   # Chapter 3  Colligative Properties

Colligative properties depend on the number of solute particles found in solution.

Various colligative properties require different expressions of the solute particle concentration in solution to account for the property.

## 3.1  Concentration Units

There are many units used to express the concentration of a solute in a solvent.

Those listed with a * are mildly temperature dependent because they depend on a volume; the rest are temperature independent.

1. Molarity*, M = moles of solute/liters of solution = mol/L
2. Molality, m = moles of solute/kilograms of solvent = mol/kg
3. Mass Fraction, part by mass = mass of solute/mass of solution
• % by mass = mass fraction × 100
• part per million, ppm = mass fraction × 106; used for small mass fractions
• part per billion, ppb = mass fraction × 109; used for tiny mass fractions
4. Mole Fraction, χ = moles of solute/moles of solution particles
5. Volume Fraction, part by volume* = volume of solute/volume of solution
• % by volume = volume fraction × 100
##### Problem

What is the molarity of a 37% HCl by mass aqueous solution? The density of the solution is 1.20 g/mL.

 37 g HCl 100 g solution
×
 1.20 g solution 1 mL solution
×
 1000 mL 1 L
×
 1 mol HCl 36.461 g HCl
=
 12.2 mol HCl L solution
= 12.2 M HCl

———————————————————————–

##### Problem

What is the mole fraction of HCl in a 37% HCl by mass aqueous solution?

There are 37 g HCl in 100 g solution, meaning the remaining mass is 100 − 37 = 63 g H2O.

37 g HCl ×
 1 mol HCl 36.461 g HCl
= 1.015 mol HCl

63 g H2O ×
 1 mol H2O 18.0152 g H2O
= 3.497 mol H2O

χHCl =
 1.015 mol HCl 1.015 mol HCl + 3.497 mol H2O
= 0.225 HCl

It is now trivial to obtain the mole fraction of water in the solution: χH2O = 1 − 0.225 = 0.775 since the fractions must always add up to 1.

Similarly, the mass fraction of water in this solution is 63%.

———————————————————————–

##### Problem

What is the molality of a 37% HCl by mass aqueous solution?

37 g HCl ×
 1 mol HCl 36.461 g HCl
= 1.015 mol HCl

63 g H2O ×
 1 kg H2O 1000 g H2O
= 0.063 kg H2O

 1.015 mol HCl 0.063 kg H2O
= 16.1 m HCl

Note that the molarity and molality are similarly sized numbers.

Molality is often used instead of molarity when a temperature independent concentration is needed.

Molality is the only unit of concentration to use a unit of solvent in the denominator; all the rest use a unit of solution.

———————————————————————–

## 3.2  Vapor Pressure Lowering

Addition of a soluble solute to a solvent (making a solution) decreases the vapor pressure of the solvent.

The vapor pressure is the partial pressure of solvent gas that is in dynamic equilibrium with the solvent liquid.

The vapor pressure over pure solvent, Psolvent, is higher than the vapor pressure over a solution of that solvent, Psolvent.

The lowered vapor pressure of the solvent is calculated using Raoult’s Law:

 Psolvent = χsolventPsolvent∘     (1)

Raoult’s law assumes every solution behaves ideally (where solvent and solute behave identically) and indicates a linear relationship with slope, m = Psolvent, and y-intercept, b = 0 torr, for the plot of y=Psolvent versus x = χsolvent. Non-ideal solutions deviate from Raoult’s law based on the relative strength of solute-solvent and solvent-solvent intermolecular forces/interactions.

If the solute-solvent interactions are weaker than solvent-solvent interactions, then the solvent will be more easily vaporized because the solute-solvent interaction is easier to break than a solvent-solvent interaction.

If the solute-solvent interactions are stronger than solvent-solvent interactions, then the solvent will be less able to vaporize because the solute is pulling on it more strongly than other solvent would.

 solute-solvent weaker than solvent-solvent solute-solvent stronger than solvent-solvent Raoult’s law UNDERESTIMATES Psolvent Raoult’s law OVERESTIMATES Psolvent  The change in the vapor pressure of the solvent, Δ P, can be put in terms of the mol fraction of the solute by remembering that all the fractions of a solution must add up to 1, χsolvent + χsolute = 1:

 Δ P = Psolvent∘ − Psolvent = Psolvent∘ − χsolventPsolvent∘ = (1 − χsolvent)Psolvent∘ = χsolutePsolvent∘

Of course, the solute may be volatile as well meaning it could have its own vapor pressure lowered:

 Psolute = χsolutePsolute∘

The total pressure of gas due to the solvent and solute vapor, Psolution, in dynamic equilibrium with the liquid solution can be calculated by using Dalton’s Law of Partial (Vapor) Pressures:

 Psolution  = Psolvent + Psolute = χsolventPsolvent∘ + χsolutePsolute∘     (2)
##### Problem

What is the vapor pressure of H2O at 25C over a solution that is 10 mol H2O and 1 mol non-volatile, non-electrolyte solute (e.g. sugar)?

Non-volatile means the solute has negligible vapor pressure, Psolute = 0 torr.

Non-electrolyte means the solute will not dissociate into multiple smaller particles once dissolved in water, so 1 mol solute = 1 mol particles in solution.

All we need to solve this problem is χH2O since PH2O = 23.76 torr at 25C is found in data tables.

χH2O =
 10 mol H2O 10 mol H2O + 1 mol solute
= 0.909

 PH2O = χH2OPH2O∘ = 0.909 × 23.76  torr = 21.6  torr

———————————————————————–

### 3.2.1  van’t Hoff factors

Electrolytes (e.g. ionic compounds, "salts") dissociate into multiple particles when dissolved in water.

NaCl (s) ⇋ Na+ (aq) + Cl (aq)

The van’t Hoff factor, i, accounts for the number of dissolved particles created when a solute dissolves.

i =
 moles of solute particles in solution moles of solute dissolved

 compound itheo iexp NaCl 2 1.9 FeCl3 4 3.4 C6H12O6 (glucose) 1 1

Theoretical values of the van’t Hoff factor, itheo, are simply the number of ions contained in the empirical formula.

Experimental values of the van’t Hoff factor, iexp, are always less than itheo and represent the effective number of solute particles.

iexp is always less than itheo because of transient ion-pairing in solution which reduces the effective number of particles in solution; Na+ and Cl are still attracted to each other in solution and so could ion-pair in solution back into 1 "particle".

All non-electrolytes have i = 1 by definition since they do not dissociate into ions.
Any time the moles of solute enters into a colligative property equation through a concentration unit like molarity, molality, or mole fraction, it must be multiplied by the van’t Hoff factor, i, to convert it to the moles of particles actually found in solution.

##### Problem

What is the vapor pressure of H2O at 25C over a solution that is 10 mol H2O and 1 mol non-volatile NaCl?

NaCl certainly has no vapor pressure, PNaCl = 0 torr so our only concern it its effect on the mole fraction of water, χH2O.

χH2O =
 10 mol H2O 10 mol H2O + 1 mol solute
=
 10 mol H2O 10 mol H2O + iexp × 1 mol NaCl

χH2O =
 10 mol H2O 10 mol H2O + 1.9 × 1 mol NaCl
=
 10 11.9
= 0.840
 PH2O = χH2OPH2O∘ = 0.840 × 23.76  torr = 20.0  torr

———————————————————————–

##### Problem

Calculate χNaCl for the previous problem.

χNaCl =
 iexp × 1 mol NaCl 10 mol H2O + iexp × 1 mol NaCl
=
 1.9 11.9
= 0.160

Of course, we could have obtained this from χNaCl = 1 − χH2O = 1 − 0.840 = 0.160.

———————————————————————–

##### Problem

The total pressure over a solution of pentane (C5H12) and hexane (C6H14) is 365 torr. What is the mole fraction of hexane in the solution? Ppentane = 425 torr and Phexane = 151 torr.

Both the solute and solvent have vapor pressures, however we do not know which one is the solute or solvent; this is not a problem as can still apply equation 3.2.

 365  torr = χpentane Ppentane∘ + χhexane Phexane∘ = χpentane (425  torr) + χhexane (151  torr)

Since this solution only has 2 components, χpentane + χhexane = 1 which solved for χpentane = 1− χhexane.

Making the substitution and solving for χhexane:

 365  torr = (1 − χhexane) (425  torr) + χhexane (151  torr)
 365  torr = 425  torr − (425  torr)χhexane + (151  torr)χhexane
 −60  torr = (−274  torr)χhexane
 0.219 = χhexane  and  χpentane = 1 − 0.219 = 0.781

Hexane is the solute and pentane is the solvent.

———————————————————————–

##### Problem

What are the partial pressures of pentane and hexane in the vapor phase?

Simply use equation 3.1 for each component separately.

 Ppentane = χpentane Ppentane∘ = 0.781 × 425  torr = 331.9  torr
 Phexane = χhexane Phexane∘ = 0.219 × 151  torr = 33.1  torr

———————————————————————–

##### Problem

What are the mole fractions of pentane and hexane in the gas phase?

For this, we need to remember that partial pressure is to the total pressure as partial moles is to the total moles of gas.

χpentanegas =
 npentane npentane+nhexane
=
 Ppentane Ppentane+Phexane
=
 331.9  torr 365  torr
= 0.9093

This gives χhexanegas = 1 − .9093 = 0.0907.

Note that the mole fraction of pentane in the solution, χpentane = 0.781 is lower than the mole fraction of pentane in the gas phase above the solution, χpentanegas=0.9093.

This difference in composition between the solution and gas phases is often exploited to separate mixtures; pentane could be separated from hexane by sequentially separating the gas phase from the solution phase and then condensing the gas phase with its new set of mole fractions.

———————————————————————–

## 3.3  Boiling Point Elevation

Boiling point, Tb is the temperature at which the vapor pressure equals the atmospheric/external pressure.

Assuming the normal atmosphere is at 1 atm, the vapor pressure of H2O is PH2O = 760 torr = 1 atm at 100C, so the normal boiling point of H2O is Tb = 100C.
Boiling point elevation occurs because the presence of a solute reduce’s a liquid’s vapor pressures as evidenced by Raoult’s law: Psolvent = χsolventPsolvent. Higher temperatures are then needed to reattain a vapor pressure equal to the external/atmospheric pressure.

• The volatile solvent liquid is a smaller fraction of the molecules at the surface so there are fewer solvent molecules capable of entering the gas phase from the liquid phase, but there is no restriction on the condensation of the gas phase back to the solution upon coming into contact with either solute or solvent molecule.
• Thus, the rate of vaporization is restricted, limiting the amount of solvent liquid in the gas phase, whereas the rate of condensation is not; the result being reduced vapor pressures at all temperatures.

The change in boiling point, Δ Tb, requires the molality of the solute, m, and the solvent’s boiling point elevation constant, Kb, and is given by:

 Δ Tb = m Kb

The elevated boiling point is obtained from:

 Tbsolution = Tbsolvent + Δ Tb
##### Problem

What is the boiling point of a 0.210 m NaCl solution? From data tables, KbH2O = 0.512 C/m.

This sounds easy since I was given the molality, just plug and chug.

 Δ Tb = (0.210 m NaCl) (0.512 ∘C/m)

This is wrong because molality has moles of solute in the numerator, therefore the van’t Hoff factor needs to be applied giving us the real equation:

 Δ Tb = i m Kb     (3)
 Δ Tb = iNaCl  (0.210 m NaCl) (0.512 ∘C/m)
 Δ Tb = (1.9)  (0.210 m NaCl) (0.512 ∘C/m) = (0.399 m solute) (0.512 ∘C/m) = 0.204∘C
 Tbsolution =  100∘C + 0.204∘C = 100.204∘C

This is a small effect.

———————————————————————–

## 3.4  Freezing Point Depression

Freezing point depression occurs because the solute is now in the way of the solvent molecules trying to freeze to each other which lends some extra fluidity to the solution.

The change in the freezing point, Δ Tf, requires the molality of the solute, m, it’s van’t Hoff factor, i, and the solvent’s freezing point depression constant, Kf, to give

 Δ Tf = i m Kf     (4)

The new freezing point is then:

 Tfsolution = Tfsolvent − Δ Tf
##### Problem

If 20.5 g solute in 802 g H2O gives Tfsolution = −0.264 C, what is the molar mass of the solute assuming i = 1? KfH2O = 1.86 C/m.

We have the mass of the solute, so all we need to figure out is how many moles of solute would cause the freezing point of H2O to fall by 0.264C and then divide the two quantities to get units of g/mol, the units of molar mass.

 Δ Tf = 0.0 − −0.264 = +0.264∘C
 0.264∘C = i m Kf = (1) m (1.86 ∘C/m)
m =
 0.264∘C 1.86 ∘C/m
= 0.142 m =

 0.142 mol solute kg solvent

Cross multiply and divide with the information we know from the problem.

 0.142 mol solute kg solvent
=
 x mol solute 0.802 kg H2O

 x mol solute = 0.142 m × .802 kg H2O = 0.114 mol solute

Assemble the grams and moles of solute to obtain its molar mass assuming i = 1.

 20.5 g solute 0.114 mol solute
= 180 g/mol solute

It just so happens this is the molar mass of glucose, C6H12O6, which would indeed have i = 1.

If you were to consider other possibilities for i, the molar mass above would be (g solute) per (mol of particles) instead of (mol of solute), 180 g/(mol particles).

i = 2, the molar mass of the solute would be:

 180 g mol particles
×
 2 mol particles 1 mol solute
= 360 g/mol solute

———————————————————————–

## 3.5  Osmosis

Osmosis is the flow of solvent through a semi-permeable membrane from a region of low solute concentration to a region of high solute concentration.

The membrane is semi-permeable in that it allows only the solvent to flow through while excluding the solute. The height difference, h, is indicative of the solution’s osmotic pressure, P:

P =
 F A
=
 ma A
=
 Vdg A
=
 Ahdg A
= hdg

Where F is force exerted by the extra liquid on the solution side of the osmosis cell, A is cross sectional area the force is applied to, m is the mass of the extra liquid on the solution side of the osmosis cell, a=g is acceleration due to gravity, V is the volume of the extra liquid on the solution side of the osmosis cell, and d is the density of the solution.

The osmotic pressure is related to the weight (m × a) of extra solution volume as indicated by h.

This is similar to the operation of a U-tube manometer which measures pressure differences between two separate gases by measuring the height difference of the liquid the separates the two gases.

The use of height as a measure of pressure is reflected in the use of such units as mmHg.

Since we’re talking about pressure, might as well pull out the ideal gas law:

 PV = nRT     (5)

Replacing P with the symbol for osmotic pressure, Π (capital pi), and then solving for Π on the left hand side with the realization that n/V = M, molarity:

Π =
 nRT V
=
 n V
RT = MRT

Don’t forget to account for the any potential van’t Hoff factor and you have the osmotic pressure equation:

 Π = iMRT     (6)

Another perspective of osmotic pressure, Π, is the amount of external gas pressure needed on the solution side of the osmosis cell to push the solvent back through the membrane until the liquid levels are even again, h = 0.

Reverse osmosis is the process of applying even more pressure than Π to the solution side to force the solvent (e.g. water) to go the opposite way through the semi-permeable membrane to produce purified solvent.

##### Problem

What is the osmotic pressure of a 300. mL solution of glycerin containing 20.0 g of glycerin, C3H8O3, at 25C?

Glycerin is a derivatized hydrocarbon (being composed of C, H, and O) and as such as i = 1.

The universal gas constant, R = 0.08206 L· atm/mol· K.

The temperature must be expressed in Kelvin, T = 25C + 273.15 = 298.15 K.

x mol C3H8O3 =
 20.0 g C3H8O3 92.0944 g/mol
= 0.217 mol C3H8O3

Π =
(1)

 0.217 mol C3H8O3 0.300 L

0.08206 L· atm/mol· K
(298.15 K) = 17.7 atm

Osmotic pressures can be very large for even dilute solutions, and so osmotic pressure is a good way to get a rough idea of a solution concentration because its values will be large enough to measure accurately whereas boiling point elevation or freezing point depression could be difficult to measure accurately since the magnitude of their effect is usually small.

———————————————————————–   