1. Which response lists all the following pairs of liquids that are ...

1. Which response lists all the following pairs of liquids that are mutually soluble.
Pair # 1. octane (C 8 H 18 ) and water.
Pair # 2. acetic acid (CH 3 COOH) and water.
Pair # 3. octane (C 8 H 18 ) and benzene (C 6 H 6 ).
One is polar, the other non-polar.
Both are polar and capable of hydrogen bonding
Both are not polar with only dispersion forces
A. 1, 3
B. 1, 2
C. 3
D. 2, 3
E. 2
2. Which of the following liquids would make a good solvent for iodine, I 2 ?
A. HCl
B. H 2 O
C. CH 3 OH
D. NH 3
E. C 6 H 14 This is not polar (dispersion forces only) like I 2 . The others are all polar.
3. What is the molarity of a 17.0% solution of sodium acetate, CH 3 COONa (0 = 82.0 g/mol) in water? Given the density is 1.09
g/mL.
17% indicates 17 g sodium acetate (NaOAc) in 100 g solution (17g NaOAc and 83 g H 2 O)
A. 2.26 x 10 -2 M
1 mol.
1 mL 1 L
B. 0.207 M mol. NaOAc = 17 g x = 0.207 mol. L sol. = 100 g x x = 0.0917 L
C. 2.07 M
18 g
1.09 g 1000 mL
D. 2.26 M
0.207 mol.
E. 2.72 M M = = 2.26
0.0917 L
4. What is the percent CdSO 4 (0 = 208.5 g/mol) by mass in a 1.0 molal aqueous solution?
A. 0.001 %
B. 0.10 %
C. 17.2 %
D. 20.8 %
E. 24.4 %
1.0 m indicates that 1.0 mol. CdSO 4 is dissolved in 1 kg (1000 g) water.
With a molecular mass of 208.5 g/mol. indicates that that 1.0 mol. equals 208.5 g
g CdSO4
208.5 g CdSO4
% mass = x 100% =
x 100% = 17.2 %
g total
208.5 g CdSO + 1000 g H O
4
2
5. Calculate the percent by mass of potassium nitrate in a solution made from 45.0 g KNO 3 and 295 mL of water. Density of
water is 1.00 g/mL.
A. 1.51%
B. 7.57%
C. 13.3%
D. 15.2%
E. none of them
g KNO3
45.0 g KNO3
% mass = x 100% =
x 100% = 13.2 %
g total
45.0 g KNO + 295 g H O
3
2
6. How many grams of water are needed to dissolve 27.8 g of ammonium nitrate (0 = 80.05 g/mol) in order to prepare a
0.452 m solution?
A. 0.157 g
B. 36.2 g
C. 100 g
D. 769 g
E. 157 g
1 mol.
mol. NH
=
80.05 g
4NO3
= 27.8 g x 0.247 mol. NH 4NO3
0.247 mol. NH 4NO3
0.452 m = ⇒ x = 0.769 g H 2O
= 769 g H 2O
x kg H O
2

7. Calculate the molality of a solution containing 14.3 g of NaCl (0 = 58.4 g/mol) in 42.2 g of water.
A. 2.45 x 10 4 m
B. 5.80 x 10 4 m
C. 24.5 m
D. 5.79 m
E. 103 m
1 mol.
mol. NaCl = 14.3g x = 0.245 mol. NaCl
58.4 g
mol. NaCl 0.245 mol. NaCl
m = =
= 5.79 m
kg H O 0.0422 kg H O
2
2
8. The solubility of CO 2 gas in water
A. increases with increasing temperature.
B. decreases with increasing temperature.
C. is not dependent on temperature.
9. The solubility of CO 2 gas in water
A. increases with increasing gas pressure. Recall Henry’s Law: c = k· P (where c is solubility and P is the pressure)
B. decreases with increasing gas pressure.
C. is not dependent on pressure.
10. Which of the following statements are false?
A. The vapor pressure of a solvent over a solution decreases as its mole fraction increases.
B. The solubility of a gas increases as the temperature decreases. See question 8.
C. The vapor pressure of a solvent over a solution is less than that of pure solvent. See question 12.
D. The greater the pressure of a gas over a solution the greater its solubility. See question 9.
E. Ionic solutes dissociate in solution causing an enhancement of all colligative properties. See question 14.
11. Consider a solution made from a nonvolatile solute and a volatile solvent. Which statement is true?
A. The vapor pressure of the solution is always greater than the vapor pressure of the pure solvent. See question 12
B. The boiling point of the solution is always greater than the boiling point of the pure solvent. See question 14
C. The freezing point of the solution is always greater than the freezing point of the pure solvent. See question 17
D. The osmotic pressure of the solution is always less than the osmotic pressure of the pure solvent. See questions 15, 16
12. The vapor pressure of water at 20°C is 17.5 mmHg. What is the vapor pressure of water over a solution prepared from 0.584
mol. of sucrose and 19.4 mol. water?
A. 0.53 mmHg
B. 5.81 mmHg
C. 17.0 mmHg
D. 18.0 mmHg
E. 33.2 mmHg
χ =
mol. solventl
mol. total
19.4 mol. H 2O
=
= 0.971
19.4 H O + 0.584 mol.sucrose
P = χP° = (0.971)(17.5 mm Hg) = 17.0 mm Hg
2
13. Dissolving a solute such as KOH in a solvent such as water results in
A. an increase in the melting point of the liquid.
B. a decrease in the boiling point of the liquid.
C. a decrease in the vapor pressure of the liquid. See question 12.
D. no change in the boiling point of the liquid.
14. Which of the following aqueous solutions has the highest boiling point? Given K b = 0.52°C/m
A. 0.2 m KCl
B. 0.2 m Na 2 SO 4
C. 0.2 m Al(NO 3 ) 3
D. pure water
E. all are the same
∆T b = i· K b· m Since the boiling point constant and molality are the same for A-C, the determining
factor is i, which is 2 for A, 3 for B and 4 for C. Recall that I gives the number of dissolved
species, and that the ionic compounds all dissociate.

15. Which of the following solutions has the highest osmotic pressure at 25°C?
A. 0.2 M LiBr
B. 0.2 M ethanol
C. 0.2 M Na 2 SO 4
D. 0.2 M KCl
Π = i· MRT Since M, R and T are the same for all solutions, the determining factor must be i,
which gives the number of dissolved species in solution. i = 3 for Na 2 SO 4 (2 Na + and 1 SO 2- 4 ).
16. During osmosis
A. pure solvent diffuses through a membrane but solutes do not.
B. pure solutes diffuse through a membrane but solvent does not.
C. pure solvent and a solution both diffuse at the same time through a membrane.
D. gases diffuse through a membrane into a solution and build up pressure.
17. Calculate the freezing point of a solution made from 22.0 g of octane (0 = 114.2 g/mol) dissolved in 148.0 g of benzene.
Benzene freezes at 5.50°C and its K f value is 5.12°C/m.
A. -1.16°C
B. 0.98°C
C. 6.66°C
D. 12.2°C
E. 5.49°C
1mol.
0.193 mol. octane
22.0 g x = 0.193 mol. octane m =
= 1.30 m
114.2 g
0.148 kg
∆T f = i· K f· m =(1)(5.12°C/m)(1.30 m) = 6.66°C
T f = T f ° - ∆T f = 5.50°C – 6.66°C = -1.16°C
18. Consider a 0.90 M Al(NO 3 ) 3 solution. This solution has a nitrate ion concentration of
A. 0.30 M
B. 0.90 M
C. 0.0 M
D. 8.1 M
E. 2.7 M
Al(NO 3 ) 3 → Al 3+ (aq) + 3NO - 3 (aq)
Thus 0.90 M Al(NO 3 ) 3 will dissociate, giving 3 times that concentration of NO - 3 .
19. Alpha particles are identical to
A. protons.
B. helium atoms.
C. hydrogen atoms.
D. helium nuclei.
E. electrons.
Consider the nuclear symbols for electrons ( 4 4
2 α ) and beta particles ( 2 He
) and the fact that the
particle originates from the nucleus and is devoid of electrons.
20. Beta particles are identical to
A. protons.
B. helium atoms.
C. hydrogen atoms.
D. helium nuclei.
E. electrons.
Consider the nuclear symbols for electrons ( - ) and beta particles (
e
0 1
- 0 1 )
33
21. How many neutrons and protons (nucleons) does an atom with the symbol: 16 S
A. 33
B. 16
C. 49
D. 16
E. none of them
33 is the mass number and is equal to the number of protons + neutrons.

22. A radioisotope decays to give an alpha particle and Pb-208. What was the original element?
A. Se
B. Bi
C. Po
D. Hg
E. Rn
A 4 208
Z X → 2α
+ 82
Pb
A = 4 + 208 = 212, Z = 82 + 2 = 84
Thus:
A
X 212
Z = 84
Po
23. When atoms of beryllium-9 are bombarded with alpha particles, neutrons are produced. What new isotope is also formed?
4 9
1
He + Be → n X
2 4
0 +
12
A. 6 C
5
B. 3 Li
8
C. 3 Li
10
D. 5 B
12
E. 5 B
4 9 1 A
2 He + 4Be
→ 0n
+ Z
A = 4 + 9 - 1 = 12, Z = 2 + 4 - 0 = 6
Thus:
A
Z X =
12 6
C
X
24. When atoms of aluminum-27 are bombarded with alpha particles, neutrons are produced. What new isotope is also formed?
4 27 1
He + Al → n X
2 13 0 +
31
A. 14 Si
30
B. 14 Si
31
C. 16 S
30
D. 15 P
31
E. 15 P
4 27 1 A
2 He + 13 Al → 0n
+ Z
A = 4 + 27 - 1 = 30, Z = 2 + 13 - 0 = 15
Thus:
A
X 30
Z = 15
P
X
25. Sulfur-35 decays by beta emission. The decay product is:
35
A. 15 P
36
B. 16 S
34
C. 16 S
34
D. 17 Cl
35
E. 17 Cl
35 0 A
16 S → -1β
+ Z
X
A = 35 - 0 = 35, Z = 16 - -1 = 17
Thus:
A
Z = 17
X 35 Cl
232
26. In the 90 Th decay series there are six radioisotopes (including the Th-232 itself) that decay be alpha emission, and four that
decay by beta emission. What is the final (stable) isotope?
204
A. 74 W
208
B. 74 W
204
C. 82 Pb
208
D. 82 Pb
208
E. 78 Pt
232
4 0 A
90 Th → 6 2α
+ 4 -1β
+ Z
X
A = 232 – 6(4) – 4(0) = 208, Z = 90 – 6(2) – 4(-1) = 82
Thus:
A
Z X =
208
82
Pb

27. Calculate the energy released in joules when one mole of Po-214 decays according to the equation
214
210 4
Po → Pb He
84 82 + 2
Given the atomic masses: Pb-210 = 209.98284 amu, Po-214 = 213.99519 amu, and He-4 = 4.00260 amu.
(c = 3.00 x 10 8 m/s, 1000 amu· m 2 /s 2 = 1 J/mol.)
A. 8.78 x 10 14 J/mol.
B. 7.2 x 10 14 J/mol.
C. 8.78 x 10 11 J/mol.
D. -9.75 x 10 -3 J/mol.
E. 1.46 x 10 -9 J/mol.
∆m = 213.99519 – 209.98284 – 4.00260 = 0.00975 amu
E = mc 2 = (0.00975 amu)(3.00 x 10 8 m/s) = 8.78 x 10 14 amu· m 2 /s 2
E = 8.78 x 10 11 J/mol.
28. What fraction of radioactive atoms remains in a sample after six half-lives?
A. zero
B. 1/6
C. 1/16
D. 1/32
E. 1/64
After one half-life, only ½ the original amount remains. After a second half-life, ½ of that is
left (leaving ¼ of the original). After a third half-life, ½ of that is left (leaving C of the
original). Thus: (½)(½)(½)(½)(½)(½) = [ is left after 6 half-lives.
29. Carbon-11 is a radioactive isotope of carbon. Its half-life is 20.3 minutes. What fraction of the initial number of C-11 atoms
in a sample will remain after 81 minutes?
A. 1/16
B. 1/4
C. 1/2
D. 1/32
E. 1/8
81 minutes represents 4 half-lives.
Thus: (½)(½)(½)(½) = L is left after 4 half-lives.
30. The 14 C activity of some ancient Peruvian corn was found to be 10 disintegrations per minute per gram of carbon. If presentday
plant life shows 15 dpm/g, how old is the Peruvian corn? The half-life of 14 C is 5730 yr.
A. 1,455 yr
B. 1,910 yr
C. 3,350 yr
D. 3,820 yr
E. 9,080 yr
0.693
λ
-4 -1
1/2 = 5730 = ⇒ λ = 1.21x
10 yr
t
⎛ Ao
⎞ ⎛ 15 ⎞
ln⎜
⎟ = ln⎜
⎟ = λ ⋅ t ⇒ t = 3350 yr
⎝ A ⎠ ⎝10
⎠
31. The radioisotope potassium-40 decays to argon-40 with a half-life of 1.3 x 10 9 yr. A sample of moon rock was found to
contain 78 40 Ar atoms for every 22 40 K atoms.
40
0 40
K → + β Ar
19 1 + 19
The age of the rock is:
A. 8.1 x 10 -10 yr
B. 2.4 x 10 9 yr
C. 2.8 x 10 9 yr
D. 4.6 x 10 9 yr
E. 6.8 x 10 9 yr
0.693
λ
9
-10 -1
1/2 = 1.3 x 10 = ⇒ λ = 5.3 x 10 yr
t
⎛ N o ⎞ ⎛ 78 + 22 ⎞
ln⎜
⎟ = ln⎜
⎟ = λ ⋅ t ⇒
⎝ N ⎠ ⎝ 22 ⎠
t = 2.8 x 10
9
yr
32. What role does cadmium metal (Cd) play in a nuclear reactor?
A. slows down the fission neutrons (moderator)
B. transfers heat from the reactor to the heat exchanger (primary coolant)
C. controls chain reaction (control rods)
D. transfers heat from the condenser to the environment (cooling tower)
E. undergoes fission (fuel rods)

33. Which of the following is an advantage of nuclear power plants over coal-burning plants?
A. form numerous radioactive fission products.
B. do not pollute the air with SO 2 , soot, and fly-ash.
C. produce more thermal pollution than coal plants.
D. use more fuel.