AP Chemistry 1985 Free Response Answers and Grading Criteria

AP Chemistry 1985 Free Response Answers and Grading Criteria

AP Chemistry 1985 Free Response Answers provide detailed solutions and grading criteria for the exam. This resource is essential for students preparing for the AP Chemistry exam, offering insights into the grading process and common pitfalls. The answers cover various topics, including solubility, chemical reactions, and thermodynamics, helping students understand key concepts and improve their problem-solving skills. Ideal for AP Chemistry students looking to enhance their exam performance, this guide breaks down complex problems into understandable solutions.

Key Points

  • Includes detailed solutions for AP Chemistry free response questions from 1985.
  • Covers key topics such as solubility product constants and thermodynamic principles.
  • Provides grading criteria to help students understand how their answers will be evaluated.
  • Offers insights into common mistakes made by students in AP Chemistry exams.
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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 topics are covered in the AP Chemistry 1985 Free Response Answers?
The AP Chemistry 1985 Free Response Answers cover a variety of topics essential for understanding the exam material. Key areas include solubility equilibria, thermodynamics, and electrochemistry. Each answer is accompanied by a detailed explanation, helping students grasp the underlying concepts. This comprehensive approach ensures that students can tackle similar problems in their own exams.
How are the AP Chemistry free response questions graded?
Grading for AP Chemistry free response questions is based on a rubric that evaluates the accuracy of the answers and the clarity of the explanations. Each part of a question is assigned points, and students can lose points for significant errors in calculations or for not following proper scientific conventions. Understanding the grading criteria helps students focus their study efforts on areas that will maximize their scores.
What is the significance of the solubility product constant (Ksp) in AP Chemistry?
The solubility product constant (Ksp) is crucial in AP Chemistry as it quantifies the solubility of sparingly soluble salts. It allows students to predict whether a precipitate will form in a given solution based on the concentrations of the ions involved. Understanding Ksp is essential for solving equilibrium problems and is frequently tested in AP Chemistry exams.
What types of chemical reactions are analyzed in the AP Chemistry 1985 responses?
The AP Chemistry 1985 Free Response Answers analyze various types of chemical reactions, including redox reactions, precipitation reactions, and acid-base neutralizations. Each response provides a step-by-step breakdown of the reaction mechanisms and the calculations involved. This thorough analysis helps students develop a deeper understanding of reaction types and their applications.
How can students use the grading criteria to improve their AP Chemistry scores?
Students can use the grading criteria outlined in the AP Chemistry 1985 Free Response Answers to identify areas where they may lose points. By reviewing the common mistakes highlighted in the grading rubric, students can focus their study efforts on improving their problem-solving techniques and clarity of explanations. This targeted approach can lead to higher scores on future exams.
What are common pitfalls students face in AP Chemistry free response questions?
Common pitfalls in AP Chemistry free response questions include miscalculating concentrations, failing to include units, and not clearly explaining the reasoning behind answers. Students often overlook significant figures and proper notation, which can lead to point deductions. By being aware of these issues, students can take steps to avoid them in their own responses.