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3
School of Science and Engineering (SSE)
Spring 2017
General Chemistry II (CHE 1402)
Homework 3 - CORRECTION
I
1. Which of the following choices has the compounds correctly arranged in order of
increasing solubility in water? (Justify your answer)
A) CH3OH < CH4 < LiF
B) LiF < NaNO3 < CHCl3
C) CCl4 < CHCl3 < NaNO3
D) CH3OH < CCl4 < CHCl3
E) CH4 < NaNO3 < CHCl3
CCl4 is non polar and consequently has a very low solubility in water
CHCl3 is polar and soluble in water via London dispersion and dipole-dipole interaction
forces.
NaNO3 is an ionic compound and more soluble in water (ion-dipole interactions)
Answer is C.
Note: “like dissolves like".
2. At 25 °C the vapor pressure of pure benzene (C6H6) is 95.1 torr and that of pure
toluene (C7H8) is 28.4 torr.
A solution, is made by mixing equal number of moles of benzene and toluene, has a
vapor pressure of 49.4 torr at 25 °C.
a. If the solution exhibits ideal behavior, what would be its vapor pressure?
b. Explain in terms of intermolecular interactions whether the mixing of benzene and
toluene is endothermic or exothermic.
a. If the solution exhibits an ideal behavior, then Raoult’s law can be written as:
PTOTAL = PC6H6 + PC7 H8 = X C6H6  PCo6H6 + X C7 H8  PCo7 H8
Since the solution is made by mixing equal number of moles of benzene and toluene then
n C6 H 6
n C6 H 6
1
X C7 H8  X C6 H 6 

  0.5
n C6 H 6  n C7 H8
2  n C6 H 6 2


PTOTAL = X C6H6  PCo6H6 + X C7 H8  PCo7 H8  X C6H6 PCo6H6  PCo7 H8  0.5  95.1  28.4  61.7 torr
c. The real vapor pressure is 49.5 torr which is lower than the 61.7 torr (calculated ideal
vapor pressure); this implies that this case corresponds to a negative deviation from
the ideal solution (Raoult’s law) and consequently Hsolute-solvent > (Hsolute-solute +
Hsolvent-solvent). This means that the solute-solvent interactions are much stronger than
solute-solute interactions and the solvent-solvent interactions. Consequently
Hsolution < 0 and the process is exothermic.
II
Sodium metal Na(s) dissolves in liquid mercury (Hg(l)) to form a solution called sodium
amalgam. The density of Na(s) is 0.97g/cm3.
A solution is made by dissolving 4.85 g of Na(s) in 20 cm3 of Hg(l); if the density of the
solution is 11.07 g/cm3 determine:
a. the molality of Na(s) in the solution
b. the molarity of Na(s) in the solution
a. The molality, m, is defined as:
n solute in mol
n in mol
massNa
m

so m Na  Na
massHg in kg M Na  massHg in kg
masssolvent in kg
massHg  masssolution  massNa  d solution  Vsolution  massNa
Assuming that the volume is additive,
 massNa 
4.85 
3
  20  
Vsolution  VHg  VNa  VHg  
  25 cm  25 mL
 0.97 
 d Na 
massHg  d solution  Vsolution  massNa  11.07  25  4.85  271.9 g  0.2719 kg
m Na 
massNa
4.85

 0.776 mol / kg  0.776 m
M Na  massHg
23.0  0.2719
n solute in mol
Vsolution in L
4.85

 8.43 mol / L  8.43 M
23.0  25  10 3
c. The molarity is defined as: molarity 
so molarity Na 
massNa
M Na  Vsolution
 masssolvent in kg 

Note: molarity  molality  
 Vsolution in L 
 massHg 
0.2719 
  0.776  
 molarity Na  mNa  
 8.43M
3 
 25 10 
 Vsolution 
III
1) Which of the following, a 0.1 m aqueous solution will have the lowest decrease in
freezing point (Tf) (Justify your answer)? The Kf for water is 1.86 °C
A) K2CrO4
B) Al(NO3)3
C) Fe3(SO4)2
D) HCl
E) C6H12O6
2) What is the principal reason for the extremely low solubility of NaCl in benzene
(C6H6)?
1) For an electrolyte
T f = i x K f x m
Tf is the freezing point depression
i is the van’t Hoff factor, a measure of the extent to which electrolytes dissociate in
solution
Kf, is the molal freezing-point-depression constant (oC/m)
m is the molality of the solution
Since the solutions we have to compare are all aqueous (ie solvent = water) and that their
molality is identical, the comparison of their respective i is sufficient.
The ideal (or limiting) values of i for each electrolyte are tabulated below:
Electrolyte
Fe3(SO4)2
Al(NO3)3
K2CrO4
HCl
C6H12O6
van’t Hoff factor i
(limiting value)
5
4
3
2
1
The lowest freezing-point-depression should be observed by the solution of electrolyte
having the lowest van’t Hoff factor. C6H12O6 (sucrose) has the lowest van’t Hoff factor, it
therefore has the lowest freezing-point-depression.
Note: Tf (C6H12O6) < Tf (HCl) < Tf (K2CrO4 ) < Tf (Al(NO3)3 < Tf (Fe3(SO4)2)
2) The general rule is “like dissolves like”. Thus, compound having similar
intermolecular forces will interact with each other and thus be soluble in one another.
NaCl possesses the following intermolecular forces: ionic bonding and dispersion forces.
Benzene possesses the following intermolecular forces: dispersion forces.
Therefore, NaCl is polar and benzene is non-polar (non polar compounds are those that
possess exclusively dispersion forces), and that is why NaCl has a poor solubility in
benzene.
Note: NaCl (polar compound) is very soluble in water (polar compound). Benzene (nonpolar compound) is very soluble in hexane (non-polar compound).
IV
When a given amount of ethylene glycol (Molar Mass = 62 g/mol) is added at 35 °C to
1.00 kg of ethanol (Molar Mass = 46 g/mol), the vapor pressure of the solution decreases
to 86 torr. If the vapor pressure of ethanol at 35 °C is 100 torr, what is the mass of
ethylene glycol that must be added to 1.00 kg of ethanol to reduce its pressure to 86 torr?
Calculate the mole fraction, X(ETHANOL), of the solvent (ethanol):
According to Raoult’s law:
P
o
Pethanol = X ethanol Pethanol
 X ethanol ethanol
o
Pethanol

1
P


X EG  1  X ethanol  1   ethanol
 Po 
 ethanol 
n EG
n EG

and
nTOT n EG  nethanol 
X EG 
nEG
P

n EG
 
 1   ethanol
 Po 
 nethanol 
 ethanol 
nethanol

n EG
1
P


1   ethanol
 Po 
 ethanol 
 n EG
nEG  nethanol 
n EG

1
P


1   ethanol
o
P

 ethanol 
 methanol 


M ethanol 
nethanol















1
1

 1 
 1
  Pethanol  
  Pethanol  
 1   o  
 1   o  




P
  ethanol  
  Pethanol  
Finally,
n EG 
m EG
M EG
M EG
62
1  10 3 
M ethanol
46  219 g

 m EG 








1



 1
1

 1
  86  

  Pethanol  
1 


1

  100  
  o 


  Pethanol  
methanol 
 m EG  n EG  M EG
Conclusion: 219 g of ethylene glycol must be added.
V
1. Which of the following does not form a solution? Justify your answer?
(i) NaCl(s) in liquid NH3(l) ; (ii) C12H26(s) in C3H8(l) ;
(iv) H2O(l) in liquid H2S(l)
(iii) CH3Cl(l) in CCl4(l);
(iii) CH3Cl(l) in CCl4(l) does not form a solution because CH3Cl(l) is polar while CCl4(l)
is not polar.
NaCl(s) and NH3(l) can form a solution thanks to ion-dipole interactions.
C12H26(s) and C3H8(l) can form a solution because they are both non-polar; they only
have dispersion forces.
H2O(l) in liquid H2S(l) can form a solution thanks to dipole-dipole interactions.
2. A sulfuric acid solution, H2SO4, in water has a concentration of 2.45 M. The density of
water is 1.15 g/mL at 20°C. What is the molality at this temperature?
(dH2SO4=1.84, MH2SO4=98.1)
m
V

n solute in mol
C
V
C
 solution solution  solute   solution 
masssolvent in kg d solvent  Vsolvent
d solvent  Vsolvent 
What volume of concentrated H2SO4 is required to prepare a 1 L solution that is 2.45 M.
n H 2 SO4  C H 2 SO4  Vsolution 
 VH 2 SO4 
mH 2 SO4
M H 2 SO4

d
H 2 SO4
 VH 2 SO4
M H 2 SO4

 VH 2 SO4 
C H 2 SO4  Vsolution  M H 2 SO4
d H 2 SO4
2.54  1  98.1
 135 mL  0.135 L  Vwater  1  0.135  0.865 L
1.84
A 1 L of 2.45 M solution of H2SO4 contains 0.865 L of water, therefore:
C
2.45 mol.L1
1
m  solute 

 2.46 m
d solvent 0.865 1.15 kg.L1  0.865
Conclusion: At this temperature, the molality is 2.46 m.
VI
Lysozyme is an enzyme that breaks bacterial cell walls. A solution containing 0.150 g of
this enzyme in 210 mL of solution has an osmotic pressure of 0.953 torr at 25°C.
What is the molar mass of this substance?
The osmotic pressure of a solvent obeys the following law:
m
  V  nRT  
M
Therefore, M 

 RT

mRT 0.150  0.082  (25  273)

 13919 g / mol
0.953
 V
3
 210 10
760
Conclusion: The molar mass of this substance is 13919 g/mol.