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Unformatted text preview: AP® Chemistry MultipleChoice Answer Key Correct Correct
No. Answer No. Answer
1 D ' 31 A
2 B 32 C
3 A 33 B
4 E 34 B
5 A 35 B
6 C ' 36 C
7 D 37 E
8 B 38 D
9 C 39 C
10 A 40 E
11 E 41 C
12 D 42 B
13 B 43 A
14 C 44 C
15 C 45 D
16 E 46 C
17 C 47 E
18 E 48 D
19 D 49 A
20 B 50 C
21 E 51 B
22 D 52 E
23 C 53 B
24 C 54 D
25 C 55 D
26 A 56 C
27 D 57 E
28 E 58 D
29 1 B 59 B
30 D 60 C 33 Correct No. Answer
' 61 D
62 B
63 B
64 D
65 E
66 D
67 E
68 C
69 A
70 E
71 C
72 D
73 C
74 B
75 C AP® Chemistry
FreeResponse Scoring Guidelines In regard to mathematical errors, a 1point deduction is made; this deduction may be applied only once
per question. In regard to errors in reporting significant ﬁgures, a 1—point deduction is made; this
deduction may be applied only once per question. A leeway of plus or minus one significant figure
different from the correct number of significant figures is allowed before a 1—point deduction is made. For questions including parts that refer back to previous parts of the same question, a wrong answer
that derives from the correct use of a previously caiculated incorrect answer should not be counted as
wrong. The emphasis of scoring is on process, and an error made early on in a multipart calculation
should not jeopardize the earning of full credit for the correct use of that incorrect value later in the
same question. AP® Chemistry
, FreesResponse. Scoring, Guidelines ,, Question 1 Answer the following questions about the solubility of the salts Li3PO4 and PbClz. Assume that hydrolysis
effects are negligible. The equation for the dissolution of Li3PO4(s) is shown below.
Li3PO4(s) (t2 3 Li+(aq) + PO43“(aq) KS? = 3.2 x 10‘9 at 25°C (21) Write the equilibriumconstant expression for the dissolution of Li3PO4(s). KS? = {If} 3{PO43‘] One point is earned for the correct expression. (13) Assuming that volume changes are negligible, calculate the maximum number of moles of Li3PO4(s) that
can dissolve in (i) 0.50 L of water at 25°C One point is earned for a . _9 correct setup using the
saturated solution of L13PO4. Then K51, = 3.2 x 10 = (3x)3(x) = 27x4, KW expression from I 9
thus x = ’ 22%?" = 3.3 x 10'3 M. Therefore the number of moles p art (20‘ of Li3P04(s) that can dissolve in 0.50 L of water is Let 3: represent the molar concentration of Li3P04 present in a One point is earned for a 3,3 x 10“3 mol Li3po __ correct calculation of
05% x “mgr—5‘ = 17 x 10 3 "ml moles of 1.131304. (ii) 0.50 L of 0.20 M LiNO3 at 25°C In 0.20 M LiN03, [LY] = 0.20 M. Assume that the amount of One point is earned for Li+(aq) contributed to the solution by the Li3PO4(s) that dissolves is recognizing that {If} is
negligibly small compared to 0.20 M. Then let y represent the molar approximately equal to
concentration of P043‘(aq) present in the solution due to the Li3P04(s) 029 M in the solution 10
that dissolves in the 0.20 M LiNO3. Then Whmh the L13P04(S) was
“9 3 3.2 x 104 7 added.
KSP = 3.2 x 10 = (0.20) (y) :3 y = W = 4.0 x10 M. So [PO43‘(aq)] = [Li3PO4(aq)] = 4.0 x 104M in 0.20 M LiNO3(aq). One point is earned for a
correct calculation of moles
of Li3PO4. Therefore, the number of moles of Li3PO4(s) that can dissolve in 4.0 x 10'7moi Li3po4 = —7
1.0L 2.0x 10 mol 0.50 L of water is 0.50 L x AP® Chemistry
,, ,,,,ﬁreegResponse,,,.$eoting.Guideiines, , , Question 1 (continued) The equation for the dissolution of PbC12 is shown below.
PbC12(s) : Pb2+(aq) + 2 Cl‘(aq) KS? = 1.6 x 10’5 at 25°C (c) Calculate the concentration of C1‘(aq) in a saturated solution of PbC12 at 25°C. Let z = [Pub] in a saturated solution of PbC12 . Then Kip = [P1320 [CH2 = (z)( 2z )2 = 423. One point is earned for a correct setup. I 1.6 x 10‘5
l. ‘5 = 4 = 3 —————— _ ,
Thus 6 X 10 2:3 :> z 4 One pomt is earned for a correct calculation 2 z = [P1324] = 1.6 x 10.2 M of the value of {Cl'}. :2 [Cl‘] = 2{Pb2+] = 2(1.6x10'2M) = 3.2x 10'2M (d) An open container holds 1000 L of 0.00400 M P1302 , which is unsaturated at 25°C. Calculate the
minimum volume of water, in mL, that must evaporate from the container before solid PbClz can
precipitate. 1.000 L of 0.00400 M PbC12(aq) contains 0.00400 mol of PbClztaq),
thus it contains 0.00400 mol Ph2+(aq) and 0.00800 moi Cl"(aq). Let V = volume of the solution at saturation, then 0.00400 mol)(0.00800)2 = 2.6 ><1()"7
V V V3 One point is earned for the 2 6 x 10.7 calculation of the saturation
:> V = 3" W = 0.25 L = 250mLat saturation volume.
. x Thus the volume of water that must evaporate = 1,000. — 250 = 750 mL Ksp=1.6><10’5= ( 0R One point is earned for
subtracting the saturation From part (c), the amount of PbClZ dissolved in l L of saturated PbC12(aq) volume from 1 000 mL is 1.6 x 10'“2 M. Let V = volume of the solution at saturation, then 0.016 mol PbClZ __ 0.00400 mol PbC12 1,000mL — V :9 V=250mL Thus the volume of water that must evaporate = 1,000.  250 = 750 mL AP® Chemistry
,,FreeResponseﬁcoring,,,,Guidelines , , Question 2 Answer the following using chemical concepts and principles of the behavior of gases. (a) A metal cylinder with a volume of 5.25 L contains 3.22 g of He(g) and 11.56 g of N2(g) at 150°C. (i) Calculate the total pressure in the cylinder. 3.22 g He x W = 0.805 mol He One point is earned for the
4.00 g He calculations of moles and adding
them together. 1.00 mol N2
28.02 g N2 11.56 g N2 x = 0.4126 mol N2 One point is earned for the correct total moles of gas = (0.805 + 0.4126) = 1.218 mol . . . .
substltutlon into the gas law equation. 5.25 L One point is eamed for the
correct answer. (ii) Calculate the partial pressure of N2(g) in the cylinder. moles N2 gas One point is earned for the calculation —————————— x total ressure in ﬂask .
total moles of gas p of the mole fraction of N2 . pressure N2 = One oint is earned for the correct
= W x 5.49 atm = 1.86 atm p answer
1.218 mol ' (b) A 1.50 L container holds a 9.62 g sample of an unknown gaseous saturated hydrocarbon at 30°C and
3.62 atm. (i) Calculate the density of the gas. One point is earned for the correct answer. AP® Chemistry
,,,,,,,Fre,esResanse,,,$,cpri,n9,,,Guidelines Question 2 (continued) (ii) Calculate the molar mass of the gas. Let M = molar mass, then PV = nRT = [331) RT 2*»
1 One point is earned for the correct answer. M =——— _ M = 44.1gmol1
PV (3.62 atm)(1.50 L) (iii) Write the formula of the hydrocarbon. Saturated hydrocarbons have the generic formula CnHZn +2 , One point is earned for
therefore let 44.1 g = 120’!) + 1(2)! +2) = 14}! + 2 I) the correct answer. 42.1:1411 :9 3=n :> C3118 (iv) Calculate the rootmean—square speed of the gas molecules in the container at 30°C.
(Note: 11 = 1kg stZ) 2 _2 _1 ‘1 One point is earned for the correct
I) = ’3RT = 3(331 kg m S “101 K X303 K) setup using the molar mass in
ms M 0.0441 kg mol"1 kilograms. One point is earned for the
= 414 m 5‘1 correct answer. 33. AP® Chemistry
,, FreeResponse,Scoring.Guidelines, Question 3
A student performs a titration in which a 10.00 mL sample of 0.0571 M HCl is titrated with a solution of
NaOH of unknown concentration. (a) Describe the steps that the student should take to prepare and ﬁll the buret for the titration given a wet
50.00 mi. buret and the materials listed below. 0.0571 M HCl solution Indicator solution
N a0H(aq) (unknown concentration) Distilled water
10.5 M NaOH solution 100 mL beaker Rinse the buret with some distilled water and then pour some
of the NaOH solution of unknown concentration into the
beaker and use it to rinse the buret two times. One point is earned for rinsing the
buret with the titrant solution. Use the beaker to carefully ﬁll the buret with the NaOH solution of unknown concentration. One point is earned for removing air
bubbles from the neck and tip of the Put the beaker under the buret and drain enough solution to bur et remove any air bubbles in the neck and tip of the buret. (b) Calculate the pH of the 0.0571 M HCl. HCI is a strong acid :> [11*] in 0.0571 M HCl = 0.0571 M One point is earned for the
pH = log [in] = —log (0.0571) = 1.243 cometanswer (c) A volume of 7.62 mL of the NaOH solution was needed to reach the end point of the titration. Calculate
the molarity of the NaOH solution used in the titration. One point is earned for
mol HCI titrated = 10.00 mL x W = 0.000571 mol; calculating the moles of HCl in the sample that was
ratio HClzNaOH in neutralization is 1:1, so 0.000571 moi NaOH reacted; titrated. 0.000571 mol NaOH One point is earned for = 0.0749 M NaOH calculating the molarity of 0.00762 L the NaOH solution. AP® Chemistry
,EreerResppnse,,$,c9rin9, Gu,i,delines,,,, , Question 3 (continued) In a different titration using a different NaOH solution, the concentration of NaOH was determined by the
student to be 0.0614 M. ((1) Given that the actual concentration of the NaOH solution was 0.0627 M, calculate the percent error of the
student’s result. @0614 — 0,0627] One point is earned for the correct percent error = X 100 = 2.1 % answer. 0.0627 (e) Calculate the volume of 10.5 M NaOH that is needed to prepare 250.0 mL of 0.0627 M NaOH. 0.0627 moi One point is earned for calculating the 250.0 mL X m = 0.0157 H101 NaOH needed moles 0f NaOII needed. 1,000 mL One point is earned for the correct 0.0157 mcl NaOH X = 1.49 mL answer. 10.5 mol NaOH AP® Chemistry
Free—Response,ScoringGuidettnes , Question 4 For each of the following three reactions, in part (i) write a balanced equation and in part (ii) answer the
question about the reaction. In part (i), coefﬁcients should be in terms of lowest whole numbers. Assume that
solutions are aqueous unless otherwise indicated. Represent substances in solutions as ions if the substances are
extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You may use the empty space at the bottom of the next page for scratch work, but only equations that are written in the
answer boxes provided will be graded. W
A strip of magnesium metal is added to a solution of silverﬂ) nitrate. (i) Balanced equation: 149+ 2/433 e—e Mg.“ 1» 2/17. {ii} Which substance is oxidizedin the reaction? (3) Equal volumes of 0.1 M hydroﬂuoric acid and 0.1 M potassium hydroxide are combined. (i) Balanced equation: HF+OH‘ —> F‘+H20 (ii) Draw the complete Lewis electrondot diagram for the reactant that is the Bronstediowry base in the
forward reaction.
Es a" (b) Solid calcium metal burns in air. (i) Balanced equation: 2Ca+02 —) 2Ca0 (ii) Predict the algebraic sign of AH” for the reaction. Explain your prediction. The sign of AH° will be negative because AG° is negative {the reaction occurs) and AS" is negative
ta solid and a gas react to form a solid). According to the GibbsHelmholtz equation, AH° = AG" + T AS".
Therefore AH" is the sum of two negative quantities and as such must be negative. ~41 AP® Chemistry
, FreerResp,¢ns,,e,,,§99rin9, ﬁnidelines ,, ,, , Question 4 (continued) (0) Samples of nitrogen monoxide gas and oxygen gas are combined. (i) Balanced equation: 2N0+02 —~> 2N02 (ii) If the reaction is second order with respect to nitrogen monoxide and first order with respect to
oxygen, what is the rate law for the reaction? rate = kiNOﬂOZ] General Scoring Notes for Question 4 Five points are earned for each of parts (a), (b), and (c), distributed as follows. Four points are earned for part (i): one point for the correct reactants, two points for the correct product(s), and one point for the correct coefﬁcients in the balanced equation. One point is earned for the correct answer in part (ii). AF” Chemistry iiFreesResponsesconngsmdalinesei Question 5 Voltmeter 1.00 Mcmnogh 1.00MAgNo3 Answer the following questions relating to the galvanic cell shown in the diagram above. (a) Write the balanced equation for the overall cell reaction. 2 Ag*(aq) + Co(s) ___, 2 Ag(s) + Cozﬂaq) ' One point is earned for the correct equation. (b) Calculate the value of 12° for the cell. One point is earned for the correct value of
$21: = 0.80  (4128) = 1.08 v , 0
cell ' (c) Is the value of AG° for the overall cell reaction positive, negative, or O ? lust' our answer.
, Y The value of AG° for the overall reaction must be negative
because the cell reaction occurs (is spontaneous) as the cell operates. One point is earned for the correct OR answer, including a valid justiﬁcation. Since E5,” is positive and AG° = —nFE°, the value of AG"
must be negative. AP® Chemistry
FreeResponseScoring Guidelines, Question 5 (continued) (d) Consider the cell as it is operating. (i) Does Ece“ increase, decrease, or remain the same? Explain. As the cell operates, the concentration of Ag+ decreases and the concentration of (302+ increases :2
{C0241 One point is earned for the correct the ratio Q = increases :2 in Q increases :> [ A g ]2 answer, including a valid justiﬁcation. Em, = 5:,”  311:an becomes smaller (decreases). nF (ii) Does AG of the overall cell reaction increase, decrease, or remain the same? Explain. The value of A0 for the system increases (becomes less ' One point is earned for the correct
negative) as the cell operates and the system approaches answer, including a valid justiﬁcation. equilibrium (when AG = 0). (iii) What would happen if the NaNO3 solution in the salt bridge was replaced with distilled water?
Explain. The cell would not operate. The voltage of the cell is too
small to overcome the electrical resistance of distilled water,
which is a poor conductor due to its very low concentration
of ions (about to"7 M H+(aq) and 10‘7 M OH“(aq)} that
could “carry” the current from one cell to the other. One point is earned for the correct
answer, including a valid justiﬁcation. (e) After a certain amount of time, the mass of the Ag electrode changes by x grams. Given that the molar
mass of Ag is 108 g moi“l and the molar mass of Co is 59 g mol‘l, write the expression for the change in
the mass of the Co electrode in terms of x. 1 mol Ag 1
A moi Ag = A mass Ag x M = x x E)8_ " One point is earned for using the correct mole ratio of Co to Ag . 1 Incl Co AmolCo = AmolAgx ~ C One point is earned for
A mass Co = A m0} C0 x 3.23.13. __ ' X __ . the COITECI answer
1 “101 C0 (negative sign is not required). AP® Chemistry
, ,FreeﬁewoneeScoring Guidelines Question 6 Answer each of the following using principles of atomic or molecular structure and/or intermolecular or
intramolecular forces. (a) Explain why the H—O—H bond angle in H20 is less than the H—N—H bond angles in NH3 , as shown in the table below.
HN~H
WBond AngleH 1n H20 Bond Angles 1n NH3 Both molecules have tetrahedral electron—domain geometries and One p.011“ 1s earned for Citing the
might be expected to have bond angles of 1095". However, electron difference 1? number Of
domains for nonbonding pairs of electrons exert a greater repulsion on nonbondmg p airs 0f electrons.
adjacent pairs of electrons than do electron domains for bonding pairs.
Thus, in the H20 molecule with its _t____wo nonbonding pairs of electrons, the electron domains of bonding pairs are tfiompressed to a greater repulsion from nonbon din g
greater extent than they are 1n the NH3 molecule, w ch has only one pairs as compared with bonding nonhonding pair of electrons. pairs. One point is earned for citing the (b) Explain why the radius of the Br atom is less than the radius of the Br‘ ion, as shown in the table below. Radars of Br Radius of Br: Olllnm Ol96nm One point is earned for recognition that
The nuclear charge (+35) is the same for both the Br and Br‘ Br and Br“ have the same nuclear species, but the “extra” electron in Br~ causes the electron charge. cloud to expand due to an increase in mutual repulsions among . . ‘ . .
the electrons that make up the cloud. One pornt is earned for Citing mcreased
repulsion among electrons. AP® Chemistry
Frees—Response,Scoring Guidelines Question 6 (continued) (c) Explain why the dipole moment of HI is less than the dipole moment of HCl , as shown in the table below. Di ole Moment of Ill D1 ole Moment of llCl 0 42 debye l 08 debyes Iodine, having a lower electronegativity than chlorine
has, forms a bond with hydrogen that is less polar than
the bond between chlorine and hydrogen in HCl . The
lower polarity of the H~l bond means that the dipole One point is earned for the comparison of the
moment of the bond is smaller than that of the H~Cl polarity of the two bonds. bond. One point is earned for citing the difference
in electronegativity between I and C1. (d) Explain why the normal boiling point of Ne is less than the normal boiling point of Kr, as shown in the
table below. NOTmal 301119 Point 91’, KT One point is earned for
mentioning that intermolecular
forces involved are London
(dispersion) forces. The intermolecular forces among atoms in liquid Ne are the same type of forces as those among atoms in liquid Kr, namely London
(dispersion) forces. However, the magnitudes of these forces are
smaller in Ne because the electron clouds of Ne atoms are
smaller and less polarizable than the electron clouds of Kr atoms. One point is earned for Note: An explanation that cites only periodic trends or only the mentioning the relative relative masses of Ne and Kr does not earn credit. polarizability 0f the electron
clouds of the atoms. ...
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