CHM116 Practice Test for Exam 3 Questions and Answers Spring 2024

Practice Test for Exam 3
Dr Cabirac Lecture


1. For PbCl2 (Ksp = 2.4 x 10-4 ), will a precipitate of PbCl2 form when 0.10 L of 3.0 x 10-2
M Pb(NO3)2 is added to 400 mL of 9.0 x 10-2 M NaCl?

A. Yes, because Q > Ksp.
B. No, because Q < Ksp
C. No, because Q = Ksp
D. Yes, because Q < Ksp.


2. The molar solubility of tin(II) iodide is 1.28 x 10-2 mol/L. What is Ksp for this compound?
A. 8.4 x 10-6
B. 1.28 x 10-2
C. 4.2 x 10-6
D. 1.6 x 10-4
E. 2.1 x 10-6


3. The Ksp for silver(I) phosphate is 1.8 x 10-18. Determine the silver ion concentration
in a saturated solution of silver(I) phosphate.
A. 1.6 x 10-5 M
B. 2.1 x 10-5 M
C. 3.7 x 10-5 M
D. 1.1 x 10-13 M
E. 4.8 x 10-5 M


4. Calculate the molar solubility of BaCO3 in a 0.10 M solution of
Na2CO3 (aq). (Ksp (BaCO3 ) = 8.1 x 10-9 )
A. 8.1 x 10-9 M
B. 9.0 x 10-5 M
C. 8.1 x 10-8 M
D. 2.8 x 10-4 M
E. 0.10 M


5. Which of the following compounds is more soluble in acidic solution than in
pure neutral water?
A. BaCO3
B. CuI
C. PbCl2
D. AgBr
E. NH4NO3




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, 6. Arrange these compounds in order of increasing standard molar
entropy at 25°C: C3H8(g), C2H4(g), ZnS(s), and H2O(l).
A. ZnS(s) < H 2O(l) < C3H8(g) < C2H4(g)
B. C2H4(g) < H2O(l) < C3H8(g) < NaCl(s)
C. ZnS(s) < C3H8(g) < C2H4(g) < H2O(l)
D. C3H8(g) < C2H4(g) < H2O(l) < ZnS(s)
E. ZnS(s) < H2O(l) < C2H4(g) < C3H8(g)


7. Without reference to a table, arrange these reactions according to increasing ∆S.
1) CH (g) + H O(g) → CO(g) + 3H (g)
4 2 2
2) C(s) + O (g) → CO (g)
2 2
3)
H2O2(l) → H2O(l) + 1/2O2(g)

A. 1 < 3 < 2
B. 2 < 3 < 1
C. 2 < 1 < 3
D. 3 < 2 < 1
E. 3 < 1 < 2


8. Which of the following is expected to have zero entropy?
I. N2(g) at 273 K
II. SiO2(s, amorphous) at 0 K
III. NaCl(s) perfectly ordered crystal at 25 K
IV. Na(s) perfectly ordered crystal at 0 K

A. I and IV
B. III and IV
C. I and II
D. I, II, and III
E. IV only



9. Aluminum forms a layer of aluminum oxide when exposed to air which
protects the bulk metal from further corrosion.
4Al(s) + 3O2(g) → 2Al2O3(s)
Calculate ∆G° for this reaction, given that ∆G°f of aluminum oxide is –1576.4 kJ/mol.

A. –3152.8 kJ/mol
B. –1576.4 kJ/mol
C. –788.2 kJ/mol
D. 1576.4 kJ/mol
E. 3152.8 kJ/mol




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