Extra Practice Problem Solutions
S O L U T I O N S T O T H E E X T R A P R A C T I C E P R O B L E M S F O R M O D U L E #11
1. The solubiUty o f gases increases with decreasing temperature. Thus, you should cool the liquid.
2. This is just a stoichiometry problem. We can tell this by the fact that we are being asked to
determine the amount o f one substance when we are given the amount o f another substance. The only
way to do that is by stoichiometry. Now, in order to do stoichiometry, we must first get our amount in
moles.
1.25 moles A1(N0,),
^
0 . 1 9 1 1 = 0.239 molesAl(N03)3
1t
Now that we have moles, we can do stoichiometry:
0.239 m o l o o A l ( N O , ) ,
1
1 moleAL(CO,),
-X, ^ '
= 0.120 moles A1,(C03)3
2 molesA1(N03)3
^'^
Now, o f course, this is not quite the answer we need. We were asked to figure out how many grams o f
aluminum carbonate were produced, so we have to convert from moles back to grams:
0.120 molesAl,(C03)3
1
X
^
234.0 grams A1,(C03)3
—
^
= 28.1 grams Al2(063)3
1 mole Al^ ( € 0 3 ) 3
3. We must first get our amount i n moles.
50.0 g K C N
1
1 mole K C N
^ 65.1 g K C N
= 0.768 moles K C N
N o w that we have moles, we can do stoichiometry:
0.768 moles K C N
1
X
1 mole H C N
;
= 0.768 moles H C N
1 mole K C N
Now, o f course, this is not quite the answer we need. We were asked to figure out how many m L o f
the H C N solufion is needed :
0.768 moloa H C N
1L
\ TT,
;—7^777 = 0.509 L = 509 m L
1
1.51 moles H C N
4. We must first get our amount i n moles.
50.0 gCtt 1 moleCu
T — X
— T T = 0-787 moles Cu
1
63.5 f - G «
Now that we have moles, we can do stoichiometry:
116
Solutions and Tests for Exploring Creation W i t h Chemistry
0.787 moles C u
1
8 moles H N O ,
X —
; — — ^ = 2.10 m o l e s H N O ,
3 moles C u
Now, of course, this is not quite the answer we need. W e were asked to figure out how many m L of
the HNO3 solution is needed :
2.10 moles I D J O ,
1 L
X—
.
3.5 m o l c a i m O j
1
= 0.600 L = 6.00 X 10' m L
5. T o calculate molality, we must have moles o f solute and kg o f solvent.
1 mole Mg(N03)2
50.0g M g ( N O ; ) ;
500.0 g
1 kg
^x
— = 0.5000 kg
1
1,000 g
^
N o w that we have moles o f solute and kg o f solvent, we can use Equation (11.1):
# moles solute
0.337 moles M g ( N 0 3 ) ,
molality = — ;
;
=
^
,—
^
= 0.674 m
# kg solvent
0.5000 kg water
6. First, we need to see how many moles o f C a C l 2 to add:
# moles solute
molality = - —
;
# kg solvent
moles C a C L
0.125 kg water
= 2.0 m
moles C a C l j = 0.25
Thus, we have to add 0.25 moles o f C a C h to 125 kg of water to make a 2.0 m solution. N o w we just
need to see how many grams that is:
0.25 m o l c a C a C L
1
111.1 g C a C L
r ^ ^,
= 28 g C a C L
1 molcCaCl^
—
7. Y o u want the solute that splits up into the most ions. Ca3{P04}2 splits into 5 ions (three calcium
ions and two phosphate ions. That is more than the other two, so it would give the lowest freezing
point to water.
Extra Practice Problem Solutions
8. I n a freezing-point depression problem, you must use Equation (11.2). However, in order to use
that equation, we must know Kf, i , and m. Right now, we only know Kf. However, we have been
given enough information to calculate both " i " and " m . " First let's calculate m:
10.0-g-K^ 1 mole KF
- T ^ X i e i : ^ = ''"2™lesKF
0.172 moles K F
m = —Tzz~,
0.1000 k g water
= 1-72 m
To figure out " i " , we just have to realize that according to its formula, K F splits up into one potassium
ion and one fluoride ion. Thus, i = 2. Now that we have all o f the components o f Equation (11.2), we
can use it:
°C
AT = - i • K f • m = - 2 • 1.86
m
• 1.72 m = - 6.40 °C
So the freezing point is 6.40 °C lower than that o f normal water, or -6.40 °C.
9. We can use Equation (11.2) to solve this. Since CaCb is made up o f one calcium ion and two
chloride ions, i = 3.
AT= - i - K f - m
m =
AT
.
=
'
-5.00''G
— - = 0.896 molal
-3-1.86—molal
10. To calculate boiling points, we must use Equation (11.3). To do that, however, we must know " i "
and " m " . To calculate " m " :
100.0 g ( N H J , S
lmole(NH4),S
0.7500 kg water
Since ammonium sulfide is an ionic compound, it dissolves by splitting up into its two ammonium ions
and its one sulfide ion. Thus, i = 3.
AT = i-K, -m = 3-0.521 — - 1 . 9 6 f f l = 3.06 "C
m
This means that the boiling point o f the solufion is 3.06 °C higher than that o f pure water. The boiling
point o f pure water is 100.0 °C, so the boiling point o f this solufion is 103.1 °C.
118
Solutions and Tests for Exploring Creation W i t h Chemistry
S O L U T I O N S T O T H E E X T R A P R A C T I C E P R O B L E M S F O R M O D U L E #12
L This problem asks you to predict how a gas w i l l change when you change some o f the conditions
under which it is stored. This means that you need to use the combined gas law (Equation 12.10).
P,V, _
P.V,
According to this problem, T i = 298.2 K, V i = 5.6 L, and T2 = 77.2 K. Also, the problem states that
the pressure does not change; thus. Pi and P2 cancel out:
P,V,
P,V,
T,
T,
We can now rearrange the equation to solve for the new volume:
T,
N o w we can put in the numbers and determine the new volume:
5.6 L-77.2 K
298.2 K
= 1.4 L
2. This is obviously another combined gas law problem, with Pi = 755 mmHg, V i = 6.65 L, T i =
298.2 K, P2 = 625 mmHg, and T2 = 288.2 K. The problem asks us to determine the new volume, so we
have to rearrange Equation (12.10) to solve for V2:
T,P.
Now we can plug i n the numbers:
7 5 5 ™ H g - 6 . 6 5 L • 288.2 K
V, = —
= 7.76 L
'
298.2 K-625-mfflHg-
3. This is obviously another combined gas law problem, with Pi = 740 torr, V i = 56.7 m L , T i = 298 K,
P2 = 1.00 atm (standard pressure), and T2 = 273 K (standard temperature). The problem asks us to
determine the new volume, so we have to rearrange Equation (12.10) to solve for V2:
PVT
' '^ = V
T,P,
We need to make the pressure units the same. We can do this by converting torr into atm or viceversa. I w i l l choose to do the latter: