was uploaded on 03/29/2008 for the course CH 101 taught by Professor Bigham during the Fall '08 term at N.C. State.

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CH101 Lec10 - CHAPTER 10: Solutions Heterogeneous- a...

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