Workbook 3 Problems for Exam 3

Chem 1A Dr. White 1 Workbook 3 Problems for Exam 3 3-1: Types of Solids 1. What type of crystal will each of the following substances form in its solid state? Choices to consider are ionic, metallic, network, noble gas, nonpolar molecular, polar molecular. (a) C 2 H 6 (b) Na 2 O (c) SiO 2 (d) CO 2 (e) NO (f) NaNO 3 (g) Al (h) C(diamond) (I) SO 2 2. Circle all the compounds in the following list which would be expected to form intermolecular hydrogen bonds in the molecular solid state: (a) CH 3 OCH 3 (b) CH 4 (c) HF (d) CH 3 COOH (e) Br 2 (f) CH 3 OH (dimethyl ether) (acetic acid) (methanol) 3. Specify the predominant Intermolecular Force of Attraction (Dispersion, Dipole-Dipole, or H-Bonding) or Bonding Force (ionic, metallic, or network covalent) involved for each solid in the space immediately following the substance. Then in the last column, indicate which member of the pair you would expect to have the higher melting point. Solid #1* Predominant Force Solid #2 Predominant Force Substance with Higher Melting Point (a) HCl I 2 (b) CH 3 F (c) H 2 O CH 3 OH H 2 S (d) SiO 2 SO 2 (e) Fe (f) CH 3 OH Kr CuO (g) NH 3 CH 4 (h) HCl (i) Diamond NaCl Cu * Note: Many of the compounds given are not solids at room temperature. But if you cool them down to a low enough temperature, eventually they will become solids. 3-2: IMFs 1. List the dominant IMFA in each of the following substances: Pentane (C 5 H 12 ) Methanol (CH 3 OH) water ammonia hydrochloric acid carbon tetrachloride CHCl 3 (chloroform) 2. If the two substances listed below in each case were interacting, what IMFA would dominate the interaction? NH 3 and HF XeF 4 and Br 2 Hexane (C 6 H 14 ) and pentane (C 5 H 12 ) HCl and H 2 O 3. Circle all of the species below that can form a hydrogen bond. Explain why the other species couldn't hydrogen bond. C 2 H 6 CH 3 NH 2 HCl CH 3 CH 2 CH 2 OH 4. Which of the following species should have the weakest intermolecular forces? Why? (rank from weakest to strongest) Cl 2 I 2 N 2 H 2 O 5. Which of the following materials should have the highest boiling point? Why? (rank the boiling points from highest to lowest) H 2 Se H 2 S H 2 Po H 2 Te 6. Which of the following materials should be the most viscous? Why? HOCH 2 CH 2 OH Br 2 CH 3 (CH 2 ) 3 CH 3

Chem 1A 2 Dr. White 7. Which of the following substances should have the highest melting point? Why? I2 H 2O F2 NaI 8. Determine the species that you would expect to have the desired property and briefly explain your choice. More viscous Property Species CH3CH2CH2CH2OH or CHCl3 Higher Boiling Pt Br2 or H2Se More likely to liquefy (or condense) Br2 or F2 Greater Capillary action in a glass (SiO2) tube H2O or Hg 3-3: Solids: Unit Cells and Semicomductors 1. Platinum, Pt, crystallizes in the unit cell structure shown to the left. a. Identify this unit cell type. b. How many net atoms are contained in one unit cell of this type? c. What would the edge length be for this unit cell type in terms of r? 3 d. If Pt has a density of 21.45 g/cm, what is the edge length of the unit cell in pm? 2. The metallic radius of manganese is 127 pm. If it crystallizes in a body3 centered cubic unit cell, what is its density in g/cm? 3 3. The density of sodium metal is 0.971 g/cm and the unit cell length is 428.5 pm. Determine the unit cell of sodium metal. (Hint: Determine the # of atoms/unit cell) 3 4. Iron adopts a BCC unit cell with an atomic radius of 0.124 nm. What is its density in g/cm? 3 5. Iridium adopts a FCC unit cell and a density of 22.56 g/cm. What is the radius of an Ir atom in nm? 6. Shown below is single unit cell To which cubic structure do the calcium ions adopt? Also, how many atoms of each are contained in a unit cell. What is the simplest whole number ratio? - 7. Calculate the edge length of the unit cell of NaCl on the basis on of the average radii of Cl + + (181 pm) and Na (98 pm) assuming that the corner Cl ions and the Na along the edges touch (see figure to the right). 8. Lithium chloride crystallizes with the same structure as NaCl except that the chloride anions touch along the face diagonal of the unit cell. If the unit cell edge is 513 pm, what is the radius of a chloride ion? (hint use the Pythagorean theorem what is the face diagonal equal to given the first sentence of this problem?) 9. Iron may crystallize in either the body-centered cubic or face-centered cubic structures. Calculate the densities of iron in these structures, given that the radius of an iron atom is 124 pm.

Chem 1A Dr. White 3 10. You are given a small bar of an unknown metal. You find the density of the metal to be 10.5 g/cm 3. An X- ray diffraction experiment measures the edge of the face-centered cubic unit cell as 4.09 Å. (1 Å = 1 x 10-10 m). Identify this metal. 11. Draw band diagrams for conductors, semiconductors, and insulators and explain how the band gap influences the conductivity using these diagrams. 12. Define p-type and n-type semiconductors, and list what elements would be appropriate to add to Si to make each type of extrinsic semiconductor. 13. Describe how the band structure of Si is affected by addition of elements to make p- and n-type semiconductors. Draw pictures of band structures to illustrate these changes. 14. In each of the following, determine which has the highest melting point. Explain your answer. a. Rb (s) or Sr (s) b. Zn (s) or Cd (s) c. SrCl 2 (s) or RbCl (s) d. MgBr 2 (s) or MgI 2 (s) 3.4 Heating Curves and Phase Diagrams 1. Liquid ammonia (boiling point = -33.4 C) can be used as a refrigerant and heat transfer fluid. How much energy, in kj, is needed to heat 25.0 g of NH 3 (l) from -65.0 C to -12.0 C?? NOTE: J/(g K) is the same as J/(g C) 2. Diethyl ether, used as a solvent for extraction of organic compounds from aqueous solutions, has a high vapor pressure which makes it a potential fire hazard in laboratories in which it is used. How much energy is released when 100.0 g is cooled from 53.0 C to 10.0 C? boiling point 34.5 C heat of vaporization 351 J/g specific heat capacity, (CH 3 ) 2 O(l) 3.74 J/(g K) specific heat capacity, (CH 3 ) 2 O(g) 2.35 J/(g K) 3. Use the phase diagram below to answer the questions below. a) What two phases are in equilibrium at the circle? 2500 b) Consider the point represented by the Solid Liquid triangle. Pressure in torr 2000 1500 1000 500 Gas At the Pressure of the triangle (keeping it constant), what is the temperature required to convert this substance into a solid? At the Temperature of the triangle (keeping it constant), what is the pressure required to convert this substance into a gas? 10 20 30 40 50 60 c) What is the approximate normal melting point of this substance? Is there one? Why or why not? 70 d) What is the approximate boiling point of this Temperature in 0 C substance at a pressure of 2 atmospheres? e) Identify the triple point and the critical point of this substance (write in on graph). f) Under standard thermodynamic conditions (1 atm, 25 C), what is the physical state of this material?

Chem 1A Dr. White 4 3-5 Properties of Solutions 1. Determine if each species listed below would be a strong electrolyte, weak electrolyte, or a nonelectrolyte. a. Glucose (C 6 H 12 O 6 ) b. KCl c. HCl d. HNO 2 2. How many moles of lithium chlorate are in 153 ml of a 1.764 M solution? 3. Determine the mass (g) of solute required to form 250.0 ml of a 0.250 M sodium cyanide. 4. A 25.0 ml sample of concentrated HCl (12.0 M) is diluted to a final volume of 750.0 ml. What is the molarity of the final solution? 5. A solution is prepared by dissolving 516.5 mg of oxalic acid (C 2 H 2 O 4 ) to make 100.0 ml of solution. A 10.00 ml portion is then diluted to 250.0 ml. What is the molarity of the final solution? 6. An experiment calls for the use of 25.0 ml of a 0.100 M glucose solution. All you have available is a 2.00 M stock solution of glucose. How would you prepare the desired solution? 3-6 Concentration Units 1. An aqueous solution of H 3 PO 4 is 25.0% by mass and has a density of 1.39 g/ml. Fill out the following table. Solute = Solvent = Relative Mass (g) # Moles Volume (ml) Solute ---- Solvent ---- Solution (total) Now, calculate the following concentrations: (a) Solution Molarity (M). (b) Solution Normality (N). (c) Solution molality (m). (d) Mole fraction of H 3 PO 4. 2. A 0.750 M solution of H 2 SO 4 in water has a density of 1.046 g/ml at 20 C. Fill out the table below. Solute = Solvent = Mass (g) # Moles Volume (ml) Solute ---- Solvent ---- Solution (total) Now, calculate the following concentrations: (a) Solution Normality (N). (b) Solution molality (m). (c) Mole fraction of H 2 SO 4. 3. An aqueous 1.06 m solution of CaCl 2 (MM = 110.9 g/mol) has a density of 1.07 g/ml. Determine the molarity and mass percent CaCl 2. (feel free to use a table!) 4. Suppose you want to make another batch of CaCl 2 solution with the same molality (m) as in the problem above. You only have 7.00 g of CaCl 2 (s) available. What mass of water (in grams) should you use? 3-7 Solubility 1. Predict whether each of the following is more likely to dissolve in CCl 4 or water. a. C 7 H 16 b. Na 2 SO 4 c. HCl d. I 2 2. Which of the following in each pair is likely to be the more soluble in water:

Chem 1A Dr. White 5 3. Which of the following in each pair is likely to be more soluble in CCl 4 : 5. State the dominant IMF between each solute and solvent pair given below, and then indicate whether or not you would predict the solute to dissolve well in the solvent. Will solute Solute Solvent Dominant IMF between solute & solvent dissolve well in solvent? NH 3 (l) H 2 O (l) Br 2 (l) H 2 O (l) C 9 H 19 OH (l) C 6 H 14 (l) b) Would you expect ammonia to dissolve in water? Why or why not? c) Would you expect iodine to dissolve in water? Why or why not? d) What would be a good solvent in which you could dissolve I 2? Explain. 3-8 Vapor Pressure and Colligative Properties 1. The vapor pressure of acetone (CH 3 COCH 3 ) at 25 C is 271 torr and that of methanol is 143 torr. What is the vapor pressure of a solution that is 27.5 % methanol by mass in acetone as the solvent? 2. A solution has a 1:4 ratio of pentane to hexane (this means that if there are 5 total moles, 1 mole would be pentane and 4 would be hexane). The vapor pressures of the pure hydrocarbons at 20 C are 441mmHg for pentane and 121mmHg for hexane. What are the partial pressures of the 2 hydrocarbons above the solution? What is the total pressure above the solution? 3. Calculate the vapor pressure of a solution prepared by dissolving 175 g of glucose (MM = 180.16 g/mol) into 350.0 ml water at 75 C. The vapor pressure of pure water st 75 C is 289.1 mm Hg and its density is 0.97489 g/ml. 4. If equal amounts of each of the following nonvolatile solutes is added to water to make a solution, which solute will result in larger changes in vapor pressure, boiling point and freezing point? Why? (You should be able to answer this question without doing any math) i) Glucose (C 6 H 12 O 6 ) ii) NaCl iii) MgCl 2 iv) K 3 PO 4 5. 0.134 moles of an unknown solute is dissolved in 50.0 g of water to make a solution with a boiling point of 104.1 C. Which of the following solutes was used to make this solution? Fructose(C 6 H 12 O 6 ), sodium bromide, calcium carbonate, potassium sulfate or aluminum chloride. Show all our work and briefly explain your choice. 6. A physician studying a type of hemoglobin formed during a fatal disease dissolves 21.5 mg of the protein in water at 5.0 C to make 1.5 ml of solution in order to measure its osmotic pressure. At equilibrium, the solution has an osmotic pressure of 3.61 torr. What is the molar mass (g/mol) of the hemoglobin? 7. Calculate the molality and experimental van t Hoff factor i for an aqueous solution with 1.00 mass % NaCl, freezing point = -0.593 C.

Chem 1A Dr. White 6 8. Use the following curves to answer the following questions: Vapor Pressure (mm Hg)! 200! 180! 160! 140! 120! 100! 80! 60! 40! 20! Vapor Pressure Curves for Water & an Aq. Solution of 280! 300! Glycerol! 320! 340! 0! Temperature (K)! a. Which curve corresponds to pure water? b. Which curve corresponds to the solution? c. What is the vapor pressure lowering ΔP at 320 K? d. What is the mole fraction of solute in this solution at 320 K? (What is vapor pressure of pure water at 320 K?) e. What is the boiling point elevation ΔT b at 130 mm Hg? Label this on the graph. f. If K b for water is 0.512 C/m, what is the molality of this solution? g. How much glycerol (C 3 H 8 O 3 ) in grams must be added to 1.00 kg of water to elevate the boiling point by 10.0 C? 6. In the phase diagram below, label the regions corresponding to the solid, liquid and gas phases of this particular solvent. Then draw in how this diagram would change upon addition of a nonvolatile solute to this solvent, labeling all the new curves you draw in and any important quantities, like ΔP, ΔT bp, ΔT fp. Pressure (atm) Temperature

Chem 1A Dr. White 7 3-1: Types of Solids 1. (a) C 2 H 6 nonpolar (b) Na 2 O ionic (c) SiO 2 network (d) CO 2 nonpolar (e) N 2 O 5 polar (f) NaNO 3 ionic (g) Al metallic (h) C(diamond) network (I) SO 2 polar 2. (a) CH 3 OCH 3 (b) CH 4 (c) HF (d) CH 3 COOH (e) Br 2 (f) CH 3 OH (dimethyl ether) (acetic acid) (methanol) 3. Solid #1* Predominant Force Solid #2 Predominant Force Substance with Higher Melting Point (a) HCl DDA I 2 LDF HCl (b) CH 3 F DDA CH 3 OH H-bonding CH 3 OH (c) H 2 O H-bonding H 2 S DDA H 2 O (d) SiO 2 network covalent SO 2 DDA SiO 2 (e) Fe Metallic Kr LDF Fe (f) CH 3 OH H-bonding CuO Ionic CuO (g) NH 3 H-bonding CH 4 LDF NH 3 (h) HCl DDA NaCl Ionic NaCl (i) Diamond network covalent Cu Metallic Diamond 3-2: IMFs 1. Pentane - LDFs Methanol H-bonds water H-bonds ammonia H-bonds hydrochloric acid DDAs carbon tetrachloride - LDFs CHCl 3 - DDAs 2. NH 3 and HF XeF 4 and Br 2 Hexane and pentane HCl and H 2 O H-bonds LDFs LDFs DDAs 3. C 2 H 6 CH 3 NH 2 HCl CH 3 CH 2 CH 2 OH C 2 H 6 and HCl do not H-bond because there are no H s bonded to N, O, or F. 4. N 2 < Cl 2 < I 2 < H 2 O The weakest IMFs are in N 2 because it is nonpolar and only has LDFs. The LDFs are weaker than that in the other nonpolar molecules because it is the smallest and has the least number of electrons. 5. Highest b.p. will be H 2 Po, then H 2 Te, then H 2 Se, then H 2 S. All have dipole-dipole interactions. So, they are ranked from largest to smallest size since the largest atoms have the most electrons and thus are the most polarizable and will have stronger LDFs. 6. HOCH 2 CH 2 OH because it has the strongest IMFs (H-bonds). 7. NaI because it has the strongest forces holding the solid together. It has ionic bonds while the rest have IMFs. 8. Property Species More viscous CH 3 CH 2 CH 2 CH 2 OH or CHCl 3 It has H-bonds and CHCl 3 has weaker DDAs. Higher Boiling Pt Br 2 or H 2 Se It has DDAs and Br 2 only has weaker LDFs More likely to liquefy (or condense) Br 2 or F 2 It is larger and forms stronger LDFs than F 2. This means that it will want to stay a liquid and not go to the gas phase Greater Capillary action in a glass (SiO 2 ) tube H 2 O or Hg Water forms strong interactions with the glass tube, Hg does not.

Chem 1A Dr. White 8 3-3 Solids, Unit Cells, and Semiconductors 1. a. FCC b. 4.00 atoms c. 2 2r d. 392 pm 2. 7.23 g/cm 3 3. BCC 4. 7.90 g/cm 3 5. 0.136 nm 6. FCC; 4 Ca 2+ :8F - ; CaF 2 7. 558 pm 8. 181 pm 9. FCC: 8.60 g/cm 3 and BCC: 7.90 g/cm 3 10. MM = 108.2 g/mol, therefore Ag 11. A semiconductor is on the left, the insulator is in the middle and the conductor is on the right. The size of the band gap dictates the conductivity. The smaller the band gap, the more conductive the material. 12. N-type semiconductors are made by adding atoms with more valence electrons to an intrinsic semiconductor. A donor level is created between the valance and conductance bands. Therefore the band gap is decreases and conductivity increases. We can add P (or any element with more valence electrons) to Si to make an n-type semiconductor. P-type semiconductors are made by adding atoms with less valence electrons. Positive holes are created in the valence band. The electrons can migrate to these sites and the conductivity increases. We can add Al (or any other element with less valance electrons) to Si to make a p-type semiconductor. 13. e - e - e - See description in #12. The figure on the left is a p-type and the figure on the right is the n-type. 14. a. Sr has the higher melting point. Sr has a stronger metallic bond since it has more valence electrons which creates stronger metallic bonds. b. Zn has the higher melting point because it is smaller than Cd. This makes the bonds stronger and thus the metallic bond is stronger. This translates to a higher melting point. c. SrCl 2 has a higher melting point because it has stronger ionic bonds due to the higher charge of the strontium ion. d. MgBr 2 has a higher melting point because it has stronger ionic bonds due to the smaller size of the bromide ion. This allows the ions to get closer and form stronger ionic bonds. 3-4 Heating Curves and Phase Diagrams 1. 39.4 kj 2. -48.6 kj 3. a. solid and gas b. ~32 C and ~1300 torr

Chem 1A Dr. White 9 c. There is no normal malting point. The solid goes directly to a gas at this point. d. ~56 C e. Triple point is where all three lines intersect and Critical point it the highest P & T on the liquid-gas line. f. gas 3-5 Properties of Solutions 1. a. Nonelectrolyte b. Strong Electrolyte c. Strong Electrolyte d. Weak electrolyte 2. 0.270 mol 3. 3.06 g 4. 0.400 M 5. 2.295 x 10-3 M 6. Mix 1.25 ml of the 2.00 M glucose solution with enough water to make 25.0 ml total. 3-6 Concentration Units 1. 3.55 M, 10,6 N, 3.40 m, 0.0577 2. 1.50 N, 0.771 m, 0.0137 3. 1.01 M, 10.5% 4. 59.5 g H 2 O 3-7 Solubility 1. a. CCl 4 b. water c. water d. CCl 4 2. B, B, B, B, A 3. A, B, A 4. a. NH 3 (g) H 2 O (l) H bonding yes Br 2 (l) H 2 O (l) LDFs No C 9 H 19 OH (l) C 6 H 14 (l) LDFs Yes b. I would expect ammonia (NH 3 ) to dissolve in water, as the interaction between ammonia and water (H bonding) is the same as the IMFs holding both water and ammonia together. Thus breaking apart both the solute and solvent is compensated for by the strong interaction between solute and solvent. c. Here I would not expect iodine to dissolve in water. Water has H-bonding and the nonpolar I 2 only has LDFs the new interaction between I 2 and water is not stronger than H bonding in water. So water will not want to mix with I 2. d. A nonpolar solvent, like hexane, would be a good solvent for I 2 like dissolves like. 3-8 Vapor Pressure and Colligative Properties 1.219 torr 2. pentane: 88.2 mmhg hexane: 96.8 mmhg Total: 185.0 mmhg 3.275 mmhg 4. The K 3 PO 4 would result in the largest changes in vapor pressure, boiling point and freezing point. The colligative properties all depend on the amount of solute present in solution. When dissolved in solution, the K 3 PO 4 dissociates into the most particles per mole of compound, 4 particles. Another way to say this is that i = 4 for K 3 PO 4. 5. since i = 3, Only potassium sulfate, K 2 SO 4, would be expected to break into 3 ions and thus have i = 3. 6. 6.9 x 10 4 g/mol 7. 0.174 m; i = 1.83 8. a. The solid curve belongs to pure water, it shows higher vapor pressure at any temperature. b. The dashed curve belongs to the solution, it shows lower vapor pressure at any temperature. c. Just estimate the difference between the two curves at 320 K, this looks like ~12 mm Hg d. From curve, vapor pressure of pure water at 320 K is ~80 mm Hg, so χ = 0.15 e. Just estimate the difference between the two curves at 130 mm Hg, this looks like ~2 K = 2 C to me f. 4 m g. 1.8 x 10 3 g glycerol h.

Chem 1A Dr. White 10 9.