Notes
1) Average score = 2.87
a) two points
SrSO4(s) <===> Sr(aq)2+ + SO42¯(aq)
at equilibrium: [SO42¯] = x = [Sr2+]
(x) (x) = Ksp = 7.6 x 10¯7
(x) = 8.7 x 10¯4 mol / liter = solubility of SrSO4
b) three ponts
SrF2(s) <===> Sr(aq)2+ + 2 F(aq)¯
at equilibrium: [Sr2+] = x, [F¯] = 2x
Ksp = [Sr2+] [F¯]2 = (x) (2x)2 = 7.9 x 10¯10
x = 5.8 x 10¯4 mol / liter = solubility of SrF2
c) two points
Solve for [Sr+ required for precipitation of each salt.
Ksp = [Sr2+][F¯]2 = 7.9 x 10 ¯10
= (x) (0.020 mole / 1.0 L)2 = 7.9 x 10¯10
x = 2.0 x 10¯6 M
Ksp = [Sr2+][SO42¯] = 7.6 x 10¯7
= (y) (0.10 mole/1.0 liter) = 7.6 x 10¯7
y = 7.6 x 10¯6 M
Since 2.0 x 10¯6 M < 7.6 x 10¯6 M, SrF2 must precipitate first.
When SrF2 precipitates, [Sr2+] = 2.0 x 10¯6 M
d) two points
The second precipitate to form is SrSO4, which appears when [Sr2+] = 7.6 x 10¯6 M (Based on calculation in Part c.)
When [Sr2+] = 7.6 x 10¯6 M, [F¯] is determined as follows:
Ksp = [Sr2+][F¯]2 = 7.9 x 10¯10
= (7.6 x 10¯6) (z)2 = 7.9 x 10 ¯10
z = 1.0 x 10¯2M
%F¯ still in solution = 1.0 x 10¯2 / 2.0 x 10¯2 x 100 = 50.%
2) Average score = 4.91
a) one point
Zn ---> Zn2+ + 2e¯
2 TiO2+ + 4H+ + 2e¯ ---> 2 Ti3+ + H2O
The sum of the equations above is:
2 TiO2+ + 4H¯ + Zn ---> Zn2+ + 2Ti3+ + H2O
b) 3 points; one for each term in solution
(0.0500 liter x 0.115 mole TiO2+ / 1 liter) x (1 mole Zn / 2 mole TiO2+) x (65.4 grams Zn / 1 mole Zn)
= 0.188 gram Zn
c) two points
0.500 liter x (0.115 mole TiO2+ / 1 liter) x (1mole e¯ / 1 mole TiO2+) x (96.500coulombs / 1 mole e¯) x (1 amp-sec / 1 coulomb) x (1 / 1.06 amp)
= 523 sec
d) two points
| Zn ---> Zn2+ + 2e¯ | E° = 0.763 V |
| 2 TiO2+ + 4 H++ 2e¯ ---> 2 Ti3++ 2 H20 | E° = 0.060V |
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 105 J
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
C1.20H1.00 = C6H5
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 (C6H5) = 77
Number of empirical units per mole= 154 / 77 = 2
molecular formula = (C6H5) x 2 = C12H10
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 + H2O ---> Na+ + OH¯ + H2
b) H+ + HCO3¯ ---> H2O + CO2 (Part credit for H2CO3)
c) C2H5OH + HCOOH ---> HCOOC2H5 + H2O
d) OH¯+ Zn(OH)2 ---> Zn(OH)42¯ (or Zn(OH)3¯ or ZnO22¯ + H2O)
e) BF3 + NH3 ---> BF3NH3
f) Sn2+ + Fe3+ ---> Sn4+ + Fe2+
g) POCl3 + H2O ---> H3PO4 + H+ + Cl¯
h) MnO4¯ + SO32¯ + H+ ---> Mn2+ + SO42¯ + H2O
HSO3¯ and HSO4¯ were accepted.
(Part credit if H+ and H2O were omitted)
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: Na2O + 2 H2O ---> 2 Na+ + 2 OH¯
or
MgO + H2O ---> Mg(OH)2
Acidic oxide: any one of the oxides of Cl, S, or P
SO2+ H2O ---> H2SO3 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.
7) Average score = 2.36
a) three points
Precipitation of insoluble BaSO4 by adding a solution of a soluble sulfate, e.g., Na2SO4
Isolate BaSO4 by filtration
Purify BaSO4 by washing and drying it
b) three points
Dissolve copper(II) carbonate with nitric acid to form copper (II) nitrate solution
Isolate by avoiding the addition of excess acid
Purify by heating to drive off CO2
c) two points
Form CaCl2 solution by treating CaBr2solution with Cl2. Or add a soluble carbonate.
Separate the precipitated CaCO3, wash, and dissolve it in HCl.
Purify CaCl2solution by extracting the Br2 with CH2Cl2or equivalent. Or heat the solution.
8) Average score = 3.52
a) thre points
Points on ordinate take into account the initial amounts of the three substances, and PCl5 line rises while others fall.
Lines curved at start and flat after equilibrium.
Concentration changes should be consistant with the fact that all coefficients in the equation are unity.
b) two points - First order in both reactants, Inclusion of constant
Rate = k [PCl3] [Cl2]
c) one point
Reaction requires effective collisions between molecules of PCl3 and Cl2 As concentrations of these molecules increase, the number of effective collisions must increase and the rate of action increases.
d) two points
The fraction of colliding molecules with the required activation evergy increases as the temperature rises.
9) Average score = 2.95
a) six points
H2 and C3H2 have low melting points because the forces involved are the weak van der Waals (or London) forces.
HF has a higher melting point because intermolecular hydrogen bonding is important.
CsI and LiF have still higher melting points because ionic lattice forces must be overcome to break up the crystals, and the ionic forces are stronger than van der Waals forces and hydrogen bonds.
SiC is an example of a macromolecular substance where each atom is held to its neighbors by a very strong covalent bond.
b) two points
C3H8and H2: There are more interactions per molecule in C3H8than in H2. Or C3H8 is weakly polar and H2 is nonpolar.
LiF and CsI: The smaller ions in LiF result in a higher lattice energy than CsI has. Lattice energy U is proportional to 1 / (r+ + r¯)
Copyright © 1985 by College Entrance Examination Board and
Educational Testing Service. All rights reserved.