1

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