1. a. Classify each of the following as molecular, ionic or other. CF2Cl2 b. CO2 KF HNCl2 MgSO4 Xe PF3 HOCl Show the Lewis structure of each substance you classified as molecular. For ionic compounds, write charges on the cation and anion. Give the total number of electrons in each compound. Draw and name the VSEPR shape for molecular compounds. Indicate whether the substance is polar or nonpolar. Name the dominant intermolecular force in each substance. Place the compounds in order of increasing predicted melting point. c. d. e. f. g. a. molecular molecular ionic molecular ionic other 1+ 1– 1 N 5 e– 1 H 1 e– 2 Cl 14 e– 20 e– 2+ 2– 54 e– b., c., d., e. 1 C 4 e– 2 F 14 e– 2 Cl 14 e– 32 e– 1 C 4 e– 2 O 12 e– 16 e– O C 28 e– F H C N 22 e C Cl F F P N C H O Cl Cl nonpolar P F N Cl C O H F F Cl 26 e– O F F trigonal pyramid trigonal pyramid tetrahedral Cl H 42 e– F 1 e– 6 e– 7 e– 14 e– F F 42 e– O linear O Cl C 1H 1O 1 Cl Cl Cl O 5 e– 21 e– 26 e– P Cl – F 58 e– 1P 3F 60 e– O F Cl molecular molecular F F Cl Cl H Cl bent (109.5o) O H Cl nonpolar polar polar polar polar f. dipoledipole London ion-ion H-bond ion-ion London dipoledipole H-bond Xe PF3 CF2Cl4 HOCl HNCl2 KF MgSO4 g. CO2 2. a. Classify each of the following as molecular, ionic or other. CF4 b. SeF4 NaBF4 H2NOH CaS Ar SO3 HOOH Show the Lewis structure of each substance you classified as molecular. For ionic compounds, write charges on the cation and anion. Give the total number of electrons in each compound. Draw and name the VSEPR shape for molecular compounds. Indicate whether the substance is polar or nonpolar. Name the dominant intermolecular force in each substance. Place the compounds in order of decreasing predicted boiling point. c. d. e. f. g. a. molecular molecular ionic molecular ionic other 5 e– 3 e– 6 e– 14 e– 2+ 2– 18 e– b., c., d., e. 1 C 4 e– 4 F 28 e– 32 e– 1 Se 6 e– 4 F 28 e– 34 e– F F 1+ 1– 52 e– 1N 3H 1O F F 36 e– H F N O H F H C Se F F F N trigonal pyramid O bent see-saw F C Se F F F F F nonpolar f. London H 1 e– 6 e– 7 e– 14 e– O O O H O O trigonal planar H both bent (109.5o) O O O H H S N O HH O nonpolar nonpolar polar polar polar dipoledipole ion-ion H-bond ion-ion London London H-bond NaBF4 H2NOH HOOH SeF4 CF4 SO3 Ar g. CaS H O H S HH O 18 e– O N F F O S 70 e F F 1H 1O 1 Cl 40 e– 18 e– – O tetrahedral O H F 42 e– molecular O Se F F 6 e– 18 e– 24 e– 1S 3O F C F molecular 3. Classify the following solids as molecular, network (covalent), metallic, ionic or atomic. H a. benzene, H C C C H b. c. d. e. f. g. h. i. j. H molecular C C C diamond H krypton Cd copper(II) chloride ice S8 Mg3(PO4)2 quartz (SiO2) P4 H network atomic metallic ionic molecular molecular ionic network molecular 4. For the two dimensional arrays shown below, draw a set of lattice points, a single primative unit cell and a centered unit cell. 5. How many whole atoms are in a primitive cubic unit cell? 1 6. How many whole atoms are in a body-centered cubic unit cell? 2 7. How many whole atoms are in a face-centered cubic unit cell? 4 8. Polonium crystallizes in a sc unit cell with a density of 9.20 g/cm3. Calculate the radius of a Po atom. 209 g Po 1 mole Po 1 atom Po 1 cm3 x x x 3.773 x 1023cm3 V 23 1 mole Po 6.022 x 10 atoms Po 1 unit cell 9.20 g Po a 3 V a r 2 3 3.773 x 1023cm3 3.354 x 108cm 3.354 x 10 8 2 1.68 x 10–8 cm = 1.68 Å 9. Copper crystallizes in a fcc unit cell, and has an atomic radius of 128 pm. Calculate the density of copper. 63.546 g Cu 1 mole Cu 4 atoms Cu x x = 4.2209 x 1022 g/unit cell 23 1 mole Cu 6.022 x 10 atoms Cu 1 unit cell a 2 4r 4(1.28 x 108 cm) r = ; a = = = 3.6204 x 108 cm 4 2 2 V = a3 = (3.6204 x 10–8 cm)3 = 4.7453 x 10–23 cm3 density = 10. mass of unit cell 4.2209 x 1022 g/unit cell = = 8.89 g/cm3 volume of unit cell 4.7453 x 1023 cm3 /unit cell The radius of iron is 124 pm, and it crystallizes in a bcc unit cell, what is the volume of the unit cell. r = a 3 4r 4(124 pm) ; a = = = 286.365 pm 4 3 3 V = a3 = (286.365 pm)3 = 2.35 x 107 pm3 11. For the heating curve below: a. b. c. d. e. f. Calculate the energy released when 12.76 g of CH2FCF3 goes from 0.0 oC to –75.0 oC. normal mp CH2FCF3(s) normal bp CH2FCF3(ℓ) Hvaporization CH2FCF3(ℓ) s CH2FCF3(ℓ) s CH2FCF3(g) -4000 -3500 -3000 –101.0 oC –26.6 oC 22.02 kJ/mole 1.423 J/goC 0.875 J/goC -2500 -2000 -1500 -1000 -500 0 -5 -15 -25 Temperature (oC) 12. (name of the energy) Hfusion (name of the energy) Hvaporization Temperature (oC) Heat (J) (name of phase change) melting (name of phase change) boiling -35 -45 -55 -65 -75 Heat (J) 1. q = msT = 12.76 g x 0.875 J/g∙oC x (–26.6 – 0.0 oC) = –296.989 J 2. H = –22020 J/mole x 12.76 g x 1 mole/102.03 g = –2753.849 J 3. q = msT = 12.76 g x 1.423 J/g∙oC x (–75.0 – (–26.6 oC)) = –878.822 J qtotal = (–296.989 J) + (–2753.849 J) + (–878.822 J) = –3929.66 J = –3.930 kJ 13. Given the phase diagram of xenon, Xe, answer the following: a. b. c. d. 14. Estimate the normal boiling point of xenon. Estimate the normal melting point of xenon. Estimate the vapor pressure of Xe(ℓ) at –110 oC. Is Xe(ℓ) or Xe(s) more dense? Explain. –107 oC –112 oC 0.8 atm Xe(s), the mp line slopes to the right. Calculate the difference in solubility of nitrogen (mole fraction in air = 0.78) in blood at sea level (P = 1.0 atm) versus 10 m underwater (P = 2.0 atm). Assume that blood is identical to water and kN2 = 7.0 x 10–4 mole/L·atm. CN2 = kPN2 = 7.0 x 10–4 mole/L·atm x (1.0 atm x 0.78) = 5.5 x 10–4 M N2 CN2 = kPN2 = 7.0 x 10–4 mole/L·atm x (2.0 atm x 0.78) = 1.1 x 10–3 M N2 15. Calculate the concentration of oxygen gas in water at sea level and 1.00 atm. The partial pressure of oxygen gas is 0.21 atm and Henry’s constant for oxygen (in water at RT) is 1.3 x 10–3 M·atm–1. CO2 = kPO2 = 1.3 x 10–3 mole/L·atm x (1.00 atm x 0.21) = 2.7 x 10–4 M O2 16. Is solid iodine (I2), more soluble in benzene (see #3), or chloroform, CH3Cl? benzene; I2 is nonpolar and so is benzene 17. Classify the following as molecular or ionic. What is the value of i for each when dissolved in water? a. b. c. d. e. 18. copper(II) chloride CH3F Mg3(PO4)2 sucrose, C12H22O11 benzene, C6H6 CuCl2, ionic, i molecular, i = ionic, i = 5 molecular, i = molecular, i = = 3 1 1 1 A solution is prepared by dissolving 5.00 g of glucose, C6H12O6, in 100.0 g of water. Calculate: a. the vapor pressure of water at 90 oC. The vapor pressure of pure water at 90 oC is 525.8 torr. 1 mole H 2O 100.0 g x 18.015 g H 2O = 0.995 1 mole C6 H12O6 1 mole H 2O + 100.0 g x 5.00 g C6 H12O6 x 180.156 g C6 H12O6 18.015 g H 2O Psol’n = waterPowater = (0.995)(525.8 torr) = 523.2 torr b. the boiling point of the solution. 1 mole C6 H12O6 180.156 g C6 H12O6 = 0.2775 m C6 H12O6 0.100 kg H 2O 5.00 g C6 H12 O6 x Tb = i Kbm = (1)(0.52 oC/m)(0.2775 m) = 0.14 oC higher Tb = 100.0 oC + 0.14 oC = 100.14 oC 19. a. Calculate the freezing point of a solution of 2.00 moles of NaCl in 100.00 mL of water (density of water = 1.00 g/mL). 2.00 mole NaCl = 20.0 m NaCl 0.10000 kg H 2 O Tf = i Kbm = (2)(1.86 oC/m)(20.0 m) = 74.4 oC lower Tf = 0.0 oC – 74.4 oC = –74.4 oC b. Calculate the freezing point of a solution of 2.00 moles of CaCl2 in 100.00 mL of water. 2.00 mole CaCl2 = 20.0 m CaCl2 0.10000 kg H 2 O Tf = i Kbm = (3)(1.86 oC/m)(20.0 m) = 112 oC lower Tf = 0.0 oC – 112 oC = –112 oC
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