Lecture 1

# Lecture 1 - 2 SO 4 = 368g 1230 – 368 = 862 g H 2 O m =...

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Chapter 12 Chapter 12 Physical Properties of Solutions

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Types of solution Types of solution
Solution Composition Solution Composition W In 161, we measured concentration in several ways % composition ppm χ M W We’re introducing a new measure: molality (m) m = moles of solute per kilogram of solvent

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Example Example W Find the molality of a 1.00% w/w solution of CH 3 CH 2 OH. 1.00g CH 3 CH 2 OH × 1mol/46.07g = 2.17×10 -2 mol 2.17×10 -2 mol/0.10000kg = 0.217m
Relationship between M and m Relationship between M and m W For dilute solutions, M m W Relationship depends on d W For example, 3.75M H 2 SO 4 has a density of 1.230g/mL. What is its molality? 1L weighs 1230g 1L has 3.75 mol H

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Unformatted text preview: 2 SO 4 = 368g 1230 – 368 = 862 g H 2 O m = 3.75 mol/0.862kg = 4.35m Enthalpy of solution Enthalpy of solution W The energetics of solution are complicated Enthalpy to overcome intermolecular forces in solute Magnitude depends on compound Magnitude depends on compound Large for polar, ionic compounds Large for polar, ionic compounds Enthalpy to overcome intermolecular forces in solvent Enthalpy of solvation (hydration when water is solvent) W Net result: like dissolves like Polar substances dissolve in polar solvents Nonpolar substances dissolve in nonpolar solvents...
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Lecture 1 - 2 SO 4 = 368g 1230 – 368 = 862 g H 2 O m =...

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