Colligative Properties Worksheet

Workshop 1: Colligative Properties:Previously you learned the most common unit of concentration used in chemistry is Molar, M,
which is defined as moles of the solute divided by the total volume of the solution in liters.
𝑴𝒐𝒍𝒆𝒔 𝒔𝒐𝒍𝒖𝒕𝒆
=𝑴
𝑳𝒊𝒕𝒆𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
Although this unit is the most commonly used unit of concentration in chemistry, it is not the
only unit used. The volume of liquids and solutions may change with temperature, and the
concentration in molar is only accurate at the specified temperature. For systems where the
temperature may need to change another concentration unit must be used. The most common
temperature independent concentration units are shown below:
Mass percent
This is the mass of the solute divided by the total mass multiplied by 100%. This unit is
commonly used in industry.
Example: If 24.5 grams of KCl is added to 250.0 grams of water, what is the mass % of KCl in
solution:
𝑀𝑎𝑠𝑠 % =
24.5 𝑔 𝐾𝐶𝑙
× 100% = 8.92% 𝑏𝑦 𝑚𝑎𝑠𝑠
250.0 𝑔 𝑤𝑎𝑡𝑒𝑟 + 24.5 𝑔 𝐾𝐶𝑙
PPM
The mass percent unit is closely related to ppm (parts per million). Parts per million is defined
as mg solute divided by kg solvent. Notice that this is also uses the mass of both the solute and
the solvent. This unit is used in environmental science, biochemistry, and medicine.
Example: 0.085 grams of glucose (C6H12O6) is dissolved 250.0 grams of water, what is the
concentration in ppm?
1000𝑚𝑔
1𝑔
= 340𝑝𝑝𝑚 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
1𝑘𝑔
250.0 𝑔 𝑤𝑎𝑡𝑒𝑟 × 1000 𝑔
0.085 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 ×
𝑔𝑙𝑢𝑐𝑜𝑠𝑒 𝑝𝑝𝑚 =
Mole fraction
This is defined as moles solute divided by total moles. This unit is commonly used for gases and
liquids in equilbrium with their vapor.
Example: 36.0 grams of methanol (CH3OH) is dissolved in 100.0 grams of water, what is the
mole fraction?
moles methanol = 36.0 g methanol ×
1 𝑚𝑜𝑙𝑒 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙
= 1.12 𝑚𝑜𝑙𝑒𝑠 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙
32.05 𝑔 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙
𝑚𝑜𝑙𝑒𝑠 𝑤𝑎𝑡𝑒𝑟 = 100.0 𝑔 𝑤𝑎𝑡𝑒𝑟 ×
1 𝑚𝑜𝑙𝑒𝑠 𝑤𝑎𝑡𝑒𝑟
= 5.55 𝑚𝑜𝑙𝑒𝑠 𝑤𝑎𝑡𝑒𝑟
18.02 𝑔 𝑤𝑎𝑡𝑒𝑟
1.12 𝑚𝑜𝑙𝑒𝑠 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙
(1.12 𝑚𝑜𝑙𝑒𝑠 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙 + 5.55 𝑚𝑜𝑙𝑒𝑠 𝑤𝑎𝑡𝑒𝑟)
= 0.146 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙
𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙 =
Molality
The abbreviation for this unit is the lower case m. This is defined as moles solute divided by
kilogram of solvent. This unit is temperature independent (like mass percent, ppm, and mole
fraction), however the amount of solute has been convert to moles.
Example: 6.0 grams of urea (MM= 60.06 g/mol) is dissolved in 1.00 liter of water at 25oC. What
is the molality of this solution?
At 25oC the density of water is 1.00 g/mL or also:
1.00
𝑔
1𝑘𝑔
1000𝑚𝐿
𝑘𝑔
×
×
= 1.00
𝑓𝑜𝑟 𝑤𝑎𝑡𝑒𝑟 𝑎𝑡 25𝑜 𝐶
𝑚𝐿 1000𝑔
1𝐿
𝐿
Therefore the molality is:
1𝑚𝑜𝑙𝑒 𝑢𝑟𝑒𝑎
6.0 𝑔 𝑢𝑟𝑒𝑎 × 60.06 𝑔 𝑢𝑟𝑒𝑎
1.00 𝑘𝑔 𝑤𝑎𝑡𝑒𝑟
= 0.100 𝑚
Table 1: Summary of Concentration Units
Unit of concentration
Molarity
Abbreviation for unit
M
Numerator is
Moles solute
Mass percent
% by mass
Mass solute
Parts per million
Mole fraction
ppm
χ
mg solute
Moles solute
molality
m
Moles solute
Denominator is
Volume of solution in
liters
Total mass (solute +
solvent)
kg solvent
Moles total (solute +
solvent)
kg solvent
Multiplier
1
100 %
1
1
1
Practice Problem 1:
45.2 grams of KNO3 is dissolved in 500.0 grams of water. Determine the concentration in mass
%, ppm, mole fraction and molality.
Using the density to convert between mass and volume
Often a solution’s concentration is given in mass percent, and the concentration in Molarity is
needed, this may be done using the density of the solution.
Example: Concentrated nitric acid is 70.0% by mass and the density is 1.40 g/mL, what is this in
M (moles/liter)
The nitric acid solution concentration in percent mass can be written as:
70.0 𝑔 𝑛𝑖𝑡𝑟𝑖𝑐 𝑎𝑐𝑖𝑑
100.0 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1.40 𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
and the density can be written as 1 𝑚𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.
Using the factor labeling method, the conversion from percent mass to molarity (M) is:
70.0 𝑔 𝑛𝑖𝑡𝑟𝑖𝑐 𝑎𝑐𝑖𝑑
1.40 𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 1000 𝑚𝐿
1 𝑚𝑜𝑙 𝐻𝑁𝑂3
×
×
×
= 15.5 𝑀
100.0 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1𝑚𝐿
1𝐿
63.02 𝑔 𝐻𝑁𝑂3
To convert mass percent to molality:
Molality is unique among the concentration units, in that the denominator is kg solvent and not
solution. Therefore the mass of the solvent must be determined. To convert the 70.0% nitric
acid solution to molality, first find the mass of the solvent.
Using the same fraction for the mass percent as above,
70.0 𝑔 𝑛𝑖𝑡𝑟𝑖𝑐 𝑎𝑐𝑖𝑑
100.0 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
The mass of the solvent is 100.0 grams of solution – 70.0 g nitric acid = 30.0 grams solvent. We
can start by writing, 70.0 g HNO3 divided by 30.0 g solvent and using the factor labeling
method, convert to moles solute per kg solvent.
70.0 𝑔 𝑛𝑖𝑡𝑟𝑖𝑐 𝑎𝑐𝑖𝑑
1000 𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
1 𝑚𝑜𝑙 𝐻𝑁𝑂3
×
×
= 37.0 𝑚
30.0 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
1𝑘𝑔
63.02 𝑔 𝐻𝑁𝑂3
Notice that the molarity and the molality are very different. Is that always the case? What if in
stead of a very concentrated solution our solution is very diliute, a 0.50% by mass glucose
solution (MM=180 g/mol). The density of a dilute solution will be very close the density of
water, therefore assume the density is 1.00 g/mL. The molarity is:
0.50 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
1.00 𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 1000 𝑚𝐿 1 𝑚𝑜𝑙 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
×
×
×
= 0.028 𝑀
100.0 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1𝑚𝐿
1𝐿
180. 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
The molality is:
0.50 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
1000 𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 1 𝑚𝑜𝑙 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
×
×
= 0.028 𝑚
99.5 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
1 𝑘𝑔
180. 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
In this case they are numerically the same within the significant figures. If the concentration is
known in any unit, and the density is also known, it is possible, using the factor labeling method,
to convert to any other concentration unit.
Example: A solution of NaCl in water is 0.25 M. The density is 1.05 g/mL. Determine the
molality, mass percent, and mole fraction.
Using a table to organize this, our initial concentration yields the first two columns, and using
density and molar mass of the solute and solvent, it is easy to complete a table, which contains
all the information necessary to convert our initial solution to any concentration unit. Remember
that for mass percent, the masses must be in the same unit.
Table 2: Data for conversion example
Moles NaCl
Volume
solution
From initial
concentration
From initial
concentration
0.25 moles
1.00 L
Mass
solution or
Mass total
volume x
density
1.00 L x
1.05kg/L
Mass solute
Mass solvent
Moles
solvent
Mole
total
Moles solute
x MM solute
0.25 mole
NaCl x 58.44
g/mol NaCl
Mass
solution –
mass solute
1.05 kg –
0.0146
Mol
solute +
mol
solvent
0.25 +
57.4
1.05 kg or
1050 g
14.6 g NaCl
or 0.0146 kg
1.035 kg or
1035 g
Mass
solvent x
MM
solvent
1035 g /
18.02
g/mol
57.4 mol
57.7
mol
0.25 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝐶𝑙
Molality is 1.035 𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 0.24 𝑚
14.6 𝑔 𝑁𝑎𝐶𝑙
Mass percent is 1050 𝑔 𝑡𝑜𝑡𝑎𝑙 × 100% = 1.4% 𝑏𝑦 𝑚𝑎𝑠𝑠
0.25 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝐶𝑙
Mole fraction is 57.7 𝑚𝑜𝑙𝑒𝑠 𝑡𝑜𝑡𝑎𝑙 = 0.0043 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛
Practice problem 2:
A solution of urea (MM= 60.62) is 0.85% by mass, and the density is 1.00 g/mL. Determine the
molarity, molality and mole fraction.
Table 3: Data table for Practice Problem 2
Moles Urea
(solute)
Volume
solution
Mass urea
divided by
molar mass
of urea
Mass solution
divided by the
density (may
convert to
liters)
Mass
solution or
Mass total
From the
initial
concentration
unit
Mass solute
Mass
solvent
Moles
solvent
Mole
total
From the
initial
concentration
unit
Mass
solution –
mass solute
Mass
solvent
divided
by molar
mass
Moles
solute +
moles
solvent
100.0 g
0.85 g
Practice Problem 3
For a 1.00 M solution of HCl with a density of 1.05 g/mL, determine the molality, mole fraction,
and mass percent.
Table 4: Data table for Practice Problem 3
Moles HCl
(solute)
Volume
solution
From initial
concentration
From initial
concentration
1.00 moles
1.00 L
Mass
solution or
Mass total
Volume x
density
Mass solute
Mass solvent
Moles
solvent
Mole
total
Moles x
Molar mass
Mass total –
mass solute
Mass
solvent
divided
by molar
mass
Moles
solute +
moles
solvent
Dilutions
When a less concentrated solution is made from a more concentrated solution this process is
called dilution. The equation used is:
𝐶1 𝑉1 = 𝐶2 𝑉2
Where C1 the concentration of initial concentrated solution, V1 is the volume used of that
solution, C2 is concentration of the new dilute solution and V2 is the final total volume of the
less dilute solution.
Example:
50.0 mL of a 2.0 M solution is diluted a new total volume of 200.0 mL. This means that 200.0
mL-50.0 mL = 150.0 mL of water will be added to the initial liquid volume of 50.0 mL. The
concentration of this diluted solution is:
𝐶2 =
𝐶1 𝑉1
𝑉2
=
2.0 𝑀 ×50.0 𝑚𝐿
200.0𝑚𝐿
Two important points:
= 0.50𝑀
1. The units for both concentration and volume will cancel out as long as the same unit is used
for both the initial and final values.
2. Always remember to use the final total volume and not the volume of water added.
A more common problem is to determine the amount of the initial concentrated solution needed
to make the desired amount of the dilute solution.
Example: To determine the amount of phosphate in sodas using the NMR instrument a set of
standards with the concentrations 0.0 ppm, 50.0 ppm, 100, ppm. 150.0 ppm and 200.0 ppm. A
volume of 2.0 total mL of each standard is required. The initial solution is 1000.0 ppm, what
volume of each solution is needed to be diluted to a total volume of 2.0 mL?
Using the dilution equation:
𝐶1 𝑉1 = 𝐶2 𝑉2
Where C1 the concentration of initial concentrated solution, V1 is the volume used of that
solution, C2 is concentration of the new dilute solution and V2 is the final total volume of the
less dilute solution. For solutions this dilute, it is acceptable to assume the new total volume, V2
is equal to the volume of the concentrated solution plus the volume of water added.
The amount of concentrated solution needed to make each standard solution is:
𝑉1 =
𝐶2 𝑉2
𝐶1
And the amount of water needed is V2-V1. In this case, V2 = 2.00 mL and C1= 1000.0 ppm
Table 5 Dilution
Concentration of standard
(C2) ppm
0.00
50.0
100.0
150.0
200.0
V1 needed 𝑽𝟏 =
𝑪𝟐 𝑽𝟐
𝑪𝟏
0.00 mL
0.10 mL
0.20 mL
0.30 mL
0.40 mL
Water needed V2-V1
2.00 mL
1.90 mL
1.80 mL
1.70 mL
1.60 mL
Practice Problem 4:
A student needs to make a set of standards for a Beers Law plot, they will need 10.0 mL of each
solution. The initial solution is 0.50 M and the standards should be 0.050 M, 0.100M, 0.150M,
0.200M, and 0.250 M, what volume of the initial 0.50 M solution should be diluted to 10.0mL to
make each standard?
Table 6 Dilution table for Practice problem 4
Concentration of standard
(C2)
V1 needed 𝑽𝟏 =
𝑪𝟐 𝑽𝟐
𝑪𝟏
Water needed V2-V1
0.050 M
0.100M
0.150 M
0.200 M
0.250 M
Example conversion and dilution:
A student needs to make a set of nitrate standards with concentrations of 0, 0.050, 0.010, 0.015,
and 0.020 ppm. There is a solution of NaNO3 with a concentration of 0.010 M. How can the
student make the needed solutions starting with this solution?
First convert the 0.010 M NaNO3 solution from M to ppm of NO3-. Ppm is mg solute/kg water.
Assume the density of the solution is 1.0 g/mL or 1.0 kg/L
Table 7
moles
Mass in Grams
NaNO3
0.050
Not needed
mg
NO30.010
0.620
solution
Not needed
1000 g
Solvent (water)
999.38 g or 0.9994
kg
620
The concentration of nitrate is
620 𝑚𝑔
= 624 𝑝𝑝𝑚
0.9994 𝑘𝑔
The student decides that an easy way to make the standards would be to make 100 ml of each
standard, and that using the 0.010 M solution as the standard would complicate the dilutions. It
would be convenient to make the 0.050 ppm solution by diluting 1.00 mL of the initial standard
to 100.0 mL of that standard. So they decide to make an intermediate standard of:
𝐶1 =
𝐶2 𝑉2 0.050 𝑝𝑝𝑚 × 100𝑚𝑙
=
= 5.00 𝑝𝑝𝑚
𝑉1
1.00 𝑚𝐿
To make 100.0 mL of this 5.00 ppm solution from the initial 624 ppm solution, the student,
setting C2 = 5.00 ppm, V2= 100.0 mL and C1=624 ppm.
𝑉1 =
𝐶2 𝑉2 5.00 𝑝𝑝𝑚 × 100𝑚𝑙
=
= 0.801 𝑚𝐿
𝐶1
624 𝑝𝑝𝑚
In summary, first the student takes 0.801 mL of the initial 0.010M solution and dilutes to a total
volume of 100.0 mL. This makes a 5.00 ppm standard. Now this solution may be diluted
according to Table 8 to make the standards.
Table 8 Dilution
Concentration of standard
(C2) ppm
0
V1 needed 𝑽𝟏 =
0.00 mL
𝑪𝟐 𝑽𝟐
𝑪𝟏
Water needed V2-V1
100.00 mL
0.050
0.10
0.150
0.200
1.00 mL
2.00 mL
3.00 mL
4.000 mL
99.0 mL
98.0 mL
97.0 mL
96.0 mL
Practice problem 5:
A student is performing an assay, which requires 20.0 mL solution of urea, which is 0.30 g/L.
The student cannot find any solid urea, but has found a solution 0.50 M urea, how should they
make the solution they need?
Colligative Properties:
Colligative properties are physical properties, which depend only the concentration of the solute,
not the identity. Pure substances have definite physical properties, melting points, boiling points,
and vapor pressure. If two substances are mixed these properties change. The fourth colligative
property, osmotic pressure, does not exist in pure substances, but only in solutions.
Raoult’s Law for a mixture of a volatile solvent and a nonvolatile solute is:
0
𝑃𝑣𝑎𝑝 = 𝑃𝑣𝑎𝑝
𝜒𝑎
The vapor pressure of the volatile solvent, Pvap is equal to the vapor pressure of the pure solvent
times the mole fraction, χa.
Example:
If a 50.0 gram sample of water is mixed with 50.0 grams of sucrose (MM=342 g/mol) and the
normal vapor pressure of pure water is 23mmHg at 25oC, what is vapor pressure of this solution?
1 𝑚𝑜𝑙𝑒
= 2.78 𝑚𝑜𝑙𝑒𝑠 𝑤𝑎𝑡𝑒𝑟
18.0 𝑔
1 𝑚𝑜𝑙𝑒
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑢𝑐𝑟𝑜𝑠𝑒 = 50.0 𝑔 ×
= 0.146 𝑚𝑜𝑙𝑒 𝑠𝑢𝑐𝑟𝑜𝑠𝑒
342 𝑔
2.78 𝑚𝑜𝑙𝑒𝑠
𝜒𝑤𝑎𝑡𝑒𝑟 =
= 0.949 𝑚𝑜𝑙𝑒𝑠
(2.78 + 0.146)𝑚𝑜𝑙𝑒𝑠
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 50.0 𝑔 ×
0
𝑃𝑣𝑎𝑝 = 𝑃𝑣𝑎𝑝
𝜒𝑎 = 23𝑚𝑚𝐻𝑔 × 0.949 = 21.8𝑚𝑚 𝐻𝑔
The vapor pressure of a mixture of two volatile components is equal to the sum of the vapor
pressures of the individual components multiplied by their respective mole fractions. That is for
compounds A and B,
𝑃𝑡𝑜𝑡 = 𝑃𝑎0 𝜒𝑎 + 𝑃𝑏0 𝜒𝑏
where Po is the normal vapor pressure of the pure component, and  is the symbol for mole
fraction. For a two-component system, which obeys Raoult’s Law a graph of Pressure versus
mole fraction should look like Figure 1.
Raoult’s Law
0.18
0.16
0.14
pressure methanol
pressure water
total pressure
Vapor Pressure (atm)
0.12
0.1
0.08
0.06
0.04
0.02
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Mole Fraction Methanol
Raoult’s Law is not obeyed by every system. If the intermolecular attraction of the two
components is stronger than the intermolecular attraction of each of the components, the system
will be less volatile, and what occurs is a negative deviation from Raoult’s Law. If the net
attraction is lower, a positive deviation from Raoult’s Law occurs. These cases are shown in
figures 2 and 3.
Negative Deviation
0.18
0.16
Pressure Methanol
Pressure Water
Total Pressure
Raoult’s Law Pressure
0.14
Pressure (atm)
0.12
0.1
0.08
0.06
0.04
0.02
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Mole Fraction
Positive Deviation from Raoult’s Law
0.18
0.16
Pressure Methanol
Pressure Water
Total Pressure
Raoult’s Law Pressure
0.14
Pressure (atm)
0.12
0.1
0.08
0.06
0.04
0.02
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Mole fraction Methanol
Example: If the vapor pressure of pure methanol is, P0methanol = 94 mmHg, and the vapor
pressure of pure water is, Powater = 23.8 mmHg. What is the total vapor pressure of a solution of
50.0 grams of water mixed with 50.0 grams of methanol?
First find the mole fraction of water and methanol.
50.0 𝑔 𝑤𝑎𝑡𝑒𝑟 ×
1𝑚𝑜𝑙𝑒 𝑤𝑎𝑡𝑒𝑟
= 2.77 𝑚𝑜𝑙𝑒 𝑤𝑎𝑡𝑒𝑟
18.02 𝑔
50.0 𝑔 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙 ×
𝜒𝑤𝑎𝑡𝑒𝑟 =
1 𝑚𝑜𝑙𝑒 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙
= 1.56 𝑚𝑜𝑙𝑒 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙
32.04 𝑔 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙
2.77 𝑚𝑜𝑙𝑒 𝑤𝑎𝑡𝑒𝑟
= 0.640
(2.77 𝑚𝑜𝑙𝑒 𝑤𝑎𝑡𝑒𝑟 + 1.56 𝑚𝑜𝑙𝑒 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙)
𝜒𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙 =
1.56 𝑚𝑜𝑙𝑒 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙
= 0.360
(2.77 𝑚𝑜𝑙𝑒 𝑤𝑎𝑡𝑒𝑟 + 1.56 𝑚𝑜𝑙𝑒 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙)
0
0
𝑃𝑡𝑜𝑡 = 𝑃𝑤𝑎𝑡𝑒𝑟
𝜒𝑤𝑎𝑡𝑒𝑟 + 𝑃𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙
𝜒𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙 = 0.640 × 23.8 𝑚𝑚𝐻𝑔 + 0.360 × 94.0 𝑚𝑚𝐻𝑔
= 49.1 𝑚𝑚𝐻𝑔
Practice problem 6:
What is the vapor pressure of a solution of 25.0 grams of glucose (MM=180 g/mol) mixed with
100.0 mL of water? Powater = 23.8 mmHg.
Practice problem 7:
What is the vapor pressure of 25.0 grams of methanol, P0methanol = 94 mmHg (CH3OH) mixed
with 125.0 grams of water?
Osmotic Pressure:
If a container of a pure volatile substance is placed in a closed system with a solution containing
a nonvolatile solute and the pure solvent, water will be transferred to the solution, and its
concentration will decrease.
This is similar to the situation if the two solutions are separated by a semipermeable membrane,
which only allows water to pass through it as in figure 2.
The osmotic pressure obeys the following relationship,  = MRT
Where, = osmotic pressure, M= molarity, R= ideal gas constant, T= Temperature. If R is
0.08201 L atm mol-1, K-1, the osmotic pressure is in atmospheres.
Example:
What is the osmotic of a solution of 25.0 grams of glucose (180.0 MM) in a total volume of
100.0 mL at 25oC?
The concentration of the solution is:
1 𝑚𝑜𝑙
25.0 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 ×
180.0 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
(
) = 1.39 𝑀
0.100 𝐿
𝜋 = 𝑀𝑅𝑇 = (1.39 𝑀) × (0.08201
𝐿 𝑎𝑡𝑚
) × 298𝐾 = 34.0 𝑎𝑡𝑚
𝑚𝑜𝑙 𝐾
For electrolyte solutions, the actual number of particles in solution is needed and the colligative
properties need to be multiplied by the Van’t Hoff factor. A reasonable estimate of the Van’t
Hoff factor is just the number of ions in an ionic compound. In practice the experimental Van’t
Hoff factor is usually somewhat lower. The theoretical versus experimental values are given in
Table 9.
Table 9 Van’t Hoff factors
Electrolyte
i (measured)
i (calculated)
Sucrose
1.0
1.0
HCl
1.9
2.0
NaCl
1.9
2.0
1.3
2.0
2.7
3.0
MgSO
4
MgCl
2
The equation for the osmotic pressure is 𝜋 = 𝑖𝑀𝑅𝑇.
Example:
What is the osmotic pressure of a 0.25M solution of NaCl at 25oC?
𝜋 = 𝑖𝑀𝑅𝑇 = 1.9 × 0.25𝑀 × 0.08201
𝐿 𝑎𝑡𝑚
× 298 = 11.6 𝑎𝑡𝑚
𝑚𝑜𝑙 𝐾
Practice Problem 8:
What is osmotic pressure of a solution of 35.0 grams of MgCl2 in water to a total volume of
450.0 mL at 25oC?