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Homework #4
Chapter 17 Properties of Solutions
13. a) HNO3(s) H+(aq) + NO3
-(aq) b) Na2SO4(s) 2Na+(aq) + SO4
2-(aq) c) Al(NO3)3(s) Al3+(aq) + 3NO3
-(aq) d) SrBr2(s) Sr2+(aq) + 2Br-(aq) e) KClO4(s) K+(aq) + ClO4
-(aq) f) NH4Br(s) NH4
+(aq) + Br-(aq) g) NH4NO3(s) NH4
+(aq) + NO3-(aq)
h) CuSO4(s) Cu2+(aq) + SO42-(aq)
i) NaOH(s) Na+(aq) + OH-(aq) 15. Need to find molarity (𝑀), molality (𝑚), and mole fraction (χ) of C2H6O2
Know Mass % C2H6O2 = 40.0%
305.1cm
g
solutiond
mol
g
OH
mol
g
OHC
M
M
02.18
08.62
2
262
All of the quantities that need to be calculated are intrinsic properties (do not depend on sample size), therefore, assume sample size of 100. g. Calculate molarity
solution
OHC
V
nM 262
Calculate the moles of C2H6O2
262
262
262
OHC
OHC
OHCM
mn
Calculate the mass of C2H6O2
2 6 2 2 6 2
2 6 2 2
40.0% 100.% 100.%100.
C H O C H O
C H O H O
m m
m m g
Note: It was assumed that the total mass was 100 g
2 6 240.0C H Om g
2 6 2
2 6 2
2 6 2
40.00.644
62.08
C H O
C H O g
C H O mol
m gn mol
M
Calculate the volume of solution
3
3100.95.2 0.0952
1.05
solutionsolution g
solution cm
m gV cm L
d
2
2 6 20.644
6.760.0952
C H O
solution
n molM M
V L
Calculate molality
OH
OHC
m
nm
2
262
Calculate the mass of solvent (H2O)
2 2 6 2100.0 40.0 60.0 0.0600H O solution C H Om m m g g g kg
2 6 2
2
0.64410.7
0.0600
C H O
H O
n molm m
m kg
Calculate the mole fraction
OHOHC
OHC
OHCnn
n
2262
262
262
Calculate moles of H2O
2
2
2
60.03.33
18.02
H O
H O g
H O mol
m gn mol
M
2 6 2
2 6 2
2 6 2 2
0.6440.162
0.644 3.33
C H O
C H O
C H O H O
n mol
n n mol mol
16. Need to find molarity (𝑀), molality (𝑚), and mole fraction (χ) of C2H6O2 In General
Know Mass % solvent
soluted
solvent
solute
M
M
of the quantities that need to be calculated are intrinsic properties (do not depend on sample size), therefore, assume sample size of 100. g. Molarity
solution
solute
V
nM
Molality
solvent
solute
m
nm
Mole Fraction
solventsolute
solute
solutenn
n
3
HCl Calculate moles of solute (HCl) using mass %
1 36.46
38% 100.% 100.%100.
38
38 1.0
solute solute
solute solvent
solute
mol HClsolute g HCl
m m
m m g
m g
n g mol
Calculate the Volume of solution using density
3
3100.84.0 0.0840
1.19
solutionsolution g
solution cm
m gV cm L
d
Calculate the moles of solvent (H2O)
2
2
1
18.02
100. 38 62
62 3.4
solvent solution solute
mol H O
solvent g H O
m m m g g g
n g mol
Molarity
1.012
0.0840
solute
solution
n molM M
V L
Molality 1.0
160.062
solute
solvent
n molm m
m kg
Mole Fraction
1.0
0.231.0 3.4
solutesolute
solute solvent
n mol
n n mol mol
HNO3 Calculate moles of solute (HNO3) using mass %
3
3
1
63.02
70.% 100.% 100.%100.
70.
70. 1.1
solute solute
solute solvent
solute
mol HNO
solute g HNO
m m
m m g
m g
n g mol
Calculate the Volume of solution using density
3
3100.70.4 0.0704
1.42
solutionsolution g
solution cm
m gV cm L
d
Calculate the moles of solvent (H2O)
2
2
1
18.02
100. 70. 30.
30. 1.7
solvent solution solute
mol H O
solvent g H O
m m m g g g
n g mol
Molarity
1.116
0.0704
solute
solution
n molM M
V L
4
Molality 1.1
370.030
solute
solvent
n molm m
m kg
Mole Fraction
1.1
0.391.1 1.7
solutesolute
solute solvent
n mol
n n mol mol
H2SO4 Calculate moles of solute (H2SO4) using mass %
2 4
2 4
1
98.09
95% 100.% 100.%100.
95.
95. 0.97
solute solute
solute solvent
solute
mol H SO
solute g H SO
m m
m m g
m g
n g mol
Calculate the Volume of solution using density
3
3100.54.3 0.0543
1.84
solutionsolution g
solution cm
m gV cm L
d
Calculate the moles of solvent (H2O)
2
2
1
18.02
100. 95 5
5 0.3
solvent solution solute
mol H O
solvent g H O
m m m g g g
n g mol
Molarity
0.9718
0.0543
solute
solution
n molM M
V L
Molality 0.97
2000.005
solute
solvent
n molm m
m kg
Mole Fraction
0.97
0.760.97 0.3
solutesolute
solute solvent
n mol
n n mol mol
HC2H3O2 Calculate moles of solute (HC2H3O2) using mass %
2 3 2
2 3 2
1
61.07
99% 100.% 100.%100.
99
99 1.6
solute solute
solute solvent
solute
mol HC H O
solute g HC H O
m m
m m g
m g
n g mol
Calculate the Volume of solution using density
3
3100.95.2 0.0952
1.05
solutionsolution g
solution cm
m gV cm L
d
5
Calculate the moles of solvent (H2O)
2
2
1
18.02
100. 99 1
1 0.06
solvent solution solute
mol H O
solvent g H O
m m m g g g
n g mol
Molarity
1.617
0.0952
solute
solution
n molM M
V L
Molality 1.6
20000.001
solute
solvent
n molm m
m kg
Mole Fraction
1.6
0.961.6 0.06
solutesolute
solute solvent
n mol
n n mol mol
NH3 Calculate moles of solute (NH3) using mass %
3
3
1
17.04 N
28% 100.% 100.%100.
28
28 1.6
solute solute
solute solvent
solute
mol NH
solute g H
m m
m m g
m g
n g mol
Calculate the Volume of solution using density
3
3100.110 0.110
0.90
solutionsolution g
solution cm
m gV cm L
d
Calculate the moles of solvent (H2O)
2
2
1
18.02
100. 28 72
72 4.0
solvent solution solute
mol H O
solvent g H O
m m m g g g
n g mol
Molarity
1.615
0.110
solute
solution
n molM M
V L
Molality 1.6
220.072
solute
solvent
n molm m
m kg
Mole Fraction
1.6
0.291.6 4.0
solutesolute
solute solvent
n mol
n n mol mol
6
17. Need to find molarity (𝑀), molality (𝑚), mole fraction (χ), and mass % of C6H5CH3
Know
6 5 3
36 5 3
6 6
36 6
50
0.867
125
0.874
C H CH
g
C H CH cm
C H
g
C H cm
V mL
d
V mL
d
Calculate mass%
6 5 3
6 5 3 % 100%C H CH
total
mmass C H CH
m
Calculate the mass of toluene
36 5 3 6 5 3
350 0.867 43.4g
C H CH C H CH cmm Vd cm g
Calculate the mass of benzene
36 6 6 6
3125 0.874 109.g
C H C H cmm Vd cm g
Calculate the total mass
6 5 3 6 643.4 109. 152total C H CH C Hm m m g g g
6 5 3
6 5 3
43.4 % 100% 100% 28.6%
152
C H CH
total
m gmass C H CH
m g
Calculate mole fraction
6 5 3
6 5 3
6 5 3 6 6
C H CH
C H CH
C H CH C H
n
n n
Calculate the moles of toluene
6 5 3
6 5 3 6 5 3
1 43.4 0.471
92.15
mol C H CHg C H CH mol C H CH
g
Calculate the moles of benzene
6 6
6 6 6 6
1 109 1.40
78.12
mol C Hg C H mol C H
g
6 5 3
6 5 3
6 5 3 6 6
0.4710.252
0.471 1.40
C H CH
C H CH
C H CH C H
n mol
n n mol mol
Calculate molality 0.471
4.320.109
solute
solvent
n molm m
m kg
Calculate molarity
0.4712.69
0.175
solute
solution
n molM M
V L
7
30. The extent of hydration increases with increasing charge density. In general, the smaller the size, the larger the charge density. Also, the larger the charge, the higher the charge density. a) Mg2+ b) Be2+ c) Fe3+ d) F- e) Cl- f) SO4
2-
31. a) NH3 NH3 can H-bond with water while PH3 cannot b) CH3CN CH3CN is polar and CH3CH3 is not c) CH3COOH CH3COOH can H-bond with water and CH3COOCH3 cannot 32. a) KrF2 (linear) nonpolar dissolves in CCl4 b) SF2 (bent) polar dissolves in H2O c) SO2 (bent) polar dissolves in H2O d) CO2 (linear) nonpolar dissolves in CCl4 e) MgF2 ionic dissolves in H2O f) CH2O polar dissolves in H2O g) CH2CH2 nonpolar dissolves in CCl4
37.
HP k C
40.790 8.21 10
962
H
L atmH mol
atm k M
k
MC
Catm
CkP
molatmL
H
00114.0
96210.1
40. solventsolventsolution PP
Calculate the mole fraction of H2O
3832
2
2
OHCOH
OH
OHnn
n
Calculate the moles of H2O
32 2
2
3
1
2 218.02
338 0.992 335
335 18.6
g
H O H O cm
mol H O
g
m Vd cm g
g H O mol H O
Calculate the moles of C3H8O3
3 8 31
3 8 3 3 8 392.11164 1.78
mol C H O
gg C H O mol C H O
2
18.60.913
18.6 1.78H O
mol
mol mol
8
Calculate the vapor pressure of H2O/C3H8O3
torrtorrPP solventsolventsolution 0.5074.54913.0
42. solventsolventsolution PP
Smallest Vapor Pressure Largest Vapor Pressure c<b<a<d
The vapor pressure of methanol is larger than that of water, therefore, solution d will have the largest vapor pressure. Pure water will have the next largest vapor pressure.
Glucose is a non-electrolyte, therefore, it will not dissociate in water 99.02OH .
When NaCl dissociates in water it will form 2 ions Na+ and Cl-, therefore, 99.02OH
causing solution c to have the smallest vapor pressure.
45. solventsolventsolution PP
968.0
930.0900.0
solvent
solvent atmatm
?66
66
66 nn
n
HC
HC
HC
Determine moles of C6H6
6 61
6 6 78.1278.11 1.00
mol C H
gg C H mol
03.0
00.1
00.1968.0
?
?
?66
66
66
n
nmol
mol
nn
n
HC
HC
HC
Determine molar mass of substance
mol
g
mol
g
n
mM 300
03.0
0.10?
47. a)
5 12 5 12 6 14 6 14solution C H C H C H C HP P P
5 12 6 14
5 12 6 14
5 12 6 14 6 14 5 12
C H C H
solution C H C H
C H C H C H C H
n nP P P
n n n n
Calculate moles of C5H12
35 12 2
5 12
3
1
5 12 5 1272.12
25 0.63 16
16 0.22
g
C H H O cm
mol C H
g
m Vd cm g
g C H mol C H
Calculate moles of C6H14
9
36 14 6 14
6 14
3
1
6 14 6 1486.20
45 0.66 30.
30. 0.35
g
C H C H cm
mol C H
g
m Vd cm g
g C H mol C H
5 12 6 14
5 12 6 14
5 12 6 14 6 14 5 12
0.22 0.35511 150. 290
0.35 0.22 0.35 0.22
C H C H
solution C H C H
C H C H C H C H
solution
n nP P P
n n n n
mol molP torr torr torr
mol mol mol mol
b) We need to calculate the mole fraction of pentane in the vapor phase. You cannot use the mole fraction that was calculated in part (a) because that is the mole fraction in the liquid phase. In the gas phase
RT
PVn
nRTPV
69.0290
100.2 2
125
125
146125
125
125
torr
torr
P
P
RT
VPRT
VP
nn
n
total
HC
total
HC
HCHC
HC
HC
49. We need to find χprop and χmeth
What we know
solution meth meth prop propP P P
174 303 44.6 (equ. a)meth proptorr torr torr
1 (equ. b)prop meth
Solve for χmeth in equation b
propmeth
methprop
1
1
Plug into equation (a) and solve for χprop
499.0
6.443031174
prop
propprop torrtorrtorr
Solve for χmeth
501.0
499.01
1
meth
meth
methprop
50. Plot a shows a positive deviation from Raoult’s law. This happens when there are
unfavorable relationships between the solvent and the solute. This is the situation for situation d in which the solvent/solute have differences in polarity CH3CH2CH2CH2CH2CH3 is nonpolar and H2O is polar. Therefore, each of the substances
10
would be more stable by themselves and not in solution. It is also the case for b. Both of these species are capable of H-bonding individually, therefore, adding them together will not make a more stable situation.
Plot b shows a negative deviation from Raoult’s law. This happens when there are favorable relationships between the solvent and the solute. This is the situation for situation a, because CH3COCH3 would not be able to H-bond on its own. However, when water is present it can participate in H-bonding making the intermolecular forces in the solution stronger than the individual intermolecular forces.
Plot c shows the ideal Raoult’s law. This happens when the intermolecular forces are
the same between the individual parts of the solutions as with the solution. This is the case for c. Both CH3CH2CH2CH2CH2CH2CH3 and CH3CH2CH2CH2CH2CH3 have nearly identical intermolecular forces resulting in an ideal situation when they are mixed.
51. Need to find solutionP , V
COCHCH 33 , and V
OHCH3 (V stands for the mole fraction in the vapor
phase)
OHCH
COCHCHOHCH
OHCH
COCHCH
OHCHCOCHCH
COCHCH
solution
ClCClCCOCHCHCOCHCHsolution
Pnn
nP
nn
nP
PPP
3
333
3
33
333
33
1461463333
Calculate moles of acetone
3 31
3 3 3 358.0950.0 0.861
mol CH COCH
gg CH COCH mol CH COCH
Calculate moles of methanol
31
3 332.0550.0 1.56
mol CH OH
gg CH OH mol CH OH
0.861 1.56
271 143 188.50.861 1.56 1.56 0.861
solution
mol molP torr torr torr
mol mol mol mol
Calculate the mole fraction of acetone in the vapor phase
512.06.188
5.9633
33
333
33
33
333
33
33
torr
torr
P
P
nn
n
RT
PVn
nRTPV
nn
n
total
COCHCH
RT
VP
RT
VP
OHCHCOCHCH
COCHCHV
COCHCH
OHCHCOCHCH
COCHCHV
COCHCH
total
COCHCH
Calculate the mole fraction of methanol in the vapor phase
488.06.188
1.923
333
3
3
torr
torr
P
P
nn
n
total
OHCH
COCHCHOHCH
OHCHV
OHCH
11
Since the actual vapor pressure is less than the total vapor pressure (negative deviation to Raoult’s Law) there must be favorable solute-solvent interaction. Solute-solvent interactions are stronger than solute-solute and solvent-solvent interactions.
52.
a) In order to be an ideal solution when you plot vapor pressure vs. mole fraction a
straight line should be observed. Therefore, the solution is not ideal. b) When there is a positive deviation from Raoult’s law there are unfavorable
interactions between solute and solvent, therefore, ΔH is positive. c) The positive deviation from Raoult’s law shows that the vapor pressure is higher
than the ideal case; this implies weaker interaction between propanol and the water than the pure substance.
d) The higher the vapor pressure, the lower the boiling point. Therefore, the
lowest boiling point is when 54.02OH
53. If the container feels warm it means that the solution released heat when it was formed. At
constant pressure Hq , therefore, ΔH is negative. This implies that there will be a negative
deviation to Raoult’s law and the solution is not ideal. Note: This problem is incorrect because
methanol and water would have a positive deviation which would make the solution cool to the
touch.
55. a) True b) True c) True d) True e) False The smaller the vapor pressure, the higher the boiling point.
70
72
74
76
78
80
82
84
0 0.2 0.4 0.6 0.8 1
Vap
or
Pre
ssu
re (
torr
)
Mole Fraction H2O
12
59. They want you to find Tb The boiling point of pure water is 100°C
bb TCT 0.100
Calculate the change in boiling point
bb imKT
Calculate the molality
m
nm
Calculate the moles of urea
2 2
1
2 260.072 227.0 0.449
mol NH CO
gg NH CO mol NH CO
mkg
mol
m
nm 00.3
1500.
449.0
bb imKT
I= 1 because non electrolyte
mol
kgC
bK
51.0
CmimKTmol
kgC
bb
5.151.000.31
CCCTCT bb 5.1015.10.1000.100
60. Need to determine 383 OHCm
Find fT
CT
TCC
TCT
f
f
ff
50.1
00.050.1
00.0
Find m
ff imKT
C3H8O3 is a non-electrolyte i=1
mm
mCmol
kgC
806.0
86.1150.1
Find 383 OHCm
3 8 3
2
3 8 3
3 8 3
3 8 3
3 8 3
92.11
3 8 3 3 8 31
0.8060.200
0.161
0.161 14.8
C H O
H O
C H O
C H O
g
mol C H O
nm
m
nm
kg
n mol C H O
mol C H O g C H O
13
61. Need to find Tf and Tb ff TCT 00.0
Calculate ΔTf
ff imKT
Ethylene glycol is a non-electrolyte, i=1 Calculate m
OH
OHC
m
nm
2
262
Assume 1.00 kg of solution 0.500 kg C2H6O2 0.500 kg H2O
2 6 21
2 6 2 2 6 262.08500. 8.05
mol C H O
gg C H O mol C H O
mkg
mol
m
nm
OH
OHC1.16
500.0
04.8
2
262
CmimKTmol
kgC
ff
0.3086.11.161 CCCTCT ff 0.300.3000.000.0
bb TCT 0.100
Calculate ΔTb
bb imKT
CmimKTmol
kgC
bb
21.851.01.161
CCCTCT bb 2.10821.80.1000.100
63. Need to find Tb
bb TCT 725.99
Calculate ΔTb
bb imKT Non-electrolyte, i=1 Calculate m
OH
solute
m
nm
2
158.0
24.0 0.414molg
g mol
2
0.4140.690
0.600
solute
H O
n molm m
m kg
1 0.690 0.51 0.35C kg
b b molT imK m C
CCCTCT bb 08.10035.0725.99725.99
14
68. Need to calculate 262 OHCV
CT
TCC
TCT
f
f
ff
0.30
00.00.30
00.0
Calculate 262 OHCm
ff imKT
Ethylene glycol is a non-electrolyte, i=1
mm
mCmol
kgC
1.16
86.110.30
Calculate 262 OHCn
OH
OHC
m
nm
2
262
Calculate OHm2
3
3
1.001000 11 1 1
15.0 15000 15.0gmL cm
L mL cmL g kg
2 6 2
2
2 6 2
2 6 2
16.115.0
242
C H O
H O
C H O
C H O
nm
m
nm
kg
n mol
Calculate mass of C2H6O2
2 6 2
62.08 4
2 6 2 2 6 21 242 1.50 10
g
mol C H Omol C H O g C H O
Calculate 262 OHCV
3
3
4 1 11.11 1
1.50 10 13500 13.5cm mLg cm
g mL L
Calculate the boiling point of the solution 100.00b bT C T
Calculate ΔTb
1 16.1 0.51 8.2C kg
b b molT imK m C
100.00 8.2 108.2bT C C C
69. Need to calculate ΔTf and π
ff imKT
Calculate m
OH
protein
m
nm
2
15
30.1cm
g
solutiond
Therefore, if you have 1.0 L of solution it will weight 1000 g In 1 L of solution there is 1.0 g of protein Therefore, in a 1.0 L solution there is 1.0 g of protein and 999 g of H2O
Calculate nprotein Assume 1 L solution
molg
molg 5
4101.1
100.9
10.1
mkg
mol
m
nm
OH
protein 55
101.1999.
101.1
2
CmimKTmol
kgC
ff 55 101.286.1101.11
iMRT Calculate M
ML
mol
V
nM
Solution
protein 55
101.10.1
101.1
torratmKMiMRTKmol
atmL 20.0107.229808206.0101.11 45
72. Need to calculate mass% C10H8, and C14H10
%10060.1
%100% 810810
810
g
m
m
mHCmass
HC
total
HC
%10060.1
%100% 10141014
1014
g
m
m
mHCmass
HC
total
HC
Find mass of C10H8 Know
CT
mmg
f
HCHC
81.2
60.11014810
Calculate ΔTf
CT
TCC
TCT
f
f
ff
70.2
51.581.2
51.5
Solve for mass of C10H8
16
10 8 14 10
10 8 14 10
10 8 14 1010 8 14 10
2
2
2.70 1 5.120.0200
178.24 128.180.0105
128.18 178.24 22846
f f
C H C H
C H C H C kg
mol
g g
C H C Hmol molC H C H
g g gmol mol
mol
T imK
m m
M MC
kg
m mm mmol
2
10 8 14 10
14 10 10 8
2
10 8 10 8
2
10 8
10 8
240. 178.24 128.18
1.60
240. 178.24 128.18 1.60
35 50.06
0.70
g g g
C H C Hmol mol mol
C H C H
g g g
C H C Hmol mol mol
g g
C Hmol mol
C H
m m
m g m
m g m
m
m g
810810 %44%10060.1
70.0%100% 810 HC
gm
mHCmass
total
HC
Find mass of C10H8 Know
gm
mgg
mmg
HC
HC
HCHC
90.0
70.060.1
60.1
1014
1014
1014810
10141014 %56%10060.1
90.0%100% 1014 HC
g
g
m
mHCmass
total
HC
75. Need to calculate χNaCl and vapor pressure of solution
Know
torrPsolution 6.19 at 25°C
torrP OH 8.232 at 25°C
torrP OH 9.712 at 45°C
17
2
2
2
2
2
19.6 23.8
0.824
1
1 0.824
0.176
solution solvent solvent
H O
H O
H O
H O Na Cl
H O Na Cl
Na Cl
Na Cl
P P
torr torr
n
n n n
NaCl Na+ + Cl- Therefore, the mole fractions of Na+ and Cl- must be equal and it must equal the mole fraction of NaCl
0.1760.0880
2NaClNa Cl
0.824 71.9 59.2solution solvent solventP P torr torr
77. 0.010 m Na3PO4 i= 4 0.040 m (for all ions/molecules) 0.020 m CaBr2 i= 3 0.060 m (for all ions/molecules) 0.020 m KCl i=2 0.040 m (for all ions/molecules) 0.020 m HF i~1 0.020 m (for all ions/molecules) Since HF is a weak acid
most of the HF will be HF and not H+ and F- a) C6H12O6 is a non-electrolyte, i=1 0.040 m (for all ions/molecules)
Therefore the Na3PO4 and the KCl solution will have the same boiling points the C6H12O6 solution.
b) The solution with the highest vapor pressure will be the solution that has the smallest molality for all ions/molecules. Therefore, the HF solution will have the highest vapor pressure.
c) The solution with the largest freezing point depression will be the solution that has the largest molality for all ions/molecules. Therefore, the CaBr2 solution will have the largest freezing point depression.
78. Pure Water χ=0.00 (all ions/molecules)
C6H12O6 (χ=0.01) i=1 χ=0.01 (all ions/molecules) NaCl (χ=0.01) i=1 χ=0.02 (all ions/molecules) CaCl2(χ=0.01) i=3 χ=0.03 (all ions/molecules) a) lowest χ pure water b) highest χ CaCl2 solution c) highest χ CaCl2 solution d) lowest χ pure water c) highest χ CaCl2 solution
18
111. Need to calculate mass% MgCl2
%100500.0
%100% 22
2
g
m
m
mMgClmass
MgCl
total
MgCl
%100500.0
%100%
g
m
m
mNaClmass NaCl
total
NaCl
Find mass of MgCl2 Know
atm
mmg NaClMgCl
3950.0
500.02
RTMiMi
iMRT
NaClNaClMgClMgCl
22
The upper case 𝑀 is the molarity
2
20.3950 3 2 0.08206 298
MgClNaCl
MgCl NaCl L atmmol K
solution solution
m m
M Matm K
V V
The upper case M is the molar mass
2
2
2 2
2
2
2
0.3950 3 2 0.08206 2981.0000 1.0000
175.33 190.440.01615 3 2
95.218 58.443 5564.8
89.87 175.33
MgClNaCl
MgCl NaCl L atmmol K
g g
MgCl MgCl NaClmol molNaCl
g g gmol mol
mol
g g
MgClmol mol
m m
M Matm K
L L
m m mmmol
m
2
190.44g
NaClmolm
Plug 2500.0 MgClNaCl mgm
for mass of MgCl2
gm
m
mgm
MgCl
MgClmol
g
mol
g
MgClmol
g
MgClmol
g
mol
g
354.0
11.1535.5
500.044.19033.17587.89
2
2
2
22
2
%8.70%100500.0
354.0%100% 2
2
gm
mMgClmass
total
MgCl