Chapter 13 Properties of Solutions

Exercise 1:

A certain soft drink is bottled so that a bottle at 25oC contains

CO2 gas at a pressure of 5.0 atm over the liquid. Assuming
that the partial pressure of CO2 in the atmosphere is 4.0 x 10-4
atm, calculate the equilibrium concentration of CO 2 in the
soda both before and after the bottle is opened. The Henry’s
law constant for CO2 in aqueous solution is 32 L.atm/mol at
25oC.

Answer:

PCO2 = kCO2CCO2 (Henry’s Law)

In the unopened bottle PCO2 = 5.0 atm and

CCO2 = PCO2/ kCO2 = (5.0 atm)/(32 L.atm/mol) = 0.16 mol/L

In the opened bottle, the CO2 in the soda eventually reaches

equilibrium with the atmospheric CO2, so PCO2 = 4.0 x 10-4
atm and

CCO2 = PCO2/ kCO2 = (4.0 x 10-4 atm)/(32 L.atm/mol)

= 1.2 x 10-5 mol/L

(Note the large change in concentration of CO2)

 2

Exercise 2:

Calculate the expected vapor pressure at 25oC for a solution

prepared by dissolving 158.0 g sugar (MM = 342.3 g/mol) in
643.5 cm3 of water. At 25oC, the density of water is 0.9971
g/cm3 and the vapor pressure is 23.76 torr.

Answer:

Psoln = XH2OPoH2O (Raoult’s Law)

To calculate mole fraction we need to determine number of

moles:

nsugar = 158.0 g/342.3 g mol-1 = 0.4616 mol sucrose

Density = mass g/Volume cm3

mH2O = dH2O x VH2O = 0.9971 g/cm3 x 643.5 cm3 = 641.6 g

nH2O = 641.6 g/18.01 g mol-1 = 35.63 mol H2O

 XH2O = nH2O/nsucrose + nH2O = 35.63/ 0.4616 + 35.63

= 0.9873

Then

Psoln = XH2OPoH2O = (0.9873)(23.76 torr)

= 23.46 torr

(Thus the vapor pressure of water has been lowered by 0.30

torr as going from pure state to solution)
 3

Exercise 3:

Predict the vapor pressure of a solution prepared by mixing

35.0 g solid Na2SO4 (MM = 142 g/mol) with 175 g water. The
vapor pressure of pure water at 25oC is 23.76 torr.

Answer:

Psoln = XH2OPoH2O (Raoult’s Law)

nH2O = 175.0 g/18.01 g mol-1 = 9.72 mol H2O

nNa2SO4 = 35.0 g/142 g mol-1 = 0.246 mol Na2SO4

But

Na2SO4  2Na+ + SO-24 (3 moles of ions per 1 mole of

solute present in solution)
Therefore

nsolute = 3 x 0.246 = 0.738 mol

XH2O = nH2O/nsolute + nH2O = 9.72/ 0.738 + 9.72

= 0.929

Then

Psoln = XH2OPoH2O = (0.929)(23.76 torr)

= 22.1 torr

Exercise 4:
 4

A solution is prepared by mixing 5.81 g acetone (MMC3H6O =

58.1 g/mol) and 11.9 g chloroform (MMHC?Cl3 = 119.4 g/mol).
At 35oC, this solution has a total vapor pressure of 260 torr. Is
this an ideal solution? The vapor pressure of pure acetone and
pure chloroform at 35oC are 345 and 293 torr, respectively.

Answer:

Ptotal = XA PoA + XB PoB

nacetone = 5.81 g/(58.1 g/mol) = 0.100 mol acetone

nchloroform = 11.9 g/(119 g/mol) = 0.100 mol chloroform

Xacetone = Xchloroform = 0.500

Then

Ptotal = (0.500)(345 torr) + (0.500)(293 torr)

= 319 torr

Since observed pressure is 260 torr (lower than expected),

solution does not behave ideally.

The dipole-dipole interaction between acetone and

chloroform lowers their tendency to escape from solution.

Exercise 5:
 5

A solution was prepared by dissolving 18.00 g glucose in

150.0 g water. The resulting solution was found to have a
boiling point of 100.34oC. Calculate the molar mass of
glucose. [Kb = 0.51oC. kg/mol and d = 1.00 g/cm3 for water]

Answer:

ΔTb = Kb msolute
= (100.34 -100.00)oC = 0.34oC

msolute = ΔTb/ Kb = (0.34oC)/(0.51 oC. kg/mol) = 0.67 mol/kg

= nsolute/kgsolvent

nglucose = mglucose x kgH2O = (0.67 mol/kg) x (0.1500 kg)

= 0.10 mol

MMglucose = massglucose /nglucose = 18.00 g /0.10 mol

= 180 g/mol (glucose)

Exercise 6:
 6

What mass of ethylene glycol (C2H6O2, MM = 62.1 g/mol)

must be added to 10.0 L water to produce an antifreeze
solution that freezes at -23.3oC? [Kf = 1.86oC. kg/mol and
d = 1g/cm3 for water]

Answer:

ΔTf = Kf msolute
= [0.0 – (-23.3)]oC = 23.3oC

msolute = ΔTf/ Kf = (23.3oC)/(1.86 oC. kg/mol) = 12.5 mol/kg

= nsolute/kgsolvent

nglycol = mglycol x kgH2O = (12.5 mol/kg) x (10.0 kg)

= 1.2 x 102 mol

nglycol = massglycol /MMglycol

massglycol = nglycol x MMglycol
= (12.5 x 102 mol) x (62.1 g/mol)
= 7.76 x 103 g
= 7.76 kg (Ethylene glycol)
 7

Exercise 7:

A sample of human hormone (controls metabolism) weighing

0.546 g was dissolved in 15.0 g benzene, and the freezing-
point depression was determined to be 0.24 oC. Calculate the
molar mass of the hormone. [Kf = 5.12oC. kg/mol for
benzene]

Answer:

ΔTf = Kf msolute
= 0.24oC

mhormone = ΔTf/Kf = (0.24oC)/(5.12 oC. kg/mol)

= 4.69 x 10-2 mol/kg

= nhormone/kgbenzene

nhormone = mhormone x kgbenzene

= (4.69 x 10-2 mol/kg) x (0.01500 kg)
= 7.04 x 10-4 mol

MMhormone = masshormone /nhormone = 0.546 g /7.04 x 10-4 mol

= 776 g/mol (hormone)
 8

Exercise 8:

Certain protein weighing 1.00 x 10-3 g was dissolved in

enough water to make 1.00 mL of solution The osmotic
pressure of this solution was found to be 1.12 torr at 25 oC.
Calculate the molar mass of the protein.[R = 0.08206
L.atm/K.mol and d = 1 g/cm3 for water]

Answer:

π = MRT
(π must be converted into atm and T into Kelvin because of
the units of R)

π = 1.12 torr/760 torr atm-1 = 1.47 x 10-3 atm and

T = 25.0 + 273 =298 K

Mprotein = π /RT
= (1.47 x 10-3 atm)/( 0.08206 L.atm/K.mol x 298 K)
= 6.01 x 10-5 mol/L

= nprotein/Vsolution

nprotein = Mprotein x Vsolution

= (6.01 x 10-5 mol/L) x (1.00 x 10-3 L)
= 6.01 x 10-8 mol

MMprotein = massprotein /nprotein = 1.00 x 10-3 g /6.01 x 10-8 mol

= 1.66 x 104 g/mol (protein)
 9

Exercise 9:

What concentration of NaCl in water is needed to produce an

aqueous solution isotonic with blood (π = 7.70 atm at 25oC)

Answer:

π = iMRT
M = (π / RT)/ i (i = 2 for NaCl)

7.70 atm
M =
[(0.08206 L atm)K mol) x 298 K]/2

= 0.315/2 mol/L = 0.158 M

But this represents the total molarity of solute particles:

0.315
M NaCl   0.158M
2

NaCl  Na   Cl 

0.158M 0.158M 0.158M

(0.315M)
 10

Chem 101 Chapter 11, Selected Old Exams Questions:

(Not all Answers are A):

1. Which of the following aqueous solutions has the lowest freezing point?

A 0.10 m KCl B 0.20 m C6H12O6 (glucose)

C 0.15 m SrBr2 D 0.10 m C12H22O11 (sucrose)
E 0.050 m AlCl3

The solvent is water in all cases, and thus

ΔTf = i x m x kf(water)

Thus the lowest freezing point is where the particle concentration i x m is

largest:

0.10 m KCl: 2 ions/KCl, i = 2: 0.20 m particles

0.20 m C6H12O6 (glucose), sugar: i = 1: 0.20 m particles

0.15 m SrBr2: 3 ions/SrBr2, i = 3: 0.45 m particles

0.10 m C12H22O11 (sucrose), sugar: i = 1: 0.10 m particles

0.050 m AlCl3: 4 ions/AlCl3, i = 4: 0.20 m particles

(correct answer: C)
 11

2. A solution is prepared by dissolving 4.9 g sucrose (C12H22O11), a

nonelectrolyte, in 175 g water. Calculate the boiling point of this solution (k b
for water is 0.51 oC kg/mol).

A 114.280 oC B 125.000 oC C 100.000 oC

D 100.042 oC E 98.780 oC

ΔTb = Tb - 100.000 oC = m x kb

MMsucrose = (12 x 12.01 + 22 x 1.008 + 11 x 16.00) g/mol

= 342.4 g/mol

sucrose is a non-electrolyte, thus van't Hoff factor i = 1

n = masssucrose/MMsucrose = (4.9 g)/(342.4 g/mol)

= 0.01431 mol

m = nsolute/masssolvent = (0.01431 mol)/(175 x 10-3 kg)

= 0.0818 mol/kg

Tb = 0.0818 mol/kg x 0.51 oC kg/mol + 100.000 oC

= 100.042 oC ( choice D)
 12

3. Which of the following substances dissolves in hexane (C6H14)?

A octane (C8H18) B HF C H2O
D NaCl E (NH4)2SO4
"like" dissolves "like"

hexane: non-polar, only dispersion forces

octane: non-polar, only dispersion forces, dissolves in hexane (choice A)

HF, H2O: polar, hydrogen bonds, dispersion forces

NaCl, (NH4)2SO4: ionic solids, dissolve nothing

4. What mass of HCl is required to prepare 1.00 kg of solution containing 5.5

% by mass of HCl?

A 550 g B 2.01 g C 55 g
D 18 g E 0.018 g
The total mass of solution is 1 kg, so
% HCL = (mHCl/1 kg) x 100 %
mHCl = (% HCl x 1 kg)/100 % = (5.5 % x 1 kg)/100 % = 0.055 kg = 55g
(choice C)
 13

5. An aqueous 4.69 M cesium chloride (CsCl) solution has a density of 1.58

g/mL. Determine the molality of this solution.

A 1.00 m B 5.93 m C 5.24 m

D 4.69 m E 2.97 m

for molality, m, we need the number of moles of CsCl and the mass of water in
a sample of for example 1 L solution:

1 L CsCl solution contains 4.69 mol CsCl

MMCsCl = (132.90 (Cs) + 35.45 (Cl)) g/mol

= 168.35 g/mol

Thus 1 L CsCl solution contains

massCsCl = 4.69 mol x 168.35 g/mol = 789.56 g CsCl

1 L CsCl solution has a mass of

1000 mL x 1.58 g/mL = 1580 g solution

Thus the water in 1 L CsCl solution has a mass of

(1580 - 789.56) g = 790.44 g = 0.79044 kg water

Thus the molality is (mass of solvent in kg)

m = nCsCl/masswater = 4.69 mol/0.79044 kg = 5.93 mol/kg = 5.93 m

(choice B)

6. Calculate the molality of a 2.35 M methanol (CH3OH) solution whose

density is 0.920 g/mL.

A) 1.05 m B) 2.78 m C) 3.62 m

D) 5.40 m E) 4.73 m
 14

There are 2.35 mol CH3OH in 1 L of solution (molarity):

(mass) M1L solution = 0.920 g/mL x 1000 mL = 920 g
(mass of methanol in 1 L) Mmethanol = (12.01 + 4 x 1.008 +16.00) g/mol x 2.35
mol
= 75.2987 g

thus (920 - 75.2987) g H2O in 1 L solution = 844.7013 g H2O in 1 L solution

therefore the molality, m, is

2.35 mol solute mol
m= = 2.78 = 2.78 m
0.8447013 kg solvent kg

(choice B)
 15

7. The solubility of a gas in water at 25 oC and a partial pressure of 1.0 atm is

6.7 x 10-4 mol/L. What is the molar concentration of the gas dissolved in water
if its partial pressure is 0.80 atm?

A) 6.7 x 10-4 M B) 2.1 x 10-4 M C) 1.0 x 10-4M

D) 1.8 x 10-4 M E) 5.4 x 10-4 M

Henry's law: c = kP

c 6.7  10 -4 M M
k= = = 6.7  10-4
P 1.0 atm atm

Thus the solubility, c', at 0.80 atm is

M
c = 6.7  10-4  0.80 atm = 5.4  10-4 M
atm

(choice E)

8. At 25 oC the vapor pressure of pure water is 23.76 mmHg and that of an

aqueous glucose solution is 22.56 mmHg. Calculate the mole fraction of
glucose.

A) 0.9494 B) 0.9573 C) 0.0382

D) 0.0505 E) 0.0120

Raoult's law: P = xsolventPo = (1 - xsolute) Po

thus ΔP = Po - P = xsolute Po
o
= P - P = 23.76 - 22.56 = 0.0505
x solute o
P 23.76

(choice D)

9. The molal freezing point depression constant of water is 1.86 oC/m.

Calculate the freezing point of a solution containing 185 g NaCl in 1350 g of
water.
 16

A) 4.36 oC B) -5.23 oC C) -8.72 oC

D) -2.49 oC E) -4.36 oC

for NaCl the van't Hoff factor is i = 2.

185 g
n NaCl = = 3.1656 mol
g
(22.99 + 35.45)
mol

Thus the molality of the electrolyte is

3.1656 mol
m= = 2.345 m
1.350 kg

Therefore
o
C
 T f = i m k f = 2  2.345 m  1.86 = 8.7234 o
C
m

 T f = T of - T f thus T f = T of -  T f

o o
T f = (0 - 8.7234) C = - 8.72 C

(choice C)
 17

10. A 305 mL benzene solution containing 3.25 g of a non-electrolyte

compound has an osmotic pressure of 7.54 mmHg at 22 oC. Calculate the
molar mass of this compound.

A) 1.72 x 103 g/mol B) 2.60 x 104 g/mol C) 3.85 x 102 g/mol

D) 7.23 x 105 g/mol E) 9.45 x 106 g/mol

non-electrolyte: van't Hoff factor, i=1


 = MRT ; M =
RT

Thus the molarity, M, is

1 atm
7.54 torr 
M= 760 torr = 4.094  10-4 M
L atm
0.0821  295.15 K
K mol

Then the number of moles in 305 mL solution is

mol
n = 4.094  10-4  0.305 L = 1.24867  10-4 mol
L

and finally the molar mass, MM:

3.25 g g
MM = = 2.60  10 4
1.24867  10 mol
-4
mol

(choice B)
 18

11. Which of the following statements is correct?

A) The solubility of solids in water always increases with increasing

temperature

(not always, only when the solution is endothermic, ΔsolnH > 0)

B) The solubility of solids in water always decreases with increasing

temperature

(not always, only when the solution is exothermic, ΔsolnH < 0)

C) The solubility of a gas in water increases with increasing temperature

(no, the general observation is opposite)

D) The solubility of NH3 gas in water is well described by Henry's law

(no, Henry's law holds only if no chemical reactions occur during dissolving.
here: NH3(g) + H2O(l)  NH4OH(aq) oocurs at least to a limited amount)

E) The solubility of a gas in water decreases with increasing temperature

(yes, that is the general observation)

(choice E)
 19

12. A solution is made by mixing two volatile liquids A and B. The

intermolecular forces between A and B molecules in the solution are weaker
than those between A molecules and B molecules in the pure liquids. In these
conditions which of the following statements is correct?

A) ΔsolnH = 0
(no, ideal solution only if for the forces AA = BB = AB is true)

B) The solution process is endothermic

(yes, because energy increases, since the forces are AB < AA, BB)

C) A negative deviation from Raoult's law is observed

(No, only if ΔsolnH < 0; here a positive one would be observed)

D) A and B may be benzene and toluene respectively

(no, these, because of very similar dispersion interactions AA, BB, and AB,
form an almost ideal solution)

E) The solution process is exothermic

(no, would only be true if forces AB > AA, BB)

(choice B)
 20

13. What is the mole fraction of ethanol (C2H5OH) in a solution, which

consists of 71.0 g of ethanol for every 10 g of water present?

A) 0.735
B) 0.710
C) 0.877
D) 0.265
E) 0.221

Molar masses, MM, and mole numbers, n, are for 71.0 g ethanol:
g
MM C2 H 5OH = (2  12.01 + 6  1.008 + 16.00)
mol

g
= 46.068
mol
71.0 g
nC2 H 5OH = = 1.5412 mol
g
46.068
mol

and for 10.0 g water:

g g
MM H 2O = (2  1.008 + 16.00) = 18.016
mol mol

10.0 g
nH 2O = = 0.5551 mol
g
18.016
mol

Thus the mole fraction of ethanol in the solution is

nC2H 5OH 1.5412 mol
xC2H 5OH = =
nC2H 5OH + n H 2O 1.5412 mol + 0.5551 mol
= 0.735200 = 0.735
 21

14. Rank the following three compounds according to increasing solubility in

water:

I) CH3-CH2-CH2-CH3 II) CH3-OH III) CH3-CH2-CH2-CH2-CH2-OH

A) I < III < II

B) II < III < I
C) I < II < III
D) III < II < I
E) II < I < III

I) butane, is non-polar and has only dispersion interactions, so it has the

smallest solubility in water.

II) methanol, can form hydrogen bonds like water and has the smallest
carbohydrate chain, so it dissolves best in water, has the largest solubility.

III) pentanol, can form hydrogen bonds like water and methanol, but has a
longer carbohydrate chain than methanol, so it has a medium solubility.

Solubility in water: I < III < II

 22

15. A correct statement of Henry's law is:

A) the concentration of a gas in solution is directly proportional to the pressure

above the solution.

(correct: c = kP)

B) the concentration of a gas in solution is directly proportional to the mole

fraction of solvent.

(incorrect, see A)

C) the concentration of a gas in solution is independent of pressure.

(incorrect, see A)

D) the concentration of a gas in a solution is inversely proportional to pressure.

(incorrect, see A)

E) the concentration of a gas in solution is inversely proportional to

temperature.

(incorrect, not Henry's law, see A)

 23

16. At a given temperature, you have a mixture of benzene (vapor pressure of

pure benzene = 745 torr) and toluene (vapor pressure of pure toluene = 290.
torr). The mole fraction of benzene in the solution is 0.590. Assuming ideal
behavior, calculate the mole fraction of toluene in the vapor above the
solution.

A) 0.213
B) 0.778
C) 0.641
D) 0.859
E) 0.590

Since it is a binary mixture, the mole fraction of toluene in the solution is

xtol = 1 - xbenz

The total vapor pressure, Ptot, can be calculated with Raoult's law (ideal
behavior):

Ptot = xbenz Pobenz + xtol Potol

= 0.590 x 745 torr + (1 - 0.590) x 290 torr = 558.45 torr

The partial vapor pressure of toluene over the solution follows also from
Raoult's law:

Ptol = xtol Potol = (1 - 0.590) x 290 torr = 118.90 torr

From Dalton's law of partial pressures follows the mole fraction of toluene in
the vapor, ytol:

Ptol = 118.90 = 0.2129 = 0.213

Y tol =
Ptot 558.45
 24

17. A compound contains only carbon, hydrogen, and oxygen and has an
empirical formula of C7H4O. When 1.32 g of the compound is dissolved in
11.4 g of a solvent the freezing point is lowered by 22.3 oC. The molecular
formula of the compound is:
(Kf = 40. oC kg/mol)

A) C14H8O2
B) C7H4O
C) C5H2O
D) C7H8O
E) C12H9O3

The freezing point depression of a solution is ΔTf = Kfm, with m being the
molality of the solute, m = nsolute/masssolvent(in kg), nsolute being the mole number
of the dissolved compound.

Let M denote the mass of the dissolved compound, and MM is real molar
mass, then
nsolute = M/MM. The real molar mass is MM = MM emp x k, k being an integer
and MMemp the molar mass as calculated from the empirical formula.

Then k can be calculated from the freezing point depression:

n solute
T f = K f  m= K f 
mass solvent (in kg)

M
MM real Kf M
=Kf =
mass solvent (in kg) MM emp  k  M solvent

Solving for k yields:

Kf M
k=
MM emp   T f  M solvent
 25

g g
MM emp = (7  12.01 + 4  1.008 + 16.00) = 104.102
mol mol

and thus we obtain k:

o
C kg
40.  1.32 g
k= mol = 1.995 = 2
g
104.102  22.3 o C  11.4  10-3 kg
mol

Thus our molecular formula is (C7H4O)2 = C14H8O2

18. A liquid-liquid solution is called an ideal solution if:

I. it obeys PV = nRT
II. it obeys Raoult's law
III. solute-solute, solvent-solvent, and solute-solvent interactions are very
similar.

IV. solute-solute, solvent-solvent, and solute-solvent interactions are quite

different.

A) II, III B) I, II, III C) I, II, IV

D) II, IV E) I, IV, III

(An ideal solution is one which obeys Raoult's law (II), which is the case when
ΔsolnH is about 0, which is the case when solute-solute, solvent-solvent, and
solute-solvent interactions are very similar (III). Thus A) II, III is correct)

19. When a non-volatile solute is added to a volatile solvent, the solution vapor
pressure ________, the boiling point ________, the freezing point ________,
and the osmotic pressure across a semi-permeable membrane ________.

A) decreases, increases, decreases, increases

B) decreases, increases, decreases, decreases
 26

C) increases, increases, decreases, increases

D) increases, decreases, increases, decreases
E) decreases, decreases, increases, decreases

For the colligative properties our laws are:

vapor pressure lowering: ΔP = Po - P = xsolutePo, thus P decreases when solute is

added.

boiling point elevation: ΔTb = Tb - Tbo = Kbmsolute, thus Tb increases when solute
is added.

freezing point depression: ΔTf = Tfo - Tf = Kfmsolute, thus Tf decreases when

solute is added.

osmotic pressure: Π = MsoluteRT, thus Π increases when solute is added.

(A is correct)