2012 AP CHEMISTRY FREE-RESPONSE QUESTIONS

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1 01 AP CHEMISTRY FREE-RESPONSE QUESTIONS. A sample of a pure, gaseous hydrocarbon is introduced into a previously evacuated rigid 1.00 L vessel. The pressure of the gas is 0.00 atm at a temperature of 17C. (a) Calculate the number of moles of the hydrocarbon in the vessel. (b) O (g) is introduced into the same vessel containing the hydrocarbon. After the addition of the O (g), the total pressure of the gas mixture in the vessel is 1.40 atm at 17C. Calculate the partial pressure of O (g) in the vessel. The mixture of the hydrocarbon and oxygen is sparked so that a complete combustion reaction occurs, producing CO (g) and H O(g). The partial pressures of these gases at 17C are atm for CO (g) and atm for H O(g). There is O (g) remaining in the container after the reaction is complete. (c) Use the partial pressures of CO (g) and H O(g) to calculate the partial pressure of the O (g) consumed in the combustion. (d) On the basis of your answers above, write the balanced chemical equation for the combustion reaction and determine the formula of the hydrocarbon. (e) Calculate the mass of the hydrocarbon that was combusted. (f) As the vessel cools to room temperature, droplets of liquid water form on the inside walls of the container. Predict whether the ph of the water in the vessel is less than 7, equal to 7, or greater than 7. Explain your prediction. 01 The College Board. -7- GO ON TO THE NEXT PAGE.

2 01 SCORING GUIDELINES Question (10 points) A sample of a pure, gaseous hydrocarbon is introduced into a previously evacuated rigid 1.00 L vessel. The pressure of the gas is 0.00 atm at a temperature of 17C. (a) Calculate the number of moles of the hydrocarbon in the vessel. PV n (0.00 atm)(1.00 L) (0.081 L atm mol K )(400. K) RT 1 1 n = mol 1 point is earned for the setup. 1 point is earned for the numerical answer. (b) O (g) is introduced into the same vessel containing the hydrocarbon. After the addition of the O (g), the total pressure of the gas mixture in the vessel is 1.40 atm at 17C. Calculate the partial pressure of O (g) in the vessel. P O = 1.40 atm 0.00 atm = 1.0 atm 1 point is earned for the correct pressure. The mixture of the hydrocarbon and oxygen is sparked so that a complete combustion reaction occurs, producing CO (g) and H O(g). The partial pressures of these gases at 17C are atm for CO (g) and atm for H O(g). There is O (g) remaining in the container after the reaction is complete. (c) Use the partial pressures of CO (g) and H O(g) to calculate the partial pressure of the O (g) consumed in the combustion.... C x H y +... O... CO +... H O before rxn: 0.00 atm 1.0 atm - - after rxn: 0 atm? atm atm atm 1 atm O atm CO 1 atm CO = atm O 1 atm O atm H O atm H O = atm O Total O consumed = atm OR, based on PV = nrt and mole calculations: n n PV (0.800 atm)(1.00 L) 1 mol O mol H O (0.081 L atm mol K )(400. K) mol H O HO RT 1 1 = 0.01 mol O PV (0.600 atm)(1.00 L) 1 mol O mol CO (0.081 L atm mol K )(400. K) 1 mol CO CO RT point is earned for the correct stoichiometry in O consumption. 1 point is earned for the calculated result. = mol O Total moles O = ; P nrt V 1 1 ( mol)(0.081 L atm mol K )(400. K) 1.00 L P = 1.00 atm O 01 The College Board.

3 01 SCORING GUIDELINES Question (continued) (d) On the basis of your answers above, write the balanced chemical equation for the combustion reaction and determine the formula of the hydrocarbon. The partial pressures occur in the same proportions as the number of moles. P : P : P : P hydrocarbon O CO HO 0.00 atm : 1.00 atm : atm : atm = 1 : 5 : 3 : 4 C 3 H O 3 CO + 4 H O OR n n PV (0.800 atm)(1.00 L) mol H mol H O (0.081 L atm mol K )(400. K) 1 mol H O HO RT 1 1 = mol H PV (0.600 atm)(1.00 L) 1 mol C mol CO (0.081 L atm mol K )(400. K) 1 mol CO CO RT 1 1 = mol C CH mol H.66 mol H 3 8 mol H mol C 1 mol C 3 3 mol C C 3 H O 3 CO + 4 H O 1 point is earned for the formula of the hydrocarbon. 1 point is earned for a balanced equation with the correct proportions among reactants and products. (e) Calculate the mass of the hydrocarbon that was combusted. mass = (number of moles)(molar mass) = ( mol)(44.1 g/mol) = 0.69 g 1 point is earned for using the number of moles combusted from part (a). 1 point is earned for the calculated mass. (f) As the vessel cools to room temperature, droplets of liquid water form on the inside walls of the container. Predict whether the ph of the water in the vessel is less than 7, equal to 7, or greater than 7. Explain your prediction. The ph will be less than 7 because CO is soluble in water, with which it reacts to form H + ions. 1 point is earned for the correct choice and explanation. 01 The College Board.

4 011 AP CHEMISTRY FREE-RESPONSE QUESTIONS (Form B). An 8.55 mol sample of methanol, CH 3 OH, is placed in a 15.0 L evacuated rigid tank and heated to 37!C. At that temperature, all of the methanol is vaporized and some of the methanol decomposes to form carbon monoxide gas and hydrogen gas, as represented in the equation below. CH 3 OH(g) " # CO(g) + H (g) (a) The reaction mixture contains 6.30 mol of CO(g) at equilibrium at 37!C. (i) Calculate the number of moles of H (g) in the tank. (ii) Calculate the number of grams of CH 3 OH(g) remaining in the tank. (iii) Calculate the mole fraction of H (g) in the tank. (iv) Calculate the total pressure, in atm, in the tank at 37!C. (b) Consider the three gases in the tank at 37!C: CH 3 OH(g), CO(g), and H (g). (i) How do the average kinetic energies of the molecules of the gases compare? Explain. (ii) Which gas has the highest average molecular speed? Explain. (c) The tank is cooled to 5!C, which is well below the boiling point of methanol. It is found that small amounts of H (g) and CO(g) have dissolved in the liquid CH 3 OH. Which of the two gases would you expect to be more soluble in methanol at 5!C? Justify your answer. 011 The College Board. -7- GO ON TO THE NEXT PAGE.

5 011 SCORING GUIDELINES (Form B) Question (9 points) An 8.55 mol sample of methanol, CH 3 OH, is placed in a 15.0 L evacuated rigid tank and heated to 37 C. At that temperature, all of the methanol is vaporized and some of the methanol decomposes to form carbon monoxide gas and hydrogen gas, as represented in the equation below. CH 3 OH(g)! " CO(g) + H (g) (a) The reaction mixture contains 6.30 mol of CO(g) at equilibrium at 37!C. (i) Calculate the number of moles of H (g) in the tank. mol H 6.30 mol CO # 1 mol CO = 1.6 mol H 1 point is earned for the correct number of moles. (ii) Calculate the number of grams of CH 3 OH(g) remaining in the tank. 1 mol CH3OH 6.30 mol CO # 1 mol CO = 6.30 mol CH 3 OH reacted 8.55 mol CH 3 OH initial $ 6.30 mol CH 3 OH reacted =.5 mol CH 3 OH.5 mol # 3.04 g 1 mol = 7.1 g 1 point is earned for the correct number of grams." (iii) Calculate the mole fraction of H (g) in the tank. 1.6 mol H.5 mol CH OH mol CO mol H 1 point is earned for the correct setup. = = point is earned for the correct answer." (iv) Calculate the total pressure, in atm, in the tank at 37!C. PV = nrt % P = nrt V = L atm (1.15 mol)(0.081 )(600 K) mol K 15.0 L = 69.5 atm 1 point is earned for the correct setup. 1 point is earned for the correct answer." 011 The College Board.

6 011 SCORING GUIDELINES (Form B) Question (continued) (b) Consider the three gases in the tank at 37!C: CH 3 OH(g), CO(g), and H (g). (i) How do the average kinetic energies of the molecules of the gases compare? Explain. The average kinetic energies are the same because all three gases are at the same temperature. 1 point is earned for the correct answer and explanation.! (ii) Which gas has the highest average molecular speed? Explain. KE = 1 mv, so at a given temperature the molecules with the lowest mass have the highest average speed. Therefore the molecules in H gas have the highest average molecular speed. 1 point is earned for the correct answer and explanation.! (c) The tank is cooled to 5"C, which is well below the boiling point of methanol. It is found that small amounts of H (g) and CO(g) have dissolved in the liquid CH 3 OH. Which of the two gases would you expect to be more soluble in methanol at 5"C? Justify your answer. The only attractive forces between molecules of H and CH 3 OH would be due to weak London dispersion forces (LDFs). In contrast, the LDFs are stronger between CO molecules and CH 3 OH molecules because CO has more electrons than H. In addition CO is slightly polar; thus intermolecular dipole-dipole attractions can form between CO molecules and CH 3 OH molecules. With stronger intermolecular interactions between molecules of CO and CH 3 OH, CO would be expected to be more soluble in CH 3 OH than H. 1 point is earned for the correct answer and justification. 011 The College Board.

7 009 AP CHEMISTRY FREE-RESPONSE QUESTIONS (d) A buffer solution is prepared by dissolving some solid NaOCl in a solution of HOCl at 98 K. The ph of the buffer solution is determined to be (i) Calculate the value of [H 3 O + ] in the buffer solution. (ii) Indicate which of HOCl(aq) or OCl " (aq) is present at the higher concentration in the buffer solution. Support your answer with a calculation.. A student was assigned the task of determining the molar mass of an unknown gas. The student measured the mass of a sealed 843 ml rigid flask that contained dry air. The student then flushed the flask with the unknown gas, resealed it, and measured the mass again. Both the air and the unknown gas were at 3.0#C and 750. torr. The data for the experiment are shown in the table below. Volume of sealed flask Mass of sealed flask and dry air Mass of sealed flask and unknown gas 843 ml g g (a) Calculate the mass, in grams, of the dry air that was in the sealed flask. (The density of dry air is 1.18 g L "1 at 3.0#C and 750. torr.) (b) Calculate the mass, in grams, of the sealed flask itself (i.e., if it had no air in it). (c) Calculate the mass, in grams, of the unknown gas that was added to the sealed flask. (d) Using the information above, calculate the value of the molar mass of the unknown gas. After the experiment was completed, the instructor informed the student that the unknown gas was carbon dioxide (44.0 g mol "1 ). (e) Calculate the percent error in the value of the molar mass calculated in part (d). (f) For each of the following two possible occurrences, indicate whether it by itself could have been responsible for the error in the student s experimental result. You need not include any calculations with your answer. For each of the possible occurrences, justify your answer. Occurrence 1: The flask was incompletely flushed with CO (g), resulting in some dry air remaining in the flask. Occurrence : The temperature of the air was 3.0#C, but the temperature of the CO (g) was lower than the reported 3.0#C. (g) Describe the steps of a laboratory method that the student could use to verify that the volume of the rigid flask is 843 ml at 3.0#C. You need not include any calculations with your answer. 009 The College Board. All rights reserved. Visit the College Board on the Web: GO ON TO THE NEXT PAGE.

8 009 SCORING GUIDELINES Question (10 points) A student was assigned the task of determining the molar mass of an unknown gas. The student measured the mass of a sealed 843 ml rigid flask that contained dry air. The student then flushed the flask with the unknown gas, resealed it, and measured the mass again. Both the air and the unknown gas were at 3.0 C and 750. torr. The data for the experiment are shown in the table below. Volume of sealed flask Mass of sealed flask and dry air Mass of sealed flask and unknown gas 843 ml g g (a) Calculate the mass, in grams, of the dry air that was in the sealed flask. (The density of dry air is 1.18 g L!1 at 3.0"C and 750. torr.) m! D " V = (1.18 g L!1 )(0.843 L) = g One point is earned for the correct setup and calculation of mass. (b) Calculate the mass, in grams, of the sealed flask itself (i.e., if it had no air in it) g! g = g One point is earned for subtracting the answer in part (a) from g. (c) Calculate the mass, in grams, of the unknown gas that was added to the sealed flask g! g = 1.37 g One point is earned for subtracting the answer in part (b) from g. (d) Using the information above, calculate the value of the molar mass of the unknown gas. $ 750. atm % (0.843 L) PV & 760 ' n!!! mol RT # 1 # 1 (0.081 L atm mol K )(96 K) 1.37 g molar mass = = mol OR molar mass = DRT P 40.1 g mol!1 $ 1.37 g % # 1 # 1 (0.081 L atm mol K )(96 K) & L '! $ 750. atm % & 760 ' = 40.0 g mol!1 One point is earned for the conversion of pressure (if necessary) and temperature and the use of the appropriate R. One point is earned for the correct setup and calculation of moles of gas. One point is earned for the correct setup and calculation of molar mass. OR If calculation is done in a single step, 1 point is earned for the correct P and T, 1 point is earned for the correct density, and 1 point is earned for the correct answer. 009 The College Board. All rights reserved. Visit the College Board on the Web:

9 009 SCORING GUIDELINES Question (continued) After the experiment was completed, the instructor informed the student that the unknown gas was carbon dioxide (44.0 g mol!1 ). (e) Calculate the percent error in the value of the molar mass calculated in part (d). percent error =! 1! g mol! 40.1g mol 44.0 g mol! 1 " 100 = 8.9% One point is earned for the correct setup and answer. (f) For each of the following two possible occurrences, indicate whether it by itself could have been responsible for the error in the student s experimental result. You need not include any calculations with your answer. For each of the possible occurrences, justify your answer. Occurrence 1: The flask was incompletely flushed with CO (g), resulting in some dry air remaining in the flask. This occurrence could have been responsible. The dry air left in the flask is less dense (or has a lower molar mass) than! CO! gas at the given! T! and! P. This would result in a lower mass of gas in the flask and a lower result for the molar mass of the unknown gas. One point is earned for the correct reasoning and conclusion. Occurrence : The temperature of the air was 3.0#C, but the temperature of the CO (g) was lower than the reported 3.0#C. This occurrence could not have been responsible. The density of! CO! is greater at the lower temperature. A larger mass of! CO! would be in the flask than if the CO had been at 3.0#C, resulting in a higher calculated molar mass for the unknown gas. One point is earned for the correct reasoning and conclusion. (g) Describe the steps of a laboratory method that the student could use to verify that the volume of the rigid flask is 843 ml at 3.0#C. You need not include any calculations with your answer. Valid methods include the following: 1. Find the mass of the empty flask. Fill the flask with a liquid of known density (e.g., water at 3#C), and measure the mass of the liquid-filled flask. Subtract to find the mass of the liquid. Using the known density and mass, calculate the volume.. Measure 843 ml of a liquid (e.g., water) in a 1,000 ml graduated cylinder and transfer the liquid quantitatively into the flask to see if the water fills the flask completely. One point is earned for a valid method. 009 The College Board. All rights reserved. Visit the College Board on the Web:

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