bleiweiss (emb853) – H01: Chapt 16 17 – Mccord – (90540)
1
This printout should have 25 questions.
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beFore answering.
Chapter 16 sections 10 and 11, plus the
frst part oF Chapter 17, dissolving solutes
into solvents: Likes dissolve likes.
001
10.0 points
What is the fnal concentration oF NaOH when
335 mL oF 0.75 M NaOH are mixed with
165 mL oF 0.15 M NaOH?
1.
0.55 M
correct
2.
1.67 M
3.
0.45 M
4.
0.18 M
5.
0.82 M
Explanation:
V
1
= 335 mL
M
1
= 0.75 M
V
2
= 165 mL
M
2
= 0.15 M
Molarity is moles solute per liter oF solution.
Two solutions are being mixed together in this
problem. We fnd the moles oF NaOH in each
oF the individual solutions:
? mol NaOH = 335 mL soln
×
1 L soln
1000 mL soln
×
0
.
75 mol NaOH
1 L soln
=0
.
2512 mol NaOH
? mol NaOH = 165 mL soln
×
1 L soln
1000 mL soln
×
0
.
15 mol NaOH
1 L soln
.
0248 mol NaOH
The total moles oF NaOH in the new solu
tion will be the sum oF the moles in the two
individual solutions:
? mol NaOH = 0
.
2512 mol + 0
.
0248 mol
.
276 mol NaOH
The total volume oF the new solution will be
the combined volume oF the individual solu
tions:
? mL new soln = 335 mL + 165 mL
= 500 mL soln
To calculate the molarity oF NaOH in the new
solution, we divide the total moles oF NaOH
in the new solution by the total volume oF the
new solution:
? M NaOH =
0
.
276 mol NaOH
500 mL soln
×
1000 mL soln
1 L soln
.
552 M NaOH
002
10.0 points
What is the mass oF oxygen gas in a 16.8 L
container at 23.0
◦
C and 2.50 atm?
Correct answer: 55
.
305 g.
Explanation:
T
= 23
.
0
◦
C + 273 = 296 K
P
=2
.
5 atm
V
= 16
.
8 L
m = ?
n
=
PV
RT
=
(2
.
5 atm)(16
.
8 L)
(
0
.
0821
L
·
atm
mol
·
K
)
(296 K)
=1
.
72828 mol O
2
m = (1
.
72828 mol)
±
32 g
mol
²
= 55
.
305 g O
2
003
10.0 points
±or the vaporization oF water at some temper
ature, which oF the Following is ±ALSE?
1.
The process is endothermic.
2.
Δ
S
is positive.
3.
Δ
G
will increase with increasing temper
ature.
correct
4.
Δ
G
can be negative or positive depending
on the pressure.
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2
Explanation:
At standard phase change points, the pro
cess is in equilibrium. At equilibrium Δ
G
=
0, but if we change the pressure or temper
ature Δ
G
can be positive or negative. Va
porization requires energy to be input into
the system, so vaporization is endothermic.
Gases are more disordered than liquids, so Δ
S
for vaporization is positive. If we increase the
temperature, the liquid will spontaneously go
to gas phase. Δ
G
for spontaneous changes is
negative.
004
10.0 points
A process CANNOT be spontaneous if
1.
it is endothermic and there is a decrease
in disorder.
correct
2.
it is exothermic and there is an increase
in disorder.
3.
it is exothermic and there a decrease in
disorder.
4.
it is endothermic and there is an increase
in disorder.
Explanation:
Δ
G
= Δ
H

T
Δ
S
For a process to be spontaneous, Δ
G
must
be negative. Given Δ
G
H

T
Δ
S
, if Δ
H
is positive (endothermic) and Δ
S
is negative
(decrease in disorder), Δ
G
has to be positive
(nonspontaneous).
T
is in Kelvin so it is
always positive.
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 Summer '07
 Holcombe
 0.75 m, 80 Cal, 105 J, 136 K, 212 g

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