Cliffs AP Chemistry, 3rd Edition

Therefore, the Cr ion must have a charge of +3.

Choice (D) also yields a +3 charge for Cr, and yet does not conform to standard methods

of writing complex ion formulas.

59.(C) Raoult’s law states that the partial pressure of a solvent over a solution, P 1 , is given by

the vapor pressure of the pure solvent, P° 1 , times the mole fraction of the solvent in the so-

lution, X 1.

P 1 = X 1 P° 1

A decrease in vapor pressure is directly proportional to the concentration (measured as

mole fraction) of the solute present.

60.(E) In dilution problems, we use the formula M 1 V 1 = M 2 V 2 ; therefore, it is necessary to de-
termine the molarity of the initial solution first.
.
mL solution

g solution

liter solution

mL solution

1

150

1

1000

#

.

g solution..

g HCl

g HCl

mole HCl M

100

36 45

36 45

##^1 =15 0

Next we use the relationship M 1 V 1 = M 2 V 2 :

(15.0 M)(x liters) = (5.0 M)(9.0 liters)

x= 3.0 liters

61. (B) In examining the balanced equation, note that for each mole of N 2 O 5 gas that decom-
    poses,^1 ⁄ 2 mole of O 2 gas is formed. Therefore, the rate of formation of oxygen gas should
    be half the rate of decomposition of the N 2 O 5.

62.(B) Begin by writing the equilibrium equation.

PbCl 2 DPb^2 +()aq +() 2 Cl aq-

Next, write the equilibrium expression.

KPbClsp^2

2

= 77 +-AA

In reference to the chloride ion concentration, rewrite the expression for [C1–]:

Cl

Pb

Ksp

2

12

• = +

J

L

K

K

N

P

O

7 A 7 AO

At any value greater than this expression, PbCl 2 (s) will precipitate, removing Cl–(aq) from

solution.

63.(D) Use the equation NaVa= NbVb. Solve the equation for Na.

.

..

N NVV mL.

NmL

25 0 N

050 500

a^10

a

==bb # =

Answers and Explanations for the Practice Test