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
- (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