Properties of Solutions
Chapter 17
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Properties of Solutions – ch. 17
1. Predict the relative solubility of the following:
a. O2 in H20 vs. O2 in CCl4
b. CH3OH in H2O vs. CH3CH2CH2CH2OH in H2O
c. AgCl(s) in H2O at 25°C vs. AgCl(s) in H2O at 45°C
d. CO2(g) in C8H18 at 25°C vs. CO2(g) in C8H18 at 45°C
e. N2(g) in C6H6(g)at 1 atm vs. N2(g) in C6H6(g)at 8 atm
Properties of Solutions – ch. 17
Solubility ⇒ the maximum amount of solute that can dissolve in
a given amount of solvent at any one T and P
Solubility Considerations
1. Polarity of solute and solvent ⇒ “like dissolves like”
2. Temperature ⇒ Substances are more soluble at high T’s if
ΔSdiss > 0 (typical for solid solutes) and less soluble at
high T’s if ΔSdiss < 0 (typical for gas solutes)
3. Pressure ⇒ gases are more soluble at high P’s (Le Chatlier’s
principle)
Properties of Solutions – ch. 17
2. What is the molality of an aqueous solution that is 12.5% methanol
by weight (molar mass of CH3OH is 32.05 g/mol).
Molality ⇒ m =
nsolute
kgsolvent
Properties of Solutions – ch. 17
3. Calculate the molality and mole fraction of an aqueous solution that
is 8 M NaCl, the density of the solution is 1.18 g/ml.
Concentrations Variations
Molarity ⇒ M =
nsolute
Lsolution
Molality ⇒ m =
nsolute
kgsolvent
Mole fraction ⇒ Xa =
n𝑎
ntotal
Properties of Solutions – ch. 17
4. Calculate the heat of hydration for the following ionic solids.
a. KF
b. RbF
c. Compare the ion-dipole forces for K+ vs. Rb+ in H2O
Compound
Lattice Energy
Heat of Solution
KF
-804 kJ/mol
-15 kJ/mol
RbF
-768 kJ/mol
-24 kJ/mol
Properties of Solutions – ch. 17
5. A salt solution sits in an open beaker. Assuming constant temperature,
the vapor pressure of the solution will…
a. increases over time
b. decreases over time
c. stays the same over time
d. We need to know which salt is in the solution to answer this
e. We need to know the temperature and pressure to answer this
Properties of Solutions – ch. 17
Colligative properties ⇒ the properties of a solution with a
non-volatile solute relative to a pure solvent
As the concentration of solute particles ↑
vapor pressure ↓ “depression” ⇒ Pa = Xa Pa°
freezing point ↓ “depression” ⇒ ΔTf = - kf i m
boiling point ↑ “elevation” ⇒ ΔTb = kb i m
osmotic pressure ↑ ⇒ π = iMRT
i = vant Hoff factor = moles of solute particles
moles of solute
Properties of Solutions – ch. 17
6. Rank the following aqueous solutions in order of increasing vapor
pressure:
a. 0.1 M C6H12O6
b. 0.1 M KBr
c. 0.05 M Na2SO4
d. 0.05 M CH3COOH
Properties of Solutions – ch. 17
7. Calculate the vapor pressure of a solution (in torr) made by
dissolving 159 g of ethylene glycol (HOCH2CH2OH – 62.08g/mol)
in 500 g of water at 27 °C. At 27 °C the vapor pressure of pure
water is 26.7 torr.
Properties of Solutions – ch. 17
8. Benzene (C6H6 – 78.12 g/mol) and toluene (C7H8 – 92.15 g/mol)
form an ideal solution. What is the vapor pressure of a solution
prepared by mixing 40 g of toluene with 15 g of benzene at 25 °C?
At 25 °C the vapor pressures of pure toluene and pure benzene are
28 and 95 torr respectively. What is the mole fraction of the
benzene in the vapor above the solution?
Properties of Solutions – ch. 17
9. Pentane and hexane form an ideal solution. What composition of a
pentane and hexane solution at 25 °C would give a vapor pressure
of 350 torr? At 25 °C the vapor pressures of pure pentane and
hexane are 511 torr and 150 torr respectively.
Properties of Solutions – ch. 17
10. After substance A is mixed with substance B the solution feels hotter than before
they were mixed. What deviation from Raoult’s Law (if any) would be
expected for this solution?
a. no deviation
b. positive deviation – A and B form stronger forces than pure A and B
c. positive deviation – A and B form weaker forces than pure A and B
d. negative deviation – A and B form stronger forces than pure A and B
e. negative deviation – A and B form weaker forces than pure A and B
Properties of Solutions – ch. 17
Ideal vs. Non-Ideal Solutions
a. ΔHsol’n ~ 0 ⇒ Ideal solution – the actual vapor pressures will agree with
Raoult’s law
b. ΔHsol’n > 0 ⇒ Non-Ideal solution – the forces in the solution are weaker
than in the pure substances resulting in higher VPs than expected from
Raoult’s Law
c. ΔHsol’n < 0 ⇒ Non-Ideal solution – the forces in the solution are stronger
than in the pure substances resulting in lower VPs than expected from Raoult’s
Law
Properties of Solutions – ch. 17
11. Predict if the following solutions will be ideal or non-ideal. What type of
deviation from Raoult’s Law would you expect for the non-ideal solutions?
a. carbon tetrachloride (CCl4) and dichloromethane (CH2Cl2)
b. water and acetone (CH3COCH3)
c. carbon tetrachloride and benzene (C6H6)
Properties of Solutions – ch. 17
12. The vapor pressures of several solutions of water and butanol were
determined at various compositions and the data is given below:
XH2O
VP (torr)
0.00
47.0
0.27
52.5
0.58
58.3
0.72
46.9
1.00
40.2
a. Are the solutions of water and butanol ideal?
b. Which of the above solutions would have the lowest boiling point?
Properties of Solutions – ch. 17
13. Rank the following aqueous solutions by their boiling points, and
freezing points.
a. 0.1 M CH2O
b. 0.1 M LiF
c. 0.05 M (NH4)2SO4
Properties of Solutions – ch. 17
Properties of Solutions – ch. 17
14. Calculate the boiling point and freezing point of a solution made by
dissolving 110 g of K3PO4 (212.3g/mol) in 800 mL of water at 1
atm. For water kb = 0.51 °Ckg/mol and kf = 1.86 °Ckg/mol.
Properties of Solutions – ch. 17
15. Calculate the osmotic pressure of a solution made by dissolving 83
g of glucose (C6H12O6) in 100 mL of water at 30 °C.
Properties of Solutions – ch. 17
Osmosis ⇒ the diffusion of water thru a semipermeable membrane
Osmotic pressure (π )⇒ the minimum pressure that must be applied to keep osmosis from
occurring ⇒ π= iMRT ⇒ as the concentration gradient ↑ π↑
Reverse osmosis ⇒ increasing the conc gradient by the movement of water thru a
semipermeable membrane by applying a pressure that is > the osmotic pressure
Properties of Solutions – ch. 17
16. A solution contains 3.75 g of a nonvolatile hydrocarbon in 95 g of
acetone. The boiling points of pure acetone and the solution are
55.9 °C and 56.5 °C respectively. What is the molar mass of the
hydrocarbon? For acetone the Kb = 1.71 °CKg/mol.
Properties of Solutions – ch. 17
17. A solution that contains 29 g of non-volatile/non-ionizing solute in
126 g of water has a vapor pressure of 723.4 torr at 100 °C.
What is the molar mass of the solute?
Properties of Solutions – ch. 17 – Feldwinn
18. A solid mixture contains MgCl2 (molar mass= 95.218 𝑔 𝑚𝑜𝑙) and
NaCl (molar mass = 58.443 𝑔 𝑚𝑜𝑙). When 0.500 g of this solid
is dissolved in enough water to form 1.000 L of solution, the
osmotic pressure at 25.0˚C is observed to be 0.3950 atm. What
is the mass percent of MgCl2 in the solid? (Assume ideal
behavior for the solution.)
Properties of Solutions – Answers
1. Predict the relative solubility of the following:
a. O2 in H20 vs. O2 in CCl4
b. CH3OH in H2O vs. CH3CH2CH2CH2OH in H2O
c. AgCl(s) in H2O at 25°C vs. AgCl(s) in H2O at 45°C
d. CO2(g) in C8H18 at 25°C vs. CO2(g) in C8H18 at 45°C
e. N2(g) in C6H6(g)at 1 atm vs. N2(g) in C6H6(g)at 8 atm
2. What is the molality of an aqueous solution that is 12.5% methanol by
weight (molar mass of CH3OH is 32.05 g/mol).
𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
molality = 𝑘𝑔
𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
12.% ⇒ if there’s 100 g of solution 12.5 g of CH3OH and 87.5 g of water
12.5 𝑔
0.39 𝑚𝑜𝑙
mole of CH3OH = 32.05
=
0.39
mol
⇒
m=
= 4.46m
𝑔/𝑚𝑜𝑙
0.0875 𝑘𝑔
Properties of Solutions – Answers
3. Calculate the molality and mole fraction of an aqueous solution that
is 8 M NaCl, the density of the solution is 1.18 g/ml. (molar mass
of NaCl is 58.44 g/mol)
𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
molality = 𝑘𝑔
𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
and
𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝐶𝑙
mole fraction = 𝑡𝑜𝑡𝑎𝑙
𝑚𝑜𝑙𝑒𝑠
8 M NaCl ⇒ 8 moles of NaCl per 1 L of solution
(8 mol NaCl)(58.44 g/mol) = 467.5 g NaCl
(1 L of sol’n)(1000mL/L)(1.18 g/L) = 1180 g of sol’n – 467.5g = 712.5
g of water
712.5 𝑔 𝑤𝑎𝑡𝑒𝑟
= 39.5 mol water
18.02 𝑔/𝑚𝑜𝑙
8 𝑚𝑜𝑙
molality = 0.7125
= 11.2 m
𝑘𝑔
8 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝐶𝑙
mole fraction= 8+39.5
= 0.168
𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑙𝑒𝑠
Properties of Solutions – Answers
4. Calculate the heat of hydration for the following ionic solids.
a. KF
b. RbF
c. Compare the ion-dipole forces for K+ vs. Rb+ in H2O
ΔHhydration = ΔHLE + ΔHsol’n
KF ⇒ ΔHhydration = (-804kJ/mol) + (-15kJ/mol) = -819kJ/mol
RbF ⇒ ΔHhydration = (-768kJ/mol) + (-24kJ/mol) = -792kJ/mol
Since both species have fluoride we can compare the ion-dipole forces
of K+ vs. Rb+ ⇒ the more negative the heat of hydration the stronger
the force ⇒ K+ has the stronger ion-dipole force ⇒ this makes sense
since K+ is smaller and can form shorter/stronger bonds with water
Properties of Solutions – Answers
5. A salt solution sits in an open beaker. Assuming constant
temperature, the vapor pressure of the solution will…
b. decreases over time
as water evaporates away the concentration of the salt increases
causing the VP to decrease
6. Rank the following aqueous solutions in order of increasing vapor
pressure:
a. 0.1 M C6H12O6 ⇒ i=1
as (conc)x(i) ↑ VP ↓
b. 0.1 M KBr ⇒ i=2
b<c<a<d
c. 0.05 M Na2SO4 ⇒ i=3
d. 0.05 M CH3COOH ⇒ i=1
Properties of Solutions – Answers
7. Calculate the vapor pressure of a solution (in torr) made by
dissolving 159 g of ethylene glycol (HOCH2CH2OH – 62.08g/mol)
in 500 g of water at 27 °C. At 27 °C the vapor pressure of pure
water is 26.7 torr.
(159g)/(62.08g/mol) = 2.6 mol HOCH2CH2OH
(500g)/(18.02g/mol) = 27.7 mol water
Pwater = (27.7mol/30.3mol)(26.7torr) = 24.4torr
Relative FP or VP => b < c < a < d
Relative BP or OP => b > c > a > d
Properties of Solutions – Answers
8. Toluene and benzene form an ideal solution. What is the vapor
pressure of a solution prepared by mixing 40 g of toluene with 15 g
of benzene at 25 °C? At 25 °C the vapor pressures of pure toluene
and pure benzene are 28 and 95 torr respectively.
(40g)/(92.15g/mol) = 0.43 mol toluene
(15g)/(78.12g/mol) = 0.19 mol benzene
Ptoluene = (0.43 mol/0.62 mol)(28 torr) = 19.6 torr
Pbenzene = (0.19 mol/0.62 mol)(95 torr) = 29.1 torr
Psolution = 48.7 torr
Properties of Solutions – Answers
9. Pentane and hexane form an ideal solution. What composition of a
pentane and hexane solution at 25 °C would give a vapor pressure
of 350 torr? At 25 °C the vapor pressures of pure pentane and
hexane are 511 torr and 150 torr respectively.
Ptotal = Ppentane + Phexane
Ptotal = XpPp° + XhPh°
Since Xp + Xh = 1 => Xp = 1 – Xh
Ptotal = (1 – Xh) Pp° + XhPh°
350 torr = (1 – Xh)(511 torr) + Xh(150 torr)
Xh = 0.45 => Xp = 0.55
Properties of Solutions – Answers
10. After substance A is mixed with substance B the solution feels hotter
than before they were mixed. What deviation from Raoult’s Law
(if any) would be expected for this solution?
d. negative deviation – A and B form stronger forces than pure A
and B
A hotter solution means that the heat of solution was exothermic – the
heat given off is due to forming strong forces
Properties of Solutions – Answers
11. Predict if the following solutions will be ideal or non-ideal. What type of
deviation from Raoult’s Law would you expect for the non-ideal solutions?
a. carbon tetrachloride (CCl4) and dichloromethane (CH2Cl2)
non-ideal with positive deviations - CCl4 has LDF whereas CH2Cl2 is polar
and has dipole-dipole in addition to the LDF – upon mixing the CH2Cl2 will be
unable to have as many dipole-dipole forces and therefore weaker
b. water and acetone (CH3COCH3) non-ideal with negative deviations – water
is polar and can H-Bond however acetone is polar but can’t H-Bond with itself
but can H-Bond with water – therefore the forces in solution are stronger
c. carbon tetrachloride and benzene (C6H6) ideal - both are non-polar and will
have LDFs when pure and in solution
Properties of Solutions – Answers
12. The vapor pressures of several solutions of water and butanol were
determined at various compositions and the data is given below:
XH2O
VP (torr)
0.00
47.0
0.27
52.5
0.58
58.3
0.72
46.9
1.00
40.2
a. are the solutions of water and butanol ideal? No – since the VP’s of some of
the solutions are outside of the range of pure water and pure butanol –
since they are outside on the high side the solution will have positive
deviations
b. which of the above solutions would have the lowest boiling point? Highest
VP will result in the lowest BP => XH2O = 0.58
Properties of Solutions – Answers
13. Rank the following aqueous solutions by their boiling points, and
freezing points.
a. 0.1 M CH2O
b. 0.1 M LiF
c. 0.05 M (NH4)2SO4
as (conc)x(i) ↑ FP ↓ and BP ↑
Relative FP => b < c < a
Relative BP => a < c < b
Properties of Solutions – Answers
14. Calculate the boiling point and freezing point of a solution made by
dissolving 110 g of K3PO4 (212.3g/mol) in 800 mL of water at 1
atm. For water kb = 0.51 °Ckg/mol and kf = 1.86 °Ckg/mol.
(110 g K3PO4)/(212.3g/mol) = 0.52 mol K3PO4
m = 0.52 mol/0.8 kg = 0.65m
ΔTf = -(0.65 mol/kg)(4)(1.86 °Ckg/mol) = -4.8°C
ΔTb = (0.65 mol/kg)(4)(0.51°Ckg/mol) = 1.3°C
Since pure water has a FP = 0°C and BP = 100°C then the
solution will have a FP = -4.8°C and BP = 101.3°C
Properties of Solutions – Answers
15. A solution contains 3.75 g of a nonvolatile hydrocarbon in 95 g of
acetone. The boiling points of pure acetone and the solution are 55.9
°C and 56.5 °C respectively. What is the molar mass of the
hydrocarbon? For acetone the Kb = 1.71 °CKg/mol.
molar mass = g/mol => given grams and you can get moles from
ΔTb = kb i m => mol = ΔTb kgsolvent/ kb i
mol = (56.5 °C – 55.9 °C)(0.095kg)/(1.71°CKg/mol)(1)
mol = 0.056
molar mass = 3.75g/0.056mol = 66.8g/mol
Properties of Solutions – Answers
16. Calculate the osmotic pressure of a solution made by dissolving 83
g of glucose (C6H12O6) in 100 mL of water at 30 °C.
π = iMRT
π = (1)((83g/180g/mol)/(01.L))(0.08206atmL/molK)(303K)
π = 115 atm
Properties of Solutions – Answers
17. A solution that contains 29 g of non-volatile/non-ionizing solute in
126 g of water has a vapor pressure of 723.4 torr at 100 °C.
What is the molar mass of the solute?
molar mass = g/mol => given grams and you can get moles from
Pwater = XwaterPwater° => nsolute = (nwPw°/Pw) – (nw) =>
nsolute = (126g/18g/mol)(760torr)/(723.4torr) – (126g/18g/mol)
nsolute = 0.354 moles
molar mass = 29g/0.354mol = 82 g/mol
Properties of Solutions – Answers
18. A solid mixture contains MgCl2 (molar mass= 95.218 𝑔 𝑚𝑜𝑙) and
NaCl (molar mass = 58.443 𝑔 𝑚𝑜𝑙). When 0.500 g of this solid
is dissolved in enough water to form 1.000 L of solution, the
osmotic pressure at 25.0˚C is observed to be 0.3950 atm. What
is the mass percent of MgCl2 in the solid?
If you have X g of MgCl2 then there’s 0.50 g – X g of NaCl
moles of MgCl2 = X/95.218 = 0.0105X
moles of NaCl = (0.5 – X)/58.443 = 0.00855 – 0.0171X
π = iMRT = (iM (for MgCl2)+iM (for NaCl))RT
0.3950 = (3(0.0105X/1) + 2((0.00855 – 0.0171X)/1)(0.08206)(298)
X = 0.352 g
% MgCl2 = (0.352g/0.5g)(100) = 70.4%