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.

π΄ππππ ππππππ

=π΄

π³ππππ ππππππππ

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:

πππ π % =

24.5 π πΎπΆπ

Γ 100% = 8.92% ππ¦ πππ π

250.0 π π€ππ‘ππ + 24.5 π πΎπΆπ

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?

1000ππ

1π

= 340πππ πππ’πππ π

1ππ

250.0 π π€ππ‘ππ Γ 1000 π

0.085 π πππ’πππ π Γ

πππ’πππ π πππ =

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 Γ

1 ππππ πππ‘βππππ

= 1.12 πππππ πππ‘βππππ

32.05 π πππ‘βππππ

πππππ π€ππ‘ππ = 100.0 π π€ππ‘ππ Γ

1 πππππ π€ππ‘ππ

= 5.55 πππππ π€ππ‘ππ

18.02 π π€ππ‘ππ

1.12 πππππ πππ‘βππππ

(1.12 πππππ πππ‘βππππ + 5.55 πππππ π€ππ‘ππ)

= 0.146 ππππ πππππ‘πππ πππ‘βππππ

ππππ πππππ‘πππ πππ‘βππππ =

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

π

1ππ

1000ππΏ

ππ

Γ

Γ

= 1.00

πππ π€ππ‘ππ ππ‘ 25π πΆ

ππΏ 1000π

1πΏ

πΏ

Therefore the molality is:

1ππππ π’πππ

6.0 π π’πππ Γ 60.06 π π’πππ

1.00 ππ π€ππ‘ππ

= 0.100 π

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

Ο

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 solutionβs 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 π πππ‘πππ ππππ

100.0 πππππ ππ π πππ’π‘πππ

1.40 π π πππ’π‘πππ

and the density can be written as 1 ππΏ ππ π πππ’π‘πππ.

Using the factor labeling method, the conversion from percent mass to molarity (M) is:

70.0 π πππ‘πππ ππππ

1.40 π π πππ’π‘πππ 1000 ππΏ

1 πππ π»ππ3

Γ

Γ

Γ

= 15.5 π

100.0 πππππ ππ π πππ’π‘πππ

1ππΏ

1πΏ

63.02 π π»ππ3

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 π πππ‘πππ ππππ

100.0 πππππ ππ π πππ’π‘πππ

The mass of the solvent is 100.0 grams of solution β 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 π πππ‘πππ ππππ

1000 π π πππ£πππ‘

1 πππ π»ππ3

Γ

Γ

= 37.0 π

30.0 πππππ ππ π πππ£πππ‘

1ππ

63.02 π π»ππ3

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 π πππ’πππ π

1.00 π π πππ’π‘πππ 1000 ππΏ 1 πππ πππ’πππ π

Γ

Γ

Γ

= 0.028 π

100.0 πππππ ππ π πππ’π‘πππ

1ππΏ

1πΏ

180. π πππ’πππ π

The molality is:

0.50 π πππ’πππ π

1000 π π πππ£πππ‘ 1 πππ πππ’πππ π

Γ

Γ

= 0.028 π

99.5 πππππ ππ π πππ£πππ‘

1 ππ

180. π πππ’πππ π

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 β

mass solute

1.05 kg β

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 πππππ πππΆπ

Molality is 1.035 ππ π πππ£πππ‘ = 0.24 π

14.6 π πππΆπ

Mass percent is 1050 π π‘ππ‘ππ Γ 100% = 1.4% ππ¦ πππ π

0.25 πππππ πππΆπ

Mole fraction is 57.7 πππππ π‘ππ‘ππ = 0.0043 ππππ πππππ‘πππ

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 β

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 β

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:

πΆ1 π1 = πΆ2 π2

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:

πΆ2 =

πΆ1 π1

π2

=

2.0 π Γ50.0 ππΏ

200.0ππΏ

Two important points:

= 0.50π

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:

πΆ1 π1 = πΆ2 π2

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:

π1 =

πΆ2 π2

πΆ1

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 π½π =

πͺπ π½π

πͺπ

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 π½π =

πͺπ π½π

πͺπ

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 ππ

= 624 πππ

0.9994 ππ

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:

πΆ1 =

πΆ2 π2 0.050 πππ Γ 100ππ

=

= 5.00 πππ

π1

1.00 ππΏ

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.

π1 =

πΆ2 π2 5.00 πππ Γ 100ππ

=

= 0.801 ππΏ

πΆ1

624 πππ

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 π½π =

0.00 mL

πͺπ π½π

πͺπ

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.

Raoultβs Law for a mixture of a volatile solvent and a nonvolatile solute is:

0

ππ£ππ = ππ£ππ

ππ

The vapor pressure of the volatile solvent, Pvap is equal to the vapor pressure of the pure solvent

times the mole fraction, Οa.

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 ππππ

= 2.78 πππππ π€ππ‘ππ

18.0 π

1 ππππ

πππππ ππ π π’ππππ π = 50.0 π Γ

= 0.146 ππππ π π’ππππ π

342 π

2.78 πππππ

ππ€ππ‘ππ =

= 0.949 πππππ

(2.78 + 0.146)πππππ

πππππ ππ π€ππ‘ππ = 50.0 π Γ

0

ππ£ππ = ππ£ππ

ππ = 23πππ»π Γ 0.949 = 21.8ππ π»π

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,

ππ‘ππ‘ = ππ0 ππ + ππ0 ππ

where Po is the normal vapor pressure of the pure component, and ο£ is the symbol for mole

fraction. For a two-component system, which obeys Raoultβs 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

Raoultβs 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 Raoultβs Law. If the net

attraction is lower, a positive deviation from Raoultβs 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 π π€ππ‘ππ Γ

1ππππ π€ππ‘ππ

= 2.77 ππππ π€ππ‘ππ

18.02 π

50.0 π πππ‘βππππ Γ

ππ€ππ‘ππ =

1 ππππ πππ‘βππππ

= 1.56 ππππ πππ‘βππππ

32.04 π πππ‘βππππ

2.77 ππππ π€ππ‘ππ

= 0.640

(2.77 ππππ π€ππ‘ππ + 1.56 ππππ πππ‘βππππ)

ππππ‘βππππ =

1.56 ππππ πππ‘βππππ

= 0.360

(2.77 ππππ π€ππ‘ππ + 1.56 ππππ πππ‘βππππ)

0

0

ππ‘ππ‘ = ππ€ππ‘ππ

ππ€ππ‘ππ + ππππ‘βππππ

ππππ‘βππππ = 0.640 Γ 23.8 πππ»π + 0.360 Γ 94.0 πππ»π

= 49.1 πππ»π

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, ο° = MRT

Where, ο°= 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 πππ

25.0 π πππ’πππ π Γ

180.0 π πππ’πππ π

(

) = 1.39 π

0.100 πΏ

π = ππ
π = (1.39 π) Γ (0.08201

πΏ ππ‘π

) Γ 298πΎ = 34.0 ππ‘π

πππ πΎ

For electrolyte solutions, the actual number of particles in solution is needed and the colligative

properties need to be multiplied by the Vanβt Hoff factor. A reasonable estimate of the Vanβt

Hoff factor is just the number of ions in an ionic compound. In practice the experimental Vanβt

Hoff factor is usually somewhat lower. The theoretical versus experimental values are given in

Table 9.

Table 9 Vanβt 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 = πππ
π.

Example:

What is the osmotic pressure of a 0.25M solution of NaCl at 25oC?

π = πππ
π = 1.9 Γ 0.25π Γ 0.08201

πΏ ππ‘π

Γ 298 = 11.6 ππ‘π

πππ πΎ

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?

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