Unit 9. Liquids and Solids - ANSWERS Upon successful completion of this unit, the students should be able to: 9.1 List the various intermolecular attractions in liquids and solids (dipole-dipole, London dispersion, hydrogen bond) describe the nature and relative strength of each, and identify which compounds exhibit which intermolecular attraction. 2 NN 2, NN, OC 2 C 2 O 2. London dispersion forces 3. N δ N δ+ -bond The partial positive hydrogen from one ammonia molecule is attracted to the negative unshared pair of electrons on a neighboring ammonia molecule. 9.2 Predict and explain how intermolecular attractions affect the physical properties of liquids and solids. C 3 C 2 O because it is the only one which exhibits hydrogen bonding. The others are roughly the same size as C 3 C 2 O so none will have large dispersion forces. 2. C 3 C 2 C 2 F 3. C 5 12 4. F because it is the only one which exhibits hydrogen bonding. Since the hydrogen bonding in F is so strong, even the considerably larger Br is not large enough to have dispersion forces which are stronger than the hydrogen bonds in F. 5. True 6. True 9.3 List the characteristics of molecular, network covalent, ionic and metallic solids. ionic solid 2. covalent network solid 3. Diamond and graphite are both network covalent. Diamond has a 3-D network of covalent bonds where graphite has a sheet-like 2-D network, each sheet between held to the other sheet by weak dispersion forces. Buckyball is a C 60 molecular compound shaped like a soccer ball.
4. hydrogen chloride 5. Ionic solid would typically have a higher melting point and higher boiling point and would also tend to be harder than a molecular solid. 6. a 9.4 Describe the relationship among temperature, vapor pressure, and boiling point. Boiling occurs when the vapor pressure of the liquid equals the prevailing atmospheric pressure 2. e; if ice and liquid water remain then we know the temperature has not risen yet and therefore the vapor pressure hasn t changed 9.5 Define heat of vaporization and heat of fusion and solve related problems. The heat required to melt a given amount of a substance (or the heat released upon freezing a given amount of substance). 2. F 3. To vaporize a substance (liquid to gas) we need to completely break the intermolecular forces between molecules to allow the molecules to move far apart from one another. To melt a substance (solid to liquid) we need to only partially break the intermolecular forces between molecules because molecules in the solid and liquid phases are both closely packed. 9.6 Differentiate between the terms melting, freezing, boiling, evaporation, condensation, sublimation, and deposition. b 2. the phase change from solid to gas 9.7 Draw and interpret a heating/cooling curve for a substance. 75 o C 62 o C T -9.0 o C -20 o C time
9.8 Describe the nature of a supercooled liquid and a superheated liquid. o a. Water which is below 0 C at 1 atm but still in the liquid state. b. T supercooled liquid time 9.9 Solve quantitative problems related to heat of fusion and heat of vaporization. mol 40.7 kj 9.00 g x x = 20.3 kj 18.02 g mol 2. You need to look up the heat of fusion for water: 6.02 kj/mol. 1000 g mol 6.02 kj 2.00 kg x x x = 668 kj kg 18.02 g mol 9.10 Define normal melting point and normal boiling point. The boiling point of a substance at 1 atm.
9.11 Interpret phase diagrams. b 2. d 3. 100 atm P triple point 1 atm 9.12 Define the terms critical point and triple point as it applies to phase diagrams. The only point at which all three phases (solid, liquid, and gas) are in equilibrium. Additional Unit 9 Sample Questions: a -100-50 0 50
2. c Explanation: P = 229.2 mm g x T = 25 V = 50.0 L n =? o C + 273 = 298 K 1atm 760 mm g = 0.306 atm PV = nrt PV (0.306 atm)(50.0 L) n = = RT L atm 0.0821 mol K mol 25.0 g x = 0.430 mol 58.1g ( 298 K) = 0.625 mol Since 25.0 g is only 0.430 mol of acetone, it must be that all the acetone has evaporated. If the number of mol was 0.625 or higher, this would mean that after some of the acetone had evaporated, it would eventually reach equilibrium between liquid and gas because there would be enough mol to reach the vapor pressure. 3. f 2. c 3. a 4. b