This Exam Consists of Two Sections. You Will Have 55 Minutes to Complete It

AP Chemistry

Quarter 2

Practice Exam 1-A

December 2, 2011

This exam consists of two sections. You will have 55 minutes to complete it.

When you finish the multiple-choice section and turn in your scantron sheet, you may take out your calculator.

Name: ______Block: ______

Multiple-Choice Section – Write “A” on the Subject line of the scantron sheet.

You may use your periodic table only. You may NOT use a calculator.

Please record all answers on the scantron sheet.

You may use a MAXIMUM of 25 minutes on this section.

Questions 1-4 refer to the following substances:

(A) CH3CH2OH

(B) CH3OCH3

(C) CH3CH2CH3

(D) NaCN

(E)  CH3CH3

1) Exhibits the weakest intermolecular forces - E

2.  Has the highest boiling point - D

3.  Forms a liquid at room temperature with the greatest surface tension – A (though it has the same empirical formula as B, an –OH group can form hydrogren bonds)

4.  Consists of polar molecules that cannot hydrogen bond with one another. - B

Questions 5-9 refer to the following solids.

(A) Diamond

(B) BaSO4 (s)

(C) Fe (s)

(D) H2O (s)

(E)  Kr (s)

AP Chemistry Multiple-Choice Exam 1 – A

5.  Consists of ions held together by electrostatic attraction - B

6.  Consists of molecules held together by intermolecular forces – D

7.  Consists of atoms sharing valence electrons as a group – C (metallic bonding)

8.  Consists of a network of covalently bonded atoms - A

9.  Consists of atoms weakly held together by dispersion forces - E

10.  When a sample of oxygen gas in a closed container of constant volume is heated until absolute temperature is doubled, which of the following is also doubled?

(A) The density of the gas

(B)  The average velocity of the gas molecules

(C)  The number of molecules per cm3

(D) The potential energy of the molecules

(E)  The pressure of the gas

11.  When the actual volume occupied by a sample of gas is greater than the volume predicted by the ideal gas law, the explanation lies in the fact that the ideal gas law does NOT include a factor for molecular

AP Chemistry Multiple-Choice Exam 1 – A

(A) volume

(B)  mass

(C)  velocity

(D) attractions

(E)  shape

AP Chemistry Exam 1

12.  A rigid metal tank contains oxygen gas. Which of the following applies to the gas in the tank when additional oxygen is added at constant temperature?

(A) The volume of the gas increases.

(B)  The pressure of the gas decreases.

(C) The average speed of the gas molecules remains the same.

(D) The total number of gas molecules remains the same.

(E)  The average distance between the gas molecules increases.

13.  A gaseous mixture containing 12.0 moles of nitrogen, 6.0 moles of oxygen, and 2.0 mole of helium exerts a total pressure of 0.90 atm. What is the partial pressure of the nitrogen?

AP Chemistry Exam 1

(A) 0.27 atm

(B)  0.54 atm

(C)  0.63 atm

(D) 0.72 atm

(E)  0.90 atm

AP Chemistry Exam 1

14.  The molarity of a solution containing 10.0 grams of NaOH in 500. mL of solution is

(A) 0.25 M (B) 0.50 M (C) 0.75 M (D) 0.90 M (E) 2.0 M

15.  A student was measuring the density of an unknown liquid dispensed from a burette and recorded the data below:

Mass of liquid = 25.7832 g

Volume in burette before dispensing = 46.4 mL

Volume in burette after dispensing = 11.4 mL

The density of the liquid should be reported as

(A) 0.7367 g/mL (B) 0.73 g/mL (C) 0.74 g/mL (D) 0.737 g/mL (E) 0.7 g/mL

16.  Equal numbers of moles of H2(g), O2(g), and Ne(g) are placed in a glass vessel at room temperature. If the vessel has a pinhole-sized leak, which of the following would be true regarding the relative values of the partial pressures of the gases remaining in the vessel after some of the gas mixture has effused?

AP Chemistry Exam 1

(A) PH2< PNe < PO2

(B)  PH2 < PO2 < PNe

(C)  PNe < PO2 < PH2

(D) PO2 < PH2 < PNe

(E)  PH2 = PNe = PO2

AP Chemistry Exam 1

17.  Solid NaCl melts at a temperature of 800oC, while solid NaBr melts at 750oC. Which of the following is an explanation for the higher melting point of NaCl?

AP Chemistry Exam 1

(A) A chlorine ion has less mass than a bromine ion

(B)  A chlorine ion has a greater negative charge than a bromine ion

(C)  A chlorine ion has a lesser negative charge than a bromine ion

(D) A chlorine ion is smaller than a bromine ion

(E)  A chlorine ion is larger than a bromine ion

18.  Which of the following can be determined directly from the difference between the boiling point of a pure solvent and the boiling point of a solution of a nonionic solute, if kb for the solution is known?

I. The mass of solute in the solution

II. The molality of the solution

III. The volume of the solution

AP Chemistry Exam 1

(A) I only (B) II only (C) III only (D) I and II only (E) I and II only

19.  A gas sample is confined in a 5-L container. Which of the following will occur if the temperature is increased

I. The kinetic energy of the gas will increase

II. The pressure of the gas will increase

III. The density of the gas will increase

AP Chemistry Exam 1

(A) I only (B) II only (C) I and III only (D) I and II only (E) I, II, and III

20.  If 46 grams of MgBr2 (molar mass = 184 grams) are dissolved in water to form 0.50 L of solution, what is the concentration of bromine ions in the solution?

(A) 0.25 M (B) 0.50 M (C) 1.0 M (D) 2.0 M (E) 4.0 M

Name: ______Block: ______

Question / Score
Mult. Choice / _____/50
1 / _____/16
2+3 / _____/16
4 / _____/8
5 / _____/18
Total / _____/108

Open Response Section

You may use your periodic table, formula sheet, and calculator.

Please write all answers in the space provided.

1. Suppose you have five identical balloons, each filled to the same volume at 25°C and 1.0 atmosphere pressure with a different pure gas. (4 pt each)

Balloon # / 1 / 2 / 3 / 4 / 5
Gas / CO2 / O2 / He / N2 / CH4

(a)  Which balloon contains the greatest mass of gas? Explain clearly.

Assuming exhibit “ideal” behavior, there will be an equal number of moles of gas in each balloon. Therefore, the gas with the greatest molar mass will have the greatest mass. In the example above, the gas with the largest molar mass is CO2.

(b)  Compare the average kinetic energies of the gas molecules in the balloons. Explain clearly.

The average kinetic energy (KE) in each balloon is the same because they are all at the same temperature. Temperature is just a measure of average kinetic energy of a sample, and if the temperatures are equal, the KEs will be equal. (Note: This does not mean they have the same velocity, remember KE = ½mv2, so the gases with the higher mass will have a lower velocity).

(c)  Which balloon contains the gas that would be expected to deviate least from the behavior of an ideal gas? Explain clearly.

The balloon with He will deviate the least form the behavior of an ideal gas. He is the smallest of all the gas particles and its only IM forces are very weak dispersion forces.

(d)  Twelve hours after being filled, all the balloons have decreased in size. Predict which balloon will be the smallest. Explain your reasoning clearly.

Balloon 3 will be the smallest. The gas particles are escaping through tiny holes in the balloons. Since He atoms are the lightest, at the same temperature they will move the fastest, so they will escape the fastest.

2) Explain each of the following in terms of the bonding and interactions in the substances involved. For full credit, your answer must refer to BOTH substances.

(a)  At ordinary conditions, HF (normal boiling point = 20°C) is a liquid, whereas HC1 (normal boiling point = -114°C) is a gas. (4 pt)

Fluorine is more electronegative than chlorine, so the HF molecules are more polarized. Because of this polarization, HF molecules can hydrogen bond with one another, while HCl molecules cannot. The stronger intermolecular forces in HF make it stay a liquid at room temperature.

(b)  The heat of vaporization, DHvap, of argon is 6.5 kJ mol-1, while the heat of vaporization of xenon is 12.3 kJ mol-1. (4 pt)

Both of these liquids would be attracted by dispersion forces alone, because they are nonpolar. Since xenon is a larger, more polarizable atom than argon, xenon atoms have stronger dispersion forces between them. Therefore, it takes more energy to cause the atoms of xenon to separate than it takes to separate argon atoms.

3) Consider the two compounds H2S and H2O. The two types of intermolecular forces present in H2S are London dispersion forces and dipole-dipole forces.

(a)  Compare the strength of the London dispersion forces in liquid H2S to the strength of the London dispersion forces in liquid H2O. Explain. (4 pt)

Dispersion forces are based on the random movement of electrons. Therefore, when a molecule has more electrons that are loosely held, the dispersion forces will be stronger. Since H2S has more electrons than H2O, and its electrons are more loosely held (in energy level 3 compared to 2), H2S molecules will experience stronger dispersion forces.

(b)  Compare the strength of the dipole-dipole forces in liquid H2S to the strength of the dipole-dipole forces in liquid H2O. Explain. (4 pt)

Dipole forces are based on the partial positive and negative charges in a molecule. In H2S, the bonds are only slightly polar, since S is only slightly more electronegative than H. Therefore, the partial charges in H2S are relatively small. In H2O, on the other hand, the partial charges are much larger, because O is much more electronegative than H. Larger charges result in a stronger attraction between molecules. Therefore, the dipole forces in H2O would be much stronger than those in H2S. In fact, the dipole forces in H2O are so strong that we give them a special name: hydrogen bonding.

4) Suppose you make the following 4 solutions in water.

A: 1.5 m C6H12O6 B: 1.5 m CoCl3 C: 2 m NaI D: 1 m Ba(NO3)2

a)  Rank these 4 solutions from lowest to highest according to their expected boiling points. (6 pt)

Lowest BP: ___A______D___ __C______B___: Highest BP

b)  When you test the boiling point of solution B, you find that it is not as high as you predicted. Offer an explanation for this observation. (2 pt)

There may be ion-pairing in solution b, which would decrease the number of dissolved particles. This would decrease the predicted boiling point.

5) A student was assigned the task of experimentally determining the heat of fusion of water, ΔHfus(H2O). The student measured the mass of an empty Styrofoam cup. Then the student put some warm deionized water in the cup and massed it again. The student then added a small mass of ice at 0°C to the cup and observed the temperature. Finally, the student massed the cup again. The data recorded by the student are shown below.

Mass of cup / 1.37 g
Mass of cup + water / 38.22 g
Mass of cup + water + ice / 43.28 g
Initial temperature of water / 38.3°C
Final temperature of mixture / 25.3°C

a.  Calculate the amount of heat, in J, given off by the cooling water in the cup. The specific heat capacity of water is 4.18 J g-1 °C-1. (2 pt)

qcooling = mcΔT = (38.22g – 1.37 g)(4.18 J g-1 °C-1)(25.3 ° - 38.3 °C) = -2002 J

Since we are asked for heat given off, we ignore the negative sign (which is indicating that heat was given off). With proper sig figs, 2.00 x 103 J were given off.

1. Calculate the amount of heat, in J, absorbed by the melted ice as it warmed from 0°C to the final temperature of 25.3 °C. (2 pt)

qwarming = mcΔT = (43.28 g – 38.22 g)(4.18 J g-1 °C-1)(25.3 ° - 0 °C) = 535 J

525 J of heat were absorbed.

1. Calculate the total amount of heat, in J, absorbed by the ice as it melted. Assume that the system was perfectly insulated from the surroundings. (2 pt)

qmelting + qwarming = -qcooling

qmelting = -qcooling - qwarming

qmelting = 2002 J – 535 J = 1467 J (or 1470 J with proper sig figs)

1. Calculate the heat of fusion of water, ΔHfus(H2O), in J mol-1, according to the student’s data. (2 pt)

5.06 g ice (1 mol/18 g) = 0.281 mol ice

1470 J/0.281 mol = 5220 J/mol

1. The accepted value for the heat of fusion of water is 6020 J mol-1. Calculate the percent error in the student’s value for ΔHfus(H2O). (2 pt)

1. 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.(4 pt each)

Occurrence 1: The cup was insufficiently insulated from the surroundings, resulting in some heat from the water being transferred to the air instead of the ice.