Part 1: Multiple Choice (2 points each)Directions: Please circle the best answer for each of the following questions.13. Part 2: Short Answer14. Directions: Answer each of the following questions. Be sure to use complete sentences where appropriate. For full credit be sure to show all of your work.Grossmont College Chemistry 141 Exam 4 Fall 2015Name: ___________________________________Date: ________Instructor: Diana VanceExam 4Part 1: Multiple Choice (2 points each) Directions: Please circle the best answer for each of the following questions. 1. Which substance has the least negative lattice energy? a. MgI2b. MgBr2c. MgCl2d. MgF2e. all of the above 2. Indicate which aqueous solution has the fastestevaporation rate.a. 0.1 M KClb. 0.2 M Na2CO3c. 0.2 M NaCld. 0.1 M MgCl2e. 0.2 M MgCl23. Which statement is not correct? Determination of molarmass of an unknown sample by an osmotic pressuremeasurement requires that ________a. the solute dissolved in the solution is pure.b. the solute is a nonelectrolyte.c. the molecules of the solute do not pass through the semipermeable membrane.d. the mass of the solute dissolved in the solution is known in advance.e. the molar concentration of the solute in the solution is known in advance.4. In the cesium chlorideunit cell shown below,the cesium ions sit onthe corners of a cube.What is the name of thisunit cell?a. simple cubicb. chloride-centered cubicc. body-centered cubicd. face-centered cubice. cubic-centered5. The molecular orbital description for metal bonding isdifferent from that for diatomic molecules in that________a. there are no antibonding orbitals in the metal bonding description.b. the orbitals are so close in energy that they are referred to as bands.c. quantum theory no longer applies as the orbitals are continuous.d. the increased number of electrons results in each bond being stronger.e. All of the above are true.Page 1 of 10Grossmont College Chemistry 141 Exam 4 Fall 2015Name: ___________________________________Date: ________Instructor: Diana Vance6.Page 2 of 10Grossmont College Chemistry 141 Exam 4 Fall 2015Name: ___________________________________Date: ________Instructor: Diana Vance7. When Si is doped with P, it produces a(n) ________-typesemiconductor.a. pb. qc. npd. ne. No semiconductor will be produced.8. Two students measured an equilibrium constant for thesame chemical reaction. Ken obtained a value of 130 forthe equilibrium constant, but Barbie obtained a value of11.4. The instructor checked their results and said theywere both correct. How can that be?a. The values vary according to the way the measurement is made. Ken must have measured productconcentrations, while Barbie measured reactant concentrations.b. The values vary according to the starting conditions of the reaction prior to equilibrium. Ken must havestarted with all reactants, while Barbie must have started with all products.c. The values vary according to the stoichiometric coefficients that are used. The balancing coefficients thatKen used must have been twice those that Barbie used.d. The values vary according to direction of the reaction. Ken must have used the reverse reaction.e. The instructor must have made a mistake, because the equilibrium constant for a reaction must always bethe same.9. The equilibrium constant for the acid ionization ofmercaptoethanol (HSCH2CH2OH ) is 1.91  10–10.HSCH2CH2OH(aq) H+(aq) + SCH2CH2OH–(aq)Which of the following is true regarding this equilibrium?a. I and IIIb. I and IVc. II and IIId. II and IVe. None are true, as the concentrations of reactantsand products are comparable.10.As arule, which of the following phases are not included in theequilibrium constant? a. pure liquidsb. pure solidsPage 3 of 101. 2.3. 4.5. 6.7. 8.Grossmont College Chemistry 141 Exam 4 Fall 2015Name: ___________________________________Date: ________Instructor: Diana Vancec. aqueous solutiond. gasese. a & b only11. Inorganic waste goes in the ________a. Trash canb. Organic waste containerc. Broken glass containerd. Down the sinke. None of the above12.Page 4 of 10Grossmont College Chemistry 141 Exam 4 Fall 2015Name: ___________________________________Date: ________Instructor: Diana Vance13. Part 2: Short Answer 14. Directions: Answer each of the following questions. Besure to use complete sentences where appropriate. Forfull credit be sure to show all of your work. 1. A 10.7 m solution of NaOH has a density of 1.33 g/mL at 20oC. Calculate (15 points)a. the mole fraction of NaOH15.1kg H2O ×1000 g H2O1kg H2O×1mol H2O18.015 g H2O=55.509 mol H2O16.χNaOH=nNaOHntotal=10.7 mol NaOH55.509 mol H2O+10.7mol NaOH=10.7 mol NaOH66. 209mol total=0.16217.b. the mass percentage of NaOH18.10.7 mol NaOH1kg H2O×39.997 g NaOH1 mol NaOH×1 kg H2O1000 g H2O=427 .9679 g NaOH1000 g H2O19.mass% NaOH =mNaOHmtotal×10020.mass% NaOH =427 .9679 g NaOH1000 g H2O+427 .9679g NaOH× 10021.mass % NaOH =427.9679 g NaOH1427.9679 g soln=29.9704146% ≈30.0%22.mass% NaOH =29.9704146 %≈ 30.0%23.c. the molarity of the solution24.25.M=nNaOHLsoln26.M=30.0 g NaOH100 gsoln×1mol NaOH39.997 g NaOH×1.33 gsoln1mL soln×1000mL soln1 L sol27.M=9.965910298mol NaOHLsoln≈ 9.97 M NaOH28.29. Or Page 5 of 10Grossmont College Chemistry 141 Exam 4 Fall 2015Name: ___________________________________Date: ________Instructor: Diana Vance30.M=nNaOHLsoln=10.7mol NaOH1000g H2O+427 .9679 g NaOH1.33 g soln1 mLsoln×1000mL soln1L sol=9.97 M NaOH31.d. the osmotic pressure32.Π=iMRT =(2)(9.97molL)(0.08206Latmmol K)(20+273)K33.Π=iMRT =(2)(9.97molL)(0.08206Latmmol K)(293 K)=479 atm34.2. Use a Born-Haber cycle to calculate the lattice energy of potassium chloride from the following data (8 points):35. Ionization energy of K (g) = 425 kJ/mol36. Electron affinity of Cl (g) = -349 kJ/mol37. Energy to sublime K (s) = 89 kJ/mol38. Bond energy of Cl2 (g) = 240 kJ/mol39. ∆Hrxn for K (s) + ½ Cl2 (g) → KCl (s) = -438 kJ/mol 40. K (g) → K+ (g) + e-425 kJ/mol41. Cl (g) + e- → Cl- (g) -349 kJ/mol42. K (s) → K (g) 89 kJ/mol 43. ½ Cl2 (g) → Cl (g) 240 kJ/mol ×½ 44. + K+ (g) + Cl- (g) → KCl (s) ?45. K (s) + ½ Cl2 (g) → KCl (s) -438 kJ/mol 46.47. -438 kJ/mol = 425 kJ/mol + -349 kJ/mol + 89kJ/mol + (240 kJ/mol)(½) + ? 48. ? = -723 kJ/mol = lattice energy 49.3. Arterial blood contains about 0.25 g of oxygen per liter at 37 °C and standard atmospheric pressure. What