AP Chemistry 1985 Free Response Answers and Grading Criteria

AP Chemistry 1985 Free Response Answers and Grading Criteria

The AP Chemistry 1985 Free Response Answers provide detailed solutions to exam questions, covering key topics such as solubility, chemical reactions, and thermodynamics. This resource is essential for AP Chemistry students preparing for the exam, offering insights into scoring and grading criteria. Each problem includes step-by-step calculations and explanations, ensuring a comprehensive understanding of the material. Ideal for students seeking to improve their exam performance and grasp complex chemistry concepts.

Key Points

  • Includes detailed solutions for AP Chemistry exam questions from 1985.
  • Covers key topics like solubility products and electrochemistry.
  • Provides scoring criteria and grading insights for AP Chemistry responses.
  • Offers step-by-step calculations for complex chemistry problems.
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Advanced Placement Chemistry: 1985 Free Response Answers
[delta] and [sigma] are used to indicate the capital Greek letters.
[square root] applies to the numbers enclosed in parenthesis immediately following
All simplifying assumptions are justified within 5%.
One point deduction for a significant figure or math error, applied only once per problem.
This note was found at the end of question 3: For each of the problems, a maximum of 1 point was
subtracted for gross misuse of significant figures and for mathamatical errors if correct principles were
used.
1) Average score = 2.87
a) two points
SrSO ¯(aq)
4
(s) <===> Sr(aq) + SO
2+
4
2
at equilibrium: [SO ¯] = x = [Sr ]
4
2 2+
(x) (x) = K = 7.6 x 10¯
sp
7
(x) = 8.7 x 10¯ mol / liter = solubility of SrSO
4
4
b) three ponts
SrF
2
(s) <===> Sr(aq) + 2 F(aq)¯
2+
at equilibrium: [Sr ] = x, [F¯] = 2x
2+
K
sp
= [Sr ] [F¯] = (x) (2x) = 7.9 x 10¯
2+ 2 2 10
x = 5.8 x 10¯ mol / liter = solubility of SrF
4
2
c) two points
Solve for [Sr required for precipitation of each salt.
+
K
sp
= [Sr = 7.9 x 10 ¯
2+
][F¯]
2 10
= (x) (0.020 mole / 1.0 L) = 7.9 x 10¯
2 10
x = 2.0 x 10¯ M
6
K
sp
= [Sr ¯] = 7.6 x 10¯
2+
][SO
4
2 7
= (y) (0.10 mole/1.0 liter) = 7.6 x 10¯
7
y = 7.6 x 10¯ M
6
Since 2.0 x 10¯ M < 7.6 x 10¯ M, SrF must precipitate first.
6 6
2
When SrF precipitates, [Sr ] = 2.0 x 10¯ M
2
2+ 6
d) two points
The second precipitate to form is SrSO , which appears when [Sr ] = 7.6 x 10¯ M (Based on calculation in Part c.)
4
2+ 6
When [Sr ] = 7.6 x 10¯ M, [F¯] is determined as follows:
2+ 6
K
sp
= [Sr = 7.9 x 10¯
2+
][F¯]
2 10
= (7.6 x 10¯ ) (z) = 7.9 x 10 ¯
6 2 10
z = 1.0 x 10¯ M
2
%F¯ still in solution = 1.0 x 10¯ / 2.0 x 10¯ x 100 = 50.%
2 2
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2) Average score = 4.91
a) one point
Zn ---> Zn + 2e¯
2+
2 TiO + 4H + 2e¯ ---> 2 Ti + H
2+ + 3+
2
O
The sum of the equations above is:
2 TiO + 4H¯ + Zn ---> Zn + 2Ti + H
2+ 2+ 3+
2
O
b) 3 points; one for each term in solution
(0.0500 liter x 0.115 mole TiO / 1 liter) x (1 mole Zn / 2 mole TiO ) x (65.4 grams Zn / 1 mole Zn)
2+ 2+
= 0.188 gram Zn
c) two points
0.500 liter x (0.115 mole TiO / 1 liter) x (1mole e¯ / 1 mole TiO ) x (96.500coulombs / 1 mole e¯) x (1 amp-sec /
2+ 2+
1 coulomb) x (1 / 1.06 amp)
= 523 sec
d) two points
Zn ---> Zn + 2e¯ E° = 0.763 V
2+
2 TiO + 4 H + 2e¯ ---> 2 Ti + 2 H 0 E° = 0.060V
2+ + 3+
2
E° for total reaction: 0.763V + 0.060V = 0.823 V
[delta]G° = -nFE°
= - (2 mole e¯) (96.500 coulombs / 1 mole electrons) (0.823 / 1mole) (1 joule/1 V-coul)
= - 1.59 x 10 J
5
3) Average score = 5.15
a) two points
Assume a 100-gram sample of hydrocarbon
93.46 grams C x (1 mole C / 12.01 grams C) = 7.782 moles C
6.54 grams H x (1 mole H / 1.008 grams H) = 6.49 moles H
7.782 moles C / 6.49 moles H = 1.20
C
1.20
H H
1.00
= C
6 5
b) two points
Molality = moles solute per kilogram solvent
m = (2.53 grams solute / 25.86 grams solute) x (1 mole solute / 147.0 grams solute) x (1000 grams solute / 1 kg
solvent) = 0.665 m
c) one point
Freezing point lowering = 80.2° - 75.7° = 4.5°
Molal freezing point depression constant = (4.5° / 0.665 molal solution)
= 6.8° lowering for 1 molal solution
d) three points
Freezing point lowering = 80.2° - 76.2° = 4.0°
(6.8 x kg. solvent/mole solute) x (1 / 4.0°) x (2.43 grams solute / 26.7 grams solvent) x (1000 grams solvent / kg
solvent)
= 154 grams solute / mole solute)
e) one point
Empirical unit weight (C ) = 77
6
H
5
Number of empirical units per mole= 154 / 77 = 2
molecular formula = (C ) x 2 = C
6
H
5 12
H
10
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4) Average score = 4.05
Credit: 1 point for each set of reactants, 2 points for products. (If two products, 1 point for each product.)
a) Na + H O ---> Na + OH¯ + H
2
+
2
b) H + HCO ¯ ---> H O + CO (Part credit for H
+
3 2 2 2
CO )
3
c) C OH + HCOOH ---> HCOOC + H
2
H
5 2
H
5 2
O
d) OH¯+ Zn(OH) ---> Zn(OH) ¯ (or Zn(OH) ¯ or ZnO ¯ + H
2 4
2
3 2
2
2
O)
e) BF + NH ---> BF
3 3 3
NH
3
f) Sn + Fe ---> Sn + Fe
2+ 3+ 4+ 2+
g) POCl + H O ---> H + H + Cl¯
3 2 3
PO
4
+
h) MnO ¯ + SO ¯ + H ---> Mn + SO ¯ + H
4 3
2 + 2+
4
2
2
O
HSO
3
¯ and HSO ¯ were accepted.
4
(Part credit if H and H O were omitted)
+
2
5) Average score = 3.43
a) three points
Oxides at left are basic and become less basic: more acidic as one moves to the right.
Basic oxide: Na O + 2 H O ---> 2 Na + 2 OH¯
2 2
+
or
MgO + H O ---> Mg(OH)
2 2
Acidic oxide: any one of the oxides of Cl, S, or P
SO SO
2
+ H O ---> H
2 2 3
or equivalent for another oxide
(Both examples with no equations 1 point)
b) two points
Oxidizing strengths of halogen elements decrease down the group.
Since atoms get larger down the group, the attraction for e¯ decreases and oxidizing strength increases
c) two points
Reducing strengths of alkali metals increases down the group.
Since atoms get larger down the group, loss of outer electrons is easier and reducing strength increases.
6) Average score = 3.34
a) four points
[delta]H > 0 since heat is required to change liquid water to vapor.
[delta]S > 0 since randomness increases when a liquid changes to vapor.
[delta]G < 0 since the evaporation takes place in this situation.
[delta]T < 0 since the more rapidly moving molecules leave liquid first. The liquid remaing is cooler.
b) three points
[delta]H > 0. The system after the dissolving has a lower temperature and so the change is endothermic.
[delta]S > 0. Since the solution is less ordered than the separate substances are.
[delta]G < 0. The solution occured and so is spontaneous.
c) one point
Solubility increases. The added heat available pushes the endothermic process toward more dissolving.
Note: Both direction of change and explanation were required for full credit. If explanations for a variable were
correct but the directions of change were wrong, credit was granted only once for that variable.
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FAQs of AP Chemistry 1985 Free Response Answers and Grading Criteria

What types of problems are included in the AP Chemistry 1985 Free Response Answers?
The AP Chemistry 1985 Free Response Answers include a variety of problems related to solubility, chemical equilibrium, thermodynamics, and electrochemistry. Each question is designed to test students' understanding of key chemistry concepts and their ability to apply these concepts in problem-solving scenarios. The answers provide detailed explanations and calculations, making it easier for students to follow the reasoning behind each solution.
How does the scoring rubric work for the AP Chemistry exam?
The scoring rubric for the AP Chemistry exam is based on a point system where each problem is assigned a specific number of points. Points are awarded for correct answers, as well as for the justification of the steps taken to arrive at those answers. The rubric emphasizes the importance of showing work and applying correct principles, allowing students to earn partial credit even if their final answer is incorrect.
What is the significance of Ksp in solubility problems?
Ksp, or the solubility product constant, is a crucial concept in solubility problems as it quantifies the extent to which a compound can dissolve in solution. It is used to determine the concentrations of ions in a saturated solution and predict whether a precipitate will form when additional ions are introduced. Understanding Ksp is essential for solving equilibrium problems in AP Chemistry.
What are common themes in the AP Chemistry 1985 exam questions?
Common themes in the AP Chemistry 1985 exam questions include the application of chemical principles to real-world scenarios, the analysis of reaction mechanisms, and the understanding of thermodynamic concepts. Students are often required to integrate knowledge from different areas of chemistry, such as kinetics and equilibrium, to solve complex problems. This interdisciplinary approach helps prepare students for advanced studies in chemistry.
How can students use the 1985 Free Response Answers to prepare for the AP Chemistry exam?
Students can use the 1985 Free Response Answers as a study tool by practicing the problems and reviewing the detailed solutions provided. By working through each question, students can identify areas where they need improvement and gain a deeper understanding of the material. Additionally, reviewing the scoring criteria helps students understand how their answers will be evaluated, allowing them to focus on providing thorough justifications in their responses.

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