Solutions and Colligative Properties ❮ 193- D—This is a dilution problem:
Vafter = MV
M
()()
()before before
after()
()10.0 M HNO ()50.0 mL
4.00 M HNO3
3= 125 mL
The final volume is 125 mL. Since the original
volume was 50.0 mL, an additional 75.0 mL
must be added.
- A—Solutions cannot be separated by titra-
tions or filtrations. Electrolysis of the solu-
tion would produce hydrogen and oxygen gas.
Chromatography might achieve a minimal sepa-
ration. - C—The solubility of a gas is increased by
increasing the partial pressure of the gas and by
lowering the temperature. Pick the answer with
the highest nitrogen pressure and the lowest
temperature.
15. C—Three of the compounds are strong electro-
lytes, and one is a weak electrolyte. The weak
electrolyte, HC 2 H 3 O 2 , will not be as good a con-
ductor as the strong electrolytes (eliminating D).
The strong electrolyte that produces the highest
concentration of ions will be the best conduc-
tor. The number of ions produced by the strong
electrolytes may be found by simply counting
the ions: A, 3 (Ba^2 + + 2 OH-); B, 2 (K+ + Cl-);
C, 4 (3 Na+ + PO 43 - ). The ion concentration for
the strong electrolytes is the molarity of the solu-
tion times the number of ions present (van’t Hoff
factor). Answer C gives the highest concentration.
16. C—The ionic radii are too alike to make a sig-
nificant difference; therefore, it is necessary to
focus on the charge differences. The greater the
charge, the greater the attraction of the ion for
the polar water molecules (ion–dipole forces).
The greater the attraction is, the greater the
energy change is.
❯ Free-Response Questions
Question 1
You have 5 minutes to answer the following two-part question. You may use a calculator
and the tables in the back of the book.
Five beakers each containing 100.0 mL of an aqueous solution are on a lab bench. The
solutions are all at 25°C. Solution 1 contains 0.20 M KNO 3. Solution 2 contains 0.10 M
BaCl 2. Solution 3 contains 0.15 M C 2 H 4 (OH) 2. Solution 4 contains 0.20 M (NH 4 ) 2 SO 4.
Solution 5 contains 0.25 M KMnO 4.
(a) Which solution has the lowest pH? Explain.
(b) Which solution would be the poorest conductor of electricity? Explain.Question 2
You have 15 minutes to answer the following four-part question. You may use a calculator
and the tables in the back of the book.
Five beakers are placed in a row on a countertop. Each beaker is half-filled with
a 0.20 M aqueous solution. The solutes, in order, are (1) potassium sulfate, K 2 SO 4 ;
(2) methyl alcohol, CH 3 OH; (3) sodium carbonate, Na 2 CO 3 ; (4) ammonium chromate,
(NH 4 ) 2 CrO 4 ; and (5) barium chloride, BaCl 2. The solutions are all at 25°C. Answer the
following questions with respect to the five solutions.
(a) Which solution will form a precipitate when ammonium chromate is added to it?
(b) Which solution is the most basic? Explain.
(c) Which solution would be the poorest conductor of electricity? Explain.
(d) Which solution is colored?