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1212+Spring++08+Final+Study+Guide-2 - CHEM 1212 Spring...

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CHEM 1212 Spring Semester 2008. Study Questions to help with Revision for Final Exam. Silberberg 4 th Edn Questions. from each Chapter. Note: some of these questions are previous Homework Assignments, some are new to you. Study Give them all equal attention in your studying, and make sure you could answer them correctly if faced with them in an exam, and without your Text Book at hand! Chapter 12 12.35 Polarity refers to a permanent imbalance in the distribution of electrons in the molecule. Polarizability refers to the ability of the electron distribution in a molecule to change temporarily. The polarity affects dipole-dipole interactions, while the polarizability affects dispersion forces. 12.68 Water is a good solvent for polar and ionic substances and a poor solvent for nonpolar substances. Water is a polar molecule and dissolves polar substances because their intermolecular forces are of similar strength. Water is also able to dissolve ionic compounds and keep ions separated in solution through ion- dipole interactions. Nonpolar substances will not be very soluble in water since their dispersion forces are much weaker than the hydrogen bonds in water. A solute whose intermolecular attraction to a solvent molecule is less than the attraction between two solvent molecules will not dissolve because its attraction cannot replace the attraction between solvent molecules. 12.69 A single water molecule can form 4 hydrogen bonds. The two hydrogen atoms form a hydrogen bond each to oxygen atoms on neighboring water molecules. The two lone pairs on the oxygen atom form hydrogen bonds with hydrogen atoms on neighboring molecules. Chapter 13. 13.29 This compound would be very soluble in water. A large exothermic value in H solution (enthalpy of solution) means that the solution has a much lower energy state than the isolated solute and solvent particles, so the system tends to the formation of the solution. Entropy that accompanies dissolution always favors solution formation. Entropy becomes important when explaining why solids with an endothermic H solution (and higher energy state) are still soluble in water. 13.92 The magnitude of boiling point elevation is directly proportional to molality. a) Molality of CH 3 OH = ( 29 ( 29 3 3 3 2 3 10.0 g CH OH 1 mol CH OH 10 g 100. g H O 32.04 g CH OH 1 kg �� = 3.1210986 = 3.12 m CH 3 OH Molality of CH 3 CH 2 OH = ( 29 ( 29 3 3 2 3 2 2 3 2 20.0 g CH CH OH 1 mol CH CH OH 10 g 200. g H O 46.07 g CH CH OH 1 kg �� = 2.1706 = 2.17 m CH 3 CH 2 OH The molality of methanol, CH 3 OH, in water is 3.12 m whereas the molality of ethanol, CH 3 CH 2 OH, in water is 2.17 m. Thus, CH 3 OH/H 2 O solution has the lower freezing point. b) Molality of H 2 O = ( 29 ( 29 2 2 3 2 10.0 g H O 1 mol H O 1.00 kg CH OH 18.02 g H O = 0.5549 = 0.55 m H 2 O Molality of CH 3 CH 2 OH = ( 29 ( 29 3 2 3 2 3 3 2 10.0 g CH CH OH 1 mol CH CH OH 1.00 kg CH OH 46.07 g CH CH OH = 0.21706 = 0.217 m CH 3 CH 2 OH The molality of H 2 O in CH 3 OH is 0.555 m , whereas CH 3 CH 2 OH in CH 3 OH is 0.217 m . Therefore,
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