Chapter 13 - CHAPTER 13 PHYSICAL PROPERTIES OF SOLUTIONS...

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CHAPTER 13 PHYSICAL PROPERTIES OF SOLUTIONS 13.7 CsF is an ionic solid; the ion-ion attractions are too strong to be overcome in the dissolving process in benzene. The ion induced dipole interaction is too weak to stabilize the ion. Nonpolar naphthalene molecules form a molecular solid in which the only interparticle forces are of the weak dispersion type. The same forces operate in liquid benzene causing naphthalene to dissolve with relative ease. Like dissolves like. 13.8 Strategy: In predicting solubility, remember the saying: Like dissolves like. A nonpolar solute will dissolve in a nonpolar solvent; ionic compounds will generally dissolve in polar solvents due to favorable ion-dipole interactions; solutes that can form hydrogen bonds with a solvent will have high solubility in the solvent. Solution: Strong hydrogen bonding (dipole-dipole attraction) is the principal intermolecular attraction in liquid ethanol, but in liquid cyclohexane the intermolecular forces are dispersion forces because cyclohexane is nonpolar. Cyclohexane cannot form hydrogen bonds with ethanol, and therefore cannot attract ethanol molecules strongly enough to form a solution. 13.9 The order of increasing solubility is: O 2 < Br 2 < LiCl < CH 3 OH . Methanol is miscible with water because of strong hydrogen bonding. LiCl is an ionic solid and is very soluble because of the high polarity of the water molecules. Both oxygen and bromine are nonpolar and exert only weak dispersion forces. Bromine is a larger molecule and is therefore more polarizable and susceptible to dipole induced dipole attractions. 13.10 The longer the C C chain, the more the molecule "looks like" a hydrocarbon and the less important the OH group becomes. Hence, as the C C chain length increases, the molecule becomes less polar. Since “like dissolves like”, as the molecules become more nonpolar, the solubility in polar water decreases. The OH group of the alcohols can form strong hydrogen bonds with water molecules, but this property decreases as the chain length increases. 13.13 Percent mass equals the mass of solute divided by the mass of the solution (that is, solute plus solvent) times 100 (to convert to percentage). (a) 5.50 g NaBr 100% 78.2 g soln ×= 7.03% (b) 31.0 g KCl 100% (31.0 152)g soln + 16.9% (c) 4.5 g toluene 100% (4.5 29)g soln + 13% 13.14 Strategy: We are given the percent by mass of the solute and the mass of the solute. We can use Equation (13.2) of the text to solve for the mass of the solvent (water). Solution: (a) The percent by mass is defined as mass of solute percent by mass of solute 100% mass of solute + mass of solvent
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CHAPTER 13: PHYSICAL PROPERTIES OF SOLUTIONS 258 Substituting in the percent by mass of solute and the mass of solute, we can solve for the mass of solvent (water). 5.00 g urea 16.2% 100% 5.00 g urea + mass of water (0.162)(mass of water) = 5.00 g (0.162)(5.00g) mass of water = 25.9 g (b) Similar to part (a), 2 2 26.2 g MgCl 1.5% 100% 26.2 g MgCl + mass of water mass of water = 1.72 × 10 3 g 13.15 (a) The molality is the number of moles of sucrose (molar mass 342.3 g/mol) divided by the mass of the solvent (water) in kg.
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Chapter 13 - CHAPTER 13 PHYSICAL PROPERTIES OF SOLUTIONS...

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