Chapter Fourteen
Thinking It Through
T14.1
This is a Henry's Law calculation. Each pressure must be in the same units, either both in atm or both in
torr, before use of the following equation:
C1
C
= 2
P1
P2
where C1, P1, and P2 are given. The temperature is constant.
T14.2 The solution is 10.0 % sulfuric acid, meaning that there are 10.0 g of sulfuric acid in every 100 g of
solution. This mass is first converted to a mole amount, in order to get the necessary conversion factor:
1 mol H SO 
2
4
# mol H 2SO 4 = 10.0 g H 2SO 4 

98.0
g
H
SO

2
4 
There are thus 0.102 mol H2SO4 per 100 g of solution, and this conversion factor can now be used to solve
the problem:

 100 g solution 
# g solution = 0.100 mol H 2SO 4 

0.102 mol H 2SO 4 
To determine the number of milliliters we would need to know the density of the solution.
T14.3 The mass percent tells us that there are 12.5 g of sugar per 100 g of solution. Thus, we have 100.0 – 12.5 =
87.5 g of water. The moles of sugar are determined by dividing mass (12.5 g) by molecular mass. The
kilograms of water are found by dividing 87.5 g by 1000. Molality is then moles of sugar divided by
kilograms of water.
T14.4
The mole fraction tells us that there is the ratio of 0.0100 mole of NaCl for every mole of substance in the
solution. The mass of the NaCl is determined by multiplying the moles of NaCl (0.0100 mol) by the
formula mass of NaCl (58.44 g mol–1). The mass of the water is determined from the moles of water (1
mol of solution – 0.0100 mol NaCl = 0.99 mol H2O) and the formula mass of water (18.02 g mol–1).
Multiply the moles of water by its formula mass. Then, using the mass of the NaCl and the total mass of
the solution (mass of NaCl + mass of water) determine the mass percent of the NaCl in solution.


mass of NaCl
Mass percent = 
100%
mass of water + mass of NaCl 
T14.5
The molality indicates that there are 0.750 mole of solute per 1000 g of water in this solution. The number
of moles of water in 1000 g is:

 1 mol H O 
2
# mol H 2O = (1000 g H 2O) 
 = 55.49 mol H 2O
18.02 g H 2O 
The mole fraction of solute is then 0.750 mol divided by the total number of moles (55.49 + 0.750 = 56.24).
T14.6
The molality is 0.125, meaning 0.125 moles of solute per 1000 g of solvent, not per 1000 g of solution. We
therefore cannot simply weigh out the solution. First, it is necessary to calculate the number of grams of
solute by multiplying 0.125 moles by the formula mass:
0.125 mol  53.5 g/mol = 6.69 g
Therefore, an equivalent value to describe this solution would be to state the concentration in units of (g
solute)  (g solution). In this solution, we have
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Chapter Fourteen
6.69 g NH 4 Cl
1006.69 g solution
We can use this conversion factor, along with the molar mass in order to determine the number of grams
needed to obtain the required amount of NH 4Cl.

In order to calculate volume in mL, we first need to know the density of the solution.
T14.7
In order to calculate molarity, we need to know the number of moles of H2SO4 in each liter of solution.
There are 50.0 g of H2SO4 in every 100 g of solution. We find moles of H2SO4 by dividing 50.0 g by the
molecular mass (98.0 g/mol). The volume of the solution is found by dividing the mass (100 g) by the
density, which is the same as the specific gravity, at this temperature. Molarity is finally determined by
dividing moles of H2SO4 by the volume (in L) of the solution.
T14.8
We can use Raoult’s Law, Psolution = Xsolvent Psolvent. We know Psolution and Psolvent so we can determine
Xsolvent. We also know the mass of solvent (100 g) so we can determine the total number of moles. Subtract
the moles of solvent from the total number of moles of solute. To get mass of solute, multiply the number
of moles solute by the molar mass.
T14.9
We need the molality, to get the number of moles of the solution. For simplicity, assume you have 1000
mL of water (the solvent) (1 kg) and 1000 mL of ethylene glycol. We know the density of ethylene glycol,
so we can determine the mass and the number of moles of ethylene glycol, which are then used to calculate
m. Then using Tf = kfm and Tb = kbm we can calculate the change in freezing and boiling points.
T14.10 We must determine the Raoult's Law pressure for each volatile component separately, and then add the two
to get the total vapor pressure, since each component is volatile. Multiply the mole fraction of
tetrachloroethylene (0.450) by its pure vapor pressure (40.0 torr), multiply the mole fraction of methyl
acetate (1 – 0.450 = 0.550) by its pure vapor pressure (400 torr), and add the two partial pressures together
to get the total vapor pressure above the mixture.
T14.11 Multiply the molarity by the gas constant (0.0821 L atm/K mol) and by the temperature (293 K), as in the
equation for osmotic pressure of a solution:  = MRT
T14.12 The initial solution consisted of 15.0 g of NaCl and 135 g of water. Adding 75.0 g of water increases the
total mass to 225 g. The new mass percent is 15.0 g/225 g = 6.67 %.
T14.13 Start by assuming there is 1 kg solvent present. Use the equation Tf = ikfm. We know that Tf = 4.50C, i
= 4 (Al3+ +3Cl–) and kf for water. We can solve for m and determine the number of moles AlCl 3 and H2O.
From these we can determine XAlCl3 and XH2O. The vapor pressure of the solution can be determined from
Raoult’s Law. We need to know the vapor pressure of water at 35C to complete the calculation.
T14.14 Using Raoult’s Law, determine the mole fraction of H2O. We know the mass of water, 150 g, so we can
determine the number of moles of water, and from the mole fraction, determine the number of moles of
glycerol. To raise the vapor pressure, we have to dilute the solution.
Determine the Xsolvent for the higher vapor pressure using Raoult’s Law. Determine the total number of
moles in the new solution using the number of moles solute from the first part and the new mole fraction.
Using the total number of moles and the mole fraction of solvent, determine the number of moles solvent.
Convert this to a mass and subtract the original 150 g of water to determine the amount of water to add.
T14.15 The easiest way to solve this problem is to set up a ratio of the two experiments:
270
Chapter Fourteen
Tf (2)
m(2)

Tf (1)
m(1)
moles solute
and the amount of solvent is constant we can simplify to
# kg solvent
Tf (2)
moles (2)

Tf (1)
moles (1)
Tf (1)
moles(1)  moles(2)
Tf (2)
mass
T (1)
and since moles 
, mass(1)  mass(2) f
molar mass
Tf (2)
If we let x = mass of solute present in the initial solution, we have:
since m 

0.835 C 

x  (x  1.20) 

  (x  1.20)(0.799)
1.045 C 
x = 4.77 g
T14.16 (a)

(b)
T14.17 (a)
(b)
No. Vapor pressure is a fixed value (at a constant T) irrespective of container size or surface area
(so long as sufficient liquid is present to provide enough vapor). The surface area simply controls
the rate of evaporation.
The total energy of the two beakers is the same if the temperature is the same. The molecules that
compose the beaker on the right will exhibit the same energy distribution as the molecules that
compose the beaker on the left. However, the beaker on the right will have fewer water molecules
possessing sufficient energy to overcome the intermolecular forces of attraction. (This results
because some of the high-energy particles will be non-volatile solute particles.) As a result, the
observed vapor pressure is lower in the beaker on the right than it is in the beaker on the left.
The van’t Hoff factor for acetic acid should be slightly greater than one. Acetic acid is a weak acid
and some of the acid will dissociate in aqueous solution.
The van’t Hoff factor of 0.5 indicates that there is an association of acetic acid molecules in this
non-polar solvent. Two acetic acid molecules are attracted to each other through hydrogen
bonding. The apparent molality of the solution is reduced due to the formation of these dimers.
T14.18 See Figures 14.5 and 14.7
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