kadamati (skk558) – H01: Chapter 16.1011 – Mccord – (52450)
1
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printout
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have
21
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
How much NaNO
3
is needed to prepare 225
mL of a 1.55 M solution of NaNO
3
?
1.
12.3 g
2.
0.244 g
3.
0.132 g
4.
4.10 g
5.
29.6 g
correct
Explanation:
V
= 225 mL
M
= 1
.
55 M
? g NaNO
3
= 225 mL
×
1 L soln
1000 mL
×
1
.
55 mol NaNO
3
1 L soln
×
85 g NaNO
3
1 mol NaNO
3
= 29
.
6 g NaNO
3
002
10.0 points
What is the final concentration of NaOH when
335 mL of 0.75 M NaOH are mixed with
165 mL of 0.15 M NaOH?
1.
0.82 M
2.
0.18 M
3.
0.45 M
4.
0.55 M
correct
5.
1.67 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 find 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
= 0
.
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
= 0
.
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
= 0
.
552 M NaOH
003
10.0 points
What is the molarity of a solution containing
3.50 grams of NaCl in 500 mL of solution?
1.
1
.
20
×
10
−
1
M
correct
2.
5
.
98
×
10
−
2
M
3.
6
.
00
×
10
−
1
M
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kadamati (skk558) – H01: Chapter 16.1011 – Mccord – (52450)
2
4.
1
.
54
×
10
−
2
M
5.
7
.
69
×
10
−
2
M
Explanation:
m
NaCl
= 3
.
5 g
V
soln
= 500 mL
Molarity is mol/L. Thus we need to deter
mine the number of moles of NaCl that are
present:
? mol NaCl = 3
.
50 g
parenleftbigg
1 mol NaCl
58
.
44 g NaCl
parenrightbigg
= 0
.
0599 mol NaCl
Then, to determine the molarity:
? M =
mol solute
L soln
=
0
.
0599 mol NaCl
0
.
500 L soln
= 0
.
12 M NaCl
004
10.0 points
How many mL of a 0.75 N KOH solution
should be added to a 500 mL flask to make
500 mL of a 0.300 M KOH solution?
1.
1250 mL
2.
250 mL
3.
200 mL
correct
4.
50 mL
5.
500 mL
6.
100 mL
Explanation:
N
1
= 0.75 N
N
2
= 0.3 N
V
2
= 500 mL
This is a dilution problem.
0
.
75 N KOH = 0
.
75 M KOH
M
1
= 0.75 M
M
2
= 0.300 M
M
1
V
1
=
M
2
V
2
V
1
=
M
2
V
2
M
1
=
(0
.
3 M) (500 mL)
0
.
75 M
= 200 mL
005
10.0 points
What is the mass of oxygen gas in a 15.2 L
container at 46.0
◦
C and 4.25 atm?
Correct answer: 78
.
9312 g.
Explanation:
T
= 46
.
0
◦
C + 273 = 319 K
P
= 4
.
25 atm
V
= 15
.
2 L
m = ?
n
=
P V
R T
=
(4
.
25 atm)(15
.
2 L)
(
0
.
0821
L
·
atm
mol
·
K
)
(319 K)
= 2
.
4666 mol O
2
m = (2
.
4666 mol)
parenleftbigg
32 g
mol
parenrightbigg
= 78
.
9312 g O
2
006
10.0 points
A mixture of oxygen and helium is 92.3% by
mass oxygen.
It is collected at atmospheric
pressure (745 torr).
What is the partial pressure of oxygen in
this mixture?
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 Spring '07
 Holcombe
 Chemistry, Thermodynamics, mol

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