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DEPARTMENT OF CHEMISTRY AND CHEMICAL TECHNOLOGY CHEMISTRY OF SOLUTIONS 202-NYB-05 15/16 TEST 1 29 FEBRUARY 2012 INSTRUCTOR: I. DIONNE Answers Print your name: _____________________________________________ INSTRUCTIONS: Answer all questions in the space provided. Duration of this test is 75 minutes. No books or extra paper are permitted. Answer the questions in ink in order to preserve the right to grieve. In order to obtain full credit for your answers, you must clearly show your work. Answers to problems involving calculations must be expressed to the correct number of significant figures and proper units. 5. Calculators may not be shared. Programmable calculators are not permitted. 6. Your attention is drawn to the College policy on cheating. This policy will be enforced. 7. A Periodic Table with constants is provided. 1. 2. 3. 4. Problem 1: /2 Problem 6: /3 Sig. Fig.: /1 Problem 2: /3 Problem 7: /3 Units: /1 Problem 3: /3 Problem 8: /3 Problem 4: /3 Problem 9: /3 Problem 5: /3 Total: /28 (2 marks) PROBLEM 1 Calculate the sodium ion concentration when 70.0 mL of 3.0 M sodium carbonate (Na2CO3) is added to 30.0 mL of 1.0 M sodium bicarbonate (NaHCO3). Na2CO3: 3.0 M x 0.0700 L = 0.21 mol of Na2CO3 or 0.42 mol Na+ NaHCO3: 1.0 M x 0.0300 L = 0.030 mol of NaHCO3 or 0.030 mol Na+ Total Na+ = 0.42 + 0.030 = 0.45 mol Total volume = 0.0700 L + 0.0300 L = 0.1000 L [Na+] = 0.45 mol / 0.100 L = 4.5 M 4.5 M Answer: _______________ (3 marks) PROBLEM 2 A bottle of wine contains 12.5% ethanol by volume. The density of ethanol (C 2H5OH) is 0.789 g/mL and that of water is 1.00 g/mL. Calculate the concentration of ethanol in wine in terms of molality. [MM C2H5OH = 46.068 g/mol, MM H2O = 18.016 g/mol] 12.5% ethanol by volume implies 12.5 mL ethanol 12.5 mL ethanol + 87.5 mL water 12.5 mL ethanol x 0.789 g/mL = 9.8625 g ethanol 9.8625 g / 46.068 g/mol = 0.2140857 mol ethanol 87.5 mL water x 1.00 g/mL = 87.5 g water 87.5 g water = 0.0875 kg water molality = 0.2140857 mol / 0.0875 kg = 2.4466937 m 2.45 m Answer: _______________ Page |2 (3 marks) PROBLEM 3 A 1.37 M solution of citric acid (H3C6H5O7) in water has a density of 1.10 g/cm3. Calculate the mole fraction of the citric acid. [MM H3C6H5O7 = 192.124 g/mol, MM H2O = 18.016 g/mol] 1.37 M citric acid implies 1.37 mol citric acid 1 liter solution 1000 mL solution x 1.10 g/ml = 1100 g solution 1.37 mol citric acid x 192.124 g/mol = 263.20988 g citric acid 1100 g solution – 263.20988 g = 836.79012 g water 836.79012 g / 18.016 g/mol = 46.447054 mol water mole fraction of citric acid = mol citric acid / total mol mole fraction = 1.37 / (1.37 + 46.447054) = 0.0286509 0.029 Answer: _______________ PROBLEM 4 (3 marks) How many grams of NaBr must be added to 250. g of water to lower the vapor pressure by 1.30 mmHg at 40°C assuming complete dissociation? The vapor pressure of water at 40°C is 55.3 mmHg. [MM NaBr= 102.89 g/mol, MM H2O = 18.016 g/mol] Psolution = Χsolvent x Posolvent Psolution = 55.3 mmHg – 1.30 mmHg = 54.0 mmHg Χsolvent = 54.0 mmHg / 55.3mmHg = 0.97649 Χsolvent = 0.97649 = mol H2O / (mol H2O + mol Na+ and Br-) mol H2O = 250 g / 18.016 g/mol = 13.877 mol H2O Χsolvent = 0.97649 = 13.877 mol H2O / (13.877 mol H2O + mol Na+ and Br-) (13.877 mol H2O + mol Na+ and Br-) = 14.211 mol Na+ and Br- = 0.334 But we have 2 moles of ions for every mol of NaBr; we have 0.167 mol NaBr 0.167 mol NaBr x 102.89 g/mol = 17.18 g 20 g Answer: _______________ Page |3 (3 marks) PROBLEM 5 (a) Which pairs of liquids will be soluble in each other? yes no 1. H2O and CH3CH2CH2CH3 2. C6H6 (benzene) and CCl4 3. H2O and CH3COOH (b) A carrot is placed in a concentrated salt solution. What will most likely happen? (circle your choice). Solvent flows from low solute concentration to high solute concentration 1. Water will flow from the carrot to the solution. 2. Water will flow from the solution to the carrot. 3. Salt will flow into the carrot. 4. Salt will precipitate out. (c) Arrange the following aqueous solutions in order of increasing boiling points: 0.02 m LiBr, 0.03 m sucrose, 0.03 m MgSO4, 0.03 m CaCl2, and 0.025 m (NH4)2Cr2O7. __________ < __________ < __________ < __________ sucrose LiBr CaCl2 MgSO4 (NH4)2Cr2O7 < __________ 0.02 m LiBr = 0.04 m ions more particles = higher boiling point 0.03 m sucrose = 0.03 m ions 0.03 m MgSO4 = 0.06 m ions 0.03 m CaCl2 = 0.09 m ions 0.025 m (NH4)2Cr2O7 = 0.075 m ions Page |4 (3 marks) PROBLEM 6 An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction: 3H2(g) + N2(g) ⇌ 2NH3(g) At equilibrium, the concentrations are [H2] = 5.0 M, [N2] = 8.0 M, and [NH3] = 4.0 M. What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially? 3H2(g) + N2(g) ⇌ 2NH3(g) I initial initial 0 C −3x −x +2x E 5.0 8.0 4.0 For NH3:0 + 2x = 4.0 x = 2.0 = [NH3]0 For H2: [H2]o −3x = 5.0 [H2]o = (5.0) + (3)(2.0) = 11.0 For N2: [N2]o −x = 8.0 [N2]o = 8.0 + 2.0 = 10.0 Answers: [N2]o = _______________ and [H2]0 _______________ 11.0 M 10.0 M Page |5 (3 marks) PROBLEM 7 Exactly 4 mol sulfur trioxide is sealed in a 5.0-L container at 1529 K. Kp is 1150 for: 2SO3(g) ⇌ 2SO2(g) + O2(g) Calculate the concentrations of all species at equilibrium. [SO3]o = 4 mol / 5.0 L = 0.80 M 2SO3(g) ⇌ 2SO2(g) + O2(g) I 0.80 0 0 C −2x +2x +x E 0.80 − 2x 0 + 2x 0 + x K = Kp(RT)−Δn K = 1150(0.08206 L·atm·K−1·mol−1 x 1529 K)−(3−2) = 9.16556 K = (2x)2(x) / (0.80 − 2x)2 ≈ (2x)2(x) / (0.80)2 x = 1.136 approximation is invalid Assume that x = 0.35 M [SO3]eq = 0.80 − (2)(0.35) = 0.10 M [SO2]eq = (2)(0.35) = 0.70 M [O2]eq = x = 0.35 M 0.35 M 0.70 M 0.10 M Answers: [SO3]eq = ____________ , [SO2]eq = ____________ , [O2]eq = ____________ Page |6 (3 marks) PROBLEM 8 Choose (circle) the one alternative that best answers the following questions. (a) Consider the following equilibria involving SO2(g) and their corresponding equilibrium constants. SO2(g) + ½O2(g) ⇌ SO3(g) 2SO3(g) ⇌ 2SO2(g) + O2(g) K1 K2 Which of the following expressions relates K1 to K2? (Circle your choice). 1. K2 = K12 2. K22 = K1 3. K2 = K1 4. K2 = 1/K1 5. K2 = 1/K12 (b) The reaction quotient for a reaction has a value of 75 while the equilibrium constant has a value of 195. Which of the following statements is accurate? 1. The reaction must proceed to the left to establish equilibrium. 2. The reaction must proceed to the right to establish equilibrium. 3. The concentrations of the products will be much smaller that the concentrations of the reactants when the system is at equilibrium. 4. The concentrations of the products will be about the same as the concentrations of the reactants when the system is at equilibrium. (c) Which of the following is not a colligative property? 1. osmotic pressure 2. solubility 3. boiling point elevation 4. freezing point depression Page |7 (3 marks) PROBLEM 9 (a) Consider the reaction: Fe3+(aq) + SCN−1(aq) ⇌ FeSCN2+(aq) How will the equilibrium position shift if Shift left No change Shift right ← → 1. water is added, doubling the volume? 2. AgNO3(aq) is added? (AgSCN is insoluble). 3. NaOH(aq) is added? [Fe(OH)3 is insoluble]. 4 Fe(NO3)3(aq) is added? accept (b) This represents a system at equilibrium: O2(aq) ⇌ O2(g) In which direction will the equilibrium be shifted Shift left No change hift right ← → by an increase in pressure? by an increase in temperature? Page |8