3.Which of the following has the highest melting temperature? A) H 2 O B) CO 2 C) S 8 D) MgF 2 E) P 4

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1 2. Which if the following is the correct order of boiling points for KNO 3, CH 3 OH, C 2 H 6, Ne? A) Ne < CH 3 OH < C 2 H 6 < KNO 3 B) KNO 3 < CH 3 OH < C 2 H 6 < Ne C) Ne < C 2 H 6 < KNO 3 < CH 3 OH D) Ne < C 2 H 6 < CH 3 OH < KNO 3 E) C 2 H 6 < Ne < CH 3 OH < KNO 3 3.Which of the following has the highest melting temperature? A) H 2 O B) CO 2 C) S 8 D) MgF 2 E) P 4 You need to determine the bond strength: the stronger the bond, the higher melting point or boiling point. KNO 3, MgF 2 ionic compounds, so intermolecular bonding is the strongest among other molecules, hence higher melting and boing points. Rules: For ionic compounds: the bigger charge of ions, the stronger bond. if charges are the same the smaller ion, the stronger bond For non-ionic compounds: depends on type of bonding (hydrogen>dipole-dipole> London dispersion) if type of bonding is the same the bigger molecule, the stronger bond

2 4. A certain substance, X, has a triple-point temperature of 20 0 C at a pressure of 2.0 atm. Which one of the statements cannot possibly be true? A) X can exist as a liquid above 20 0 C. B) X can exist as a solid above 20 0 C. C) Liquid X can exist as a stable phase at 25 0 C, 1 atm. D) Both liquid and solid X have the same vapor pressure at 20 0 C. Possible two different phase diagrams A) is always true, B) is true only for first picture, but still it is possible D) is true only at pressure 2 atm, but still it is true C) cannot be true, because triple-point determines the lowest temperature when liquid can exist.

3 9. What is the molality of a solution of 49.8 g of propanol (C 3 H 7 OH) in 152 ml water, if the density of water is 1.00 g/ml? mol solute Molality = kg solvent mass 1mol mol solute propanol = = 49.8g = 0.83 mol molar mass 60g kg solvent = 152mL 1g ml Molality = 1kg 1000g = 0.152kg 0.83 mol kg = 5.46m 10. Find the mass percent of CuSO 4 in a solution whose density is 1.30 g/ml and whose molarity is 1.36 M. Let s assume, that we have 1 L of the solution. mass CuSO 4 = mol CuSO 4 Mol CuSO 4 = 1.36mol L 1L = 1.36 mol molar mass = 1.36mol 144g mol = g Mass of solution = Volume Density = 1L 1.30 g ml 1000mL 1L weigt percent = mass CuSO 4 mass of solution = 195.8g 100% = 15% 1300g = 1300g

4 13. Vapor pressure of water at C is 23.8 torr. Determine the mass of glucose (molar mass = 180 g/mol) needed to add to 50.0 g of water to change the vapor pressure to 23.0 torr. P solution = P solvent X solvent X solvent = P solution P solvent = 23.0 torr 23.8 torr = mass of solvent H 2 O = 500g 1 mol g = 27.7 mol X solvent = = mol of solvent mol solvent + mol solute = x x = mol of glucose is needed mass of glucose = mol molar mass = mol 180 g = 175 g mol

5 14. Determine the mass of a nonvolatile, nonionizing compound that must be added to 3.19 kg of water to lower the freezing point to C. The molar mass of the compound is 50.0 g/mol and K f for water is C kg/mol. T fp = m K fp i i van t Hoff Factor. For nonionizing compounds it equals 1 T fp = T fp pure solvent T fp solution = 0 0 C C = 1.3 molality = T fp K fp i = C = mol/kg C 1.86 mol/kg mass of compound = mol molar mass = molality mass of solvent water molarmass = mol kg 3.19kg H 2O 50.0g = 111 g mol

6 19. What is the expected boiling point of a solution prepared by dissolving 5.83 g of sodium iodide (NaI) in 59.2 g of water? For water, T bp = C and K bp = C m -1. T bp = molality K bp i NaI Na + + I, ence i van t Hoff factor = 2 (because we obtain two partocles from one) mol NaI molality = kg of water = mass NaI Molar mass(nai) kg water 5.83 g = g/mol = mol/kg kg T bp = 0.659m C m 2 = 0.67 T bp = T bp solution T bp pure solvent T bp solution = T bp + T bp pure solvent = =

7 22. For a reaction in which A and B react to form C, the following initial rate data were obtained: [A], mo/l [B], mol/l Rate of formation of C, mol/l sec What is the rate law?

8 22. For a reaction in which A and B react to form C, the following initial rate data were obtained: Rate = k[a] m [B] n We need to determine m and n. [A], mo/l [B], mol/l Rate of formation of C, mol/l sec

9 22. For a reaction in which A and B react to form C, the following initial rate data were obtained: [A], mo/l [B], mol/l Rate of formation of C, mol/l sec Rate = k[a] m [B] n 1) For determination of m, choose rows, where concentrations of [B] are equal rate 2 = 4 rate 3 mol L sec 8 mol L sec = m 1 2 = 1 2 m, m = 1

10 22. For a reaction in which A and B react to form C, the following initial rate data were obtained: [A], mo/l [B], mol/l Rate of formation of C, mol/l sec Rate = k[a] m [B] n 1) For determination of n, choose rows, where concentrations of [A] are equal rate 1 = 1 rate 2 mol L sec 4 mol L sec = n 1 4 = 1 2 n, n = 2

11 23. A first order reaction is 45% complete at the end of 28 min. What is the length of the half-life of this reaction? If reaction is 45% complete, it means that 55% of the reactant left. ln A A 0 = kt ln = k (28 min) k = min 1 t 1/2 = k = = 32 min

12 37. Consider the reaction H 2 + I 2 2HI for which K = 48.2 at a high temperature. If an equimolar mixture of the reactants gives the concentration of the product to be 0.50 M at equilibrium, determine the equilibrium concentration of the hydrogen. If we mix equimolar amounts of reactamts, ten at equilibrium concentration of Hydrogen will be equal to concentration of Iodine. Let sequilibrium concentration of Hydrogen H 2 = x mol L K = [HI]2 = (0.50)2 H 2 I 2 x x = 0.25 x 2 = 48.2 Solving for x, we get: x = H 2 = mol/l

13 45. If an acid, HA, is 12.2 % dissociated in a 1.0 M solution, what is the K a for this acid? HA H 2 O H 3 O + A M 12.2% M 12.2% 1.0 M 12.2% % 100% 100% = K a = H 3O + [A ] [HA] = (0.122) K a = 0.017

14 48. Silver chromate, Ag 2 CrO 4, has a K sp of 8.96* Calculate the solubility in mol/l of silver chromate. Ag 2 CrO 4(s) 2Ag + + CrO 4 2 1:1 molar ratio. That s why if we find concentration of CrO 4 2-, it will be equal to solubility of Ag 2 CrO 4. K sp = [Ag + ] 2 CrO 4 = (2x) 2 x = 4x = 4x 3 x = Solubility of Ag 2 CrO 4 = mol/l

15 ml of 0.50 M NaOH is added to a 100 ml sample of M NH 3 (K b for NH 3 = 1.8*10-5 ). What is the equilibrium concentration of NH 4 + ions? Before addition: NaOH = 15.0mL 0.50M = 7.5 mmol NH 3 = 100mL 0.376M = 37.6 mmol After addition (Total volume = 115 ml): NaOH = 7.5 mmol/115 ml = M NH 3 = 37.6 mmol/115 ml = M NH 3 H 2 O NH 4 + OH x x +x x X x K b = = x( x) x 0.065x x = NH 4 + = M

16 52. What major species is (are) present at point III? If we titrate disodium ascorbate, then we start with As 2- in starting solution. Before 1 st equivalence point (point II on scheme): As 2 + H + HAs Before 2 nd equivalence point (point III on scheme): HAs + H + H 2 As Point I. We have As 2- and HAs - in the solution. Point II. We have HAs - in the solution. Point III. We have HAs - and H 2 As in the solution. Point IV. We have H 2 As in the solution. Point V. We have H 2 As and a bunch of HCl in the solution.

17 53. What volume of M NaOH must be added to 1.00 L of M HOCl to achieve a ph of 8.00? The K a for HOCl is 3.5*10-8. Let s say, that we added x L of NaOH. Then if we find x, we ll find an answer. Before After x x mol mol Total volume: (1 + x) L x HOCl = ClO = 0.01x 1 + x 1 + x Using H H equation: ph = pk a + log C.Base = pk Acid a + log OCl HOCl 8 = log HOCl OH - H 2 O OCl L 0.05M = mol xl 0.01M = 0. 01x mol 0.01x 1 + x x 1 + x = log 0.01x 0.54 = log x x = 3.9 V NaOH = 3.9 L x x

CHM151 Quiz Pts Fall 2013 Name: Due at time of final exam. Provide explanations for your answers.

CHM151 Quiz Pts Fall 2013 Name: Due at time of final exam. Provide explanations for your answers. CHM151 Quiz 12 100 Pts Fall 2013 Name: Due at time of final exam. Provide explanations for your answers. 1. Which one of the following substances is expected to have the lowest melting point? A) BrI B)