Justin Lin
Per. 1
Chapter 11: Solution Composition
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Molarity (M)= mol solute/L solution
Molality (m)= mol solute/kg solvent
The primary difference between molarity and molality is that molarity is
dependent on temperature but molality is not.
Mass percent (%)= (mass solute/mass solution)x100
Mole fraction (χ)= mol solute/mol solution
Like dissolves like. A polar solvent is used to dissolve a polar or ionic solute. A
nonpolar solvent is used to dissolve a nonpolar solute.
The formation of a liquid solution takes place in three distinct steps:
Step 1: Separating the solute into individual components of the solute (expanding
the solute).
Step 2: Overcoming the intermolecular forces into the solvent to make room for
the solute (expanding the solvent).
Step 3: Allowing the solute and solvent to interact to form the solution.
Steps 1 and 2 are endothermic and step 3 is often exothermic.
Enthalpy (heat) of solution (∆Hsoln) is the sum of the ∆H values for the steps:
∆Hsoln= ∆H1 + ∆H2 + ∆H3
One factor that favors a process is an increase in disorder.
Processes that require large amounts of energy tend not to occur.
Since like dissolves likes, compounds containing similar structures dissolve each
other.
Hydrophobic substances are nonpolar and do not dissolve in water.
Hydrophilic substances are polar and dissolve in water.
Pressure is directly proportional to gas solubility.
Henry’s Law states that the amount of a gas dissolved in a solution is directly
proportional to the pressure of the gas above the solution. It is given by the
equation:
P=kC; where P=the pressure of the gas, k=Henry’s Law constant,
C=concentration of gas
The dissolving of a solid occurs more rapidly at higher temperatures, but the
amount of solid that can be dissolved may increase or decrease with increasing
temperature.
The solubility of gas in water typically decreases with increasing temperature.
The presence of a nonvolatile solute lowers the vapor pressure of a solvent.
Molecules are in a constant state of motion and escape. Sometimes blocking the
leaving area can slow them down.
Raoult’s Law states that the vapor pressure of a solution is directly proportional to
the mole fraction of the solvent.
Psoln= χsolventP ۫ solvent; where Psoln=the observed vapor pressure of the solution,
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χsolven=the mole fraction of the solvent, and P ۫ solvent=the vapor pressure of the pure
solvent
For liquid-liquid solutions where both components are volatile, a modified form
of Raoult’s law applies.
Ptotal= PA + PB = χAP ۫ A + χBP ۫ B
where Ptotal represents the total vapor pressure of a solution containing A and B, χA
and χB are the mole fractions of A and B, P ۫ A and P ۫ B are the vapor pressures of
pure A and pure B, and PA and PB are the partial pressures resulting from the
molecules of A and of B in the vapor above the solution.
A liquid-liquid solution that obeys Raoult’s law is called an ideal solution.
A graphical representation of Raoult’s law
A negative deviation from Raoult’s law results when the observed vapor pressure
is lower than the value predicted by Raoult’s law.
A position deviation from Raoult’s law results when the observed vapor pressure
is higher than the value predicted by Raoult’s law.
Colligative property-a property that depends on the amount of solute particles in
the solution.
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The four colligative properties are:
1) Vapor pressure lowering
2) Boiling point elevation
3) Freezing point depression
4) Osmotic Pressure
A nonvolatile solute elevates the boiling point of the solvent.
The change in boiling point can be represented by the equation:
∆T=Kbmsolute
where ∆T is the boiling point elevation, or the difference between the boiling
point of the solution and that of the pure solvent, Kb is a constant that is the molal
boiling-point elevation constant, and msolute is the molality of the solute in the
solution.
When a solute is dissolved in a solvent, the freezing point of the solution is lower
than that of the pure solvent.
The equation for freezing point depression is:
∆T=Kfmsolute
where ∆T is the boiling point elevation, or the difference between the boiling
point of the solution and that of the pure solvent, Kf is a constant that is the molal
freezing-point elevation constant, and msolute is the molality of the solute in the
solution.
Osmosis-the flow of solvent into a solution through a semipermeable membrane.
Osmotic pressure-the pressure that must be applied to a solution to stop osmosis.
π=MRT; where π=osmotic pressure, M=molarity, R=gas constant, T=temperature
Hypertonic-Having the higher osmotic pressure of two solutions. Solute
concentration is less than surroundings.
Hypotonic-Having the lower osmotic pressure of two solutions. Solute
concentration is greater than surroundings.
Isotonic-having equal osmotic pressures. Solute concentration is equal to
surroundings.
Crenation-a cell shrinks due to hypertonic conditions.
Lysis (hemolysis)-A cell bursts due to hypotonic conditions.
Reverse osmosis-a solution in contact with pure solvent across a semipermeable
membrane is subjected to an external pressure larger than its osmotic pressure.
The effect is to cause a net flow of solvent from the solution into the pure solvent.
van’t Hoff factor i:
i=(moles of particles in solution/moles of solute dissolved)
Because of ion pairing, the experimental value for i is often significantly less than
the expected value.
The colligative properties of electrolyte solutions are described by including the
van’t Hoff factor in the appropriate equation. For example, ∆T=imK and π=iMRT
Tyndall effect-the scattering of light by particles
Colloidal dispersion/colloid-A suspension of tiny particles in some medium
Particles remain suspended rather than forming larger aggregates and
precipitating out mainly because of electrostatic repulsion.
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Coagulation-the destruction of a collid. This can be usually accomplished either
by heating or by adding an electrolyte.
Multiple Choice Questions
1) Consider a mixture of H2SO4 and water. Assuming 1.00 L of solution, what is the
molality of a 5.20-molar H2SO4? The density of the solution is 1.38 grams per
millimeter.
a) 4.78 m
b) 5.78 m
c) 5.98 m
d) 6.08 m
e) 7.98 m
2) Diet Coke is bottled at 25 Celsius so that the liquid contains carbon dioxide at 5.0
atmospheres of pressure. The pressure of carbon dioxide in the atmosphere is
4.0x(10^-4) atm. What is the concentration of carbon dioxide before and after
opening the bottle, respectively. The Henry’s law constant for carbon dioxide is
32 L atm/mole.
a) 0.32 mol/L and 1.2x(10^-5) mol/L
b) 0.32 mol/L and 1.2x(10^-4) mol/L
c) 0.16 mol/L and 1.2x(10^-5) mol/L
d) 0.16 mol/L and 1.2x(10^-4) mol/L
e) 0.08 mol/L and 1.2x(10^-4) mol/L
3) What is the vapor pressure of a benzene/toluene solution if the mole fraction of
benzene is 0.30 and that of toluene is 0.70? The vapor pressure of benzene is 73
torr and that of toluene is 27 torr.
a) 50 torr
b) 41 torr
c) 46 torr
d) 35 torr
e) 47 torr
4) The molal freezing point constant for water is 1.86۫C Kg/mol. We are going to
discuss an ideal 0.050 molal solution of sand that does not dissociate. What is the
freezing point of the solution?
a) -0.093۫C
b) -0.057۫C
c) -0.027۫C
d) 0.057۫C
e) 0.093۫C
5) A solution was prepared by dissolving 18.00 g of glucose in 150.0 g of water. It
had a boiling point of 100.34۫C. What is the molecular weight of glucose?
a) 168 g/mol
b) 176 g/mol
c) 181 g/mol
d) 195 g/mol
e) 207 g/mol
Multiple Choice Answers
1) c) 5.98 m
1.00 L soln x (10^3 mL soln/1L soln) x (1.38 g soln/1 mL soln) = 1380 g soln
5.20 mol H2SO4 x (98.08 g H2SO4/1 mol H2SO4) = 510 g H2SO4
1380 g soln-510 g H2SO4=870 g water
870 g H2SO4 x (1 kg H2SO4/1000 g H2SO4)=0.870 kg water
m=mol H2SO4/kg water=5.20 mol H2SO4/0.870 kg water=5.98 m
2) c) 0.16 mol/L and 1.2x(10^-5) mol/L
P=kC
C=P/k
Before
C=P/k
C=5.00 atm/32 (L atm/mol)
C=0.16 mol/L
After
C=P/k
C=4.0x(10^-9) atm/ 32 (L atm/mol)
C=1.2x(10^-5) mol/L
3) b) 41 torr
Psoln= PB + PT
Psoln= χBP ۫ B + χTP ۫ T
Psoln= (0.30)(73 torr) + (0.70)(27 torr)
Psoln=41 torr
4) a) -0.093۫C
∆T=Kfm
∆T= (1.86 ۫C Kg/mol)(0.050 mol/kg solvent)
∆T=0.093۫C
0۫C-0.093۫C=-0.093۫C
5) c) 181 g/mol
m=∆T/Kb=0.34۫C/0.512۫C Kg/mol)=0.663 mol glucose/kg water
0.663 mol glucose/kg water x (1 kg water/1000 g water) x 150.0 g water = 0.0994
mol glucose
M.W. = 18.00 g gluc/0.0994 mol gluc
=181 g/mol
Free Response
The formula and the molecular weight of an unknown hydrocarbon compound to be
determined by the elemental analysis and the freezing-point depression method.
a) The hydrocarbon is found to contain 93.46 percent carbon and 6.54 percent
hydrogen. Calculate the empirical formula of the unknown hydrocarbon.
b) A solution is prepared by dissolving 2.53 grams of p-dichlorobenzene (molecular
weight 147.0 g/mol) in 25.86 grams of naphthalene (molecular weight 128.2
g/mol). Calculate the molality of the p-dichlorobenzene solution.
c) The freezing point of pure naphthalene is determined to be 80.2۫C. The solution
prepared in (b) is found to have an initial freezing point of 75.7۫C. Calculate the
molal freezing-point depression of naphthalene.
d) A solution of 2.42 grams of the unknown hydrocarbon dissolved in 26.7 grams of
naphthalene is found to freeze initially at 76.2۫C. Calculate the apparent
molecular weight of the unknown hydrocarbon on the basis of the freezing-point
depression experiment above.
e) What is the molecular formula of the unknown hydrocarbon?
Free Response Solution
a) Assume 100 g of substance
93.46 g C x (1 mol C/12.01 g C) = 7.782 mol C
6.54 g H x (1 mol H/1.01 g H) = 6.48 mol H
7.782 mol C/6.48 mol H = 1.20 x 5 = 6
6.48 mol H/6.48 mol H= 1.00 x 5 = 5
Emp formula C6H5
b) m=mol solute/kg solvent=(2.53 g solute x (1 mol solute/147.0 g solute))/0.02586
kg solvent
m=0.665 mol/kg
c) ∆T=Kfm
Kf=∆T/m=(80.2۫C-75.7۫C)/0.665 mol/kg
Kf=6.77 ۫C Kg/mol
d) ∆T=Kfm
∆T=Kf mol solute/kg solvent
M.W.=mass solute/mol solute
mol solute=mass solute/M.W.
∆T=Kf mass solute/M.W. kg solvent
M.W.= Kf mass solute/∆T kg solvent
M.W.=(6.77۫C Kg/mol)(2.42 g solute)/(80.2۫C-75.7۫C)(0.0267 kg)
M.W.=154 g/mol
e) C6H5 empirical mass 77.1 g/mol
M.W./emp mass=(154g/mol)/(77.1 g/mol)=2
(C6H5) x 2=C12H10