Su, Yung – Homework 6 – Due: Oct 29 2007, 11:00 pm – Inst: FAKHREDDINE
1
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printout
should
have
22
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
Each question is worth 2 points.
001
(part 1 of 1) 2 points
A one molar solution is one having
1.
one mole of solute per one mole of sol
vent.
2.
one mole of solute per one liter of solute.
3.
one mole of solute per one kilogram of
solvent.
4.
one mole of solute per one kilogram of
solution.
5.
one mole of solute per liter of solution.
correct
Explanation:
Molarity is moles of solute per liter of solu
tion. A one molar solution would have 1 mole
of solute per liter of solution.
002
(part 1 of 1) 2 points
What volume of a solution that is 40.0%
NaNO
3
by
mass
contains
0.15
mole
of
NaNO
3
? Density of the solution = 1.32 g/mL.
1.
24.1 mL
correct
2.
9.60 mL
3.
38.2 mL
4.
3.86 mL
5.
42.0 mL
Explanation:
n
NaNO
3
= 0
.
15 mol
density
soln
= 1
.
32 g
/
mL
The NaNO
3
solution is 40.0% NaNO
3
by
mass. We can write this as the ratio
40
.
0 g NaNO
3
100 g soln
The density of the solution is 1.32 g/mL. This
can be more fully written as
1
.
32 g soln
1 mL soln
We want the volume of the solution that
contains 0.15 mole NaNO
3
. We convert from
mole to grams NaNO
3
:
? g NaNO
3
= 0
.
15 mol NaNO
3
×
85 g NaNO
3
1 mol NaNO
3
= 12
.
75 g NaNO
3
We use the percent by mass to convert from
grams solution to milliliters NaNO
3
solution.
Then we use the solution density to convert
from g solution to mL solution:
? mL soln = 12
.
75 g NaNO
3
×
100 g soln
40
.
0 g NaNO
3
×
1 mL soln
1
.
32 g soln
= 24
.
1 mL soln
003
(part 1 of 1) 2 points
What is the molarity of a solution prepared
by dissolving 10.5 grams of NaF in enough
water to produce 750 mL of solution?
1.
0.333 M
correct
2.
7.9 M
3.
0.014 M
4.
3.0 M
Explanation:
m
NaF
= 10.5 g
V
= 750 mL = 0.75 L
Molarity is moles solute per liter of solution.
We use the molar mass of NaF to convert from
grams to moles NaF:
? mol NaF = 10
.
5 g NaF
×
1 mol NaF
42 g NaF
= 0
.
25 mol NaF
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Su, Yung – Homework 6 – Due: Oct 29 2007, 11:00 pm – Inst: FAKHREDDINE
2
To calculate the molarity of the NaF solution,
we divide the moles of NaF in the solution by
the total volume of the solution in liters:
? M NaF =
0
.
25 mol NaF
0
.
750 L solution
= 0
.
333 M NaF
004
(part 1 of 1) 2 points
What mass of hydrated copper(II) sulfate
(CuSO
4
·
5H
2
O) is needed to prepare one liter
of a 0.6675 molar solution of CuSO
4
?
1.
0.675 g
2.
166.7 g
correct
3.
none of these
4.
107.8 g
5.
10.8 g
Explanation:
V
= 1 L
M
= 0.6675 M
The
formula
weight
of
CuSO
4
·
5H
2
O
is
249.62 g/mol. Each liter of CuSO
4
·
5H
2
O con
tains one mole of CuSO
4
·
5H
2
O, so we need
0.6675 moles of CuSO
4
·
5H
2
O, or
0
.
6675 mol
×
249
.
62 g
/
mol
= 166
.
7 g CuSO
4
·
5H
2
O
.
005
(part 1 of 1) 2 points
Consider an aqueous solution that contains
38 % (by mass) of a hypothetical solute Z.
The formula weight of the solute Z is 247
.
3.
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 Spring '09
 Fakhreddine
 Chemistry, Mole, mol

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