Colligative Properties
These properties are the properties of solution that DEPENDS on the NUMBER of dissolved particles in the solution, but not on the IDENTITIES of the solutes. It depends on the collective effect of the CONCENTRATION of solute particles present in the solution.
It has four commonly studied properties:
1) Freezing point depression
In Freezing Point Depression, the presence of SOLUTE LOWERS THE FREEZING POINT of a realative solution to that of pure solvent. For example, a juice, salt water or sugar water has a lower freezing point than pure water.
And this depends on the CONCENTRATION of solute particles dissolved in solution.
Its formula is written as:
Δ T f = i K f m
where K f is the freezing point depression constant for the solvent (1.86°C·kg/mol for water), m is the number of moles of solute in solution per kilogram of solvent, and i is the number of ions present per formula unit.
And the Δ T f (solution)= T f (solvent) - Δ T f
This property is commonly used in situation when community put salt on a sonwy road to prevent freezing of melted snow.
Also, the antifreeze used in automobile heating and cooling systems is a solution of water and ethylene glycol (or propylene glycol); this solution has a lower freezing point than either pure water or pure ethylene glycol.
Example
A solution of 2.95 g of in 100 g cyclohexane had a freezing point of 4.18°C, pure cyclohexane has a fp of 6.50°C. What is the molecular formula of ?
DTf = 6.5 - 4.18 = 2.32°C
kf = 20.2°C kg mol-1 (look up in tables)
This is only the approximate molar mass of the since this technique is not very accurate (only 2 or three sig figs in this experiment).
atomic molar mass of is 32.06 g/mol. It takes 8 atoms of to add up to about 257 g/mol. Thus, the molecular formula for is S8 and the true molar mass is 8 × 32.06 = 256.48 g/mol
Example: How much glycol (1,2-ethanediol), C2H6O2 must be added to 1.00 L of water so the solution does not freeze above -20°C?
kf (H2O) = 1.86°C kg mol-1
DTf = kf m
since 1.0 L has a mass of 1.0 kg we need 10.8 mol of ethylene glycol
10.8 mol × 62.1 g/mol = 670 g ethylene glycol.
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2) Boiling Point Elevation
In Boiling Point Elevation,the boiling point of a solution is HIGHER than of the pure solvent.
The formula used to calculate the change in boiling point (Δ T b ) relative to the pure solvent is similar to that used for freezing point depression:
Δ T b = i K b m ,
where K b is the boiling point elevation constant for the solvent (0.52°C·kg/mol for water), and m and i have the same meanings as in the freezing point depression formula. Note that Δ T b represents an increase in the boiling point, whereas Δ T f represents a decrease in the freezing point. As with the freezing point depression formula, this one is most accurate at low solute concentrations.
*Δ T b = T b (solution) - T b(solvent)
Example:
A sample of 1.20 g of a non-volatile organic compound is dissolved in 60.0 g benzene. The BP of solution is 80.96°C. BP of pure benzene is 80.08°C. What is the molar mass of the solute.
DT = 80.96 - 80.08 = 0.88°C
but, we only have 60.0 g benzene, not 1000 g. so # moles solute = molality × #kg solvent 
NOTE: this is only an approximate molar mass, due to the inaccuracy of the measurement (small temperature effect, hard to measure accurately). |
Freezing Point Depression Constants
| Compound | | Freezing Point (oC) | | kf (oC/m) |
| water | | 0 | | 1.853 |
| acetic acid | | 16.66 | | 3.90 |
| benzene | | 5.53 | | 5.12 |
| p-xylene | | 13.26 | | 4.3 |
| naphthalene | | 80.29 | | 6.94 |
| cyclohexane | | 6.54 | | 20.0 |
| carbon tetrachloride | | -22.95 | | 29.8 |
| camphor | | 178.75 | | 37.7 |
Boiling Point Elevation Constants
| Compound | | Boiling Point (oC) | | kb (oC/m) |
| water | | 100 | | 0.515 |
| ethyl ether | | 34.55 | | 1.824 |
| carbon disulfide | | 46.23 | | 2.35 |
| benzene | | 80.10 | | 2.53 |
| carbon tetrachloride | | 76.75 | | 4.48 |
| camphor | | 207.42 | | 5.611 |
3) Osmotic Pressure
Osmosis is a process whereby liquids pass through semi-permeable membranes, spontaneously. This process is driven by CHANGES IN ENTROPY. Consider two containers of liquid, connected by a semi-permeable membrane. The semi-permeable membrane can be simply a device which holes small enough to allow the solvent to pass through but not the solute. In one container is a pure solvent and in the other is the same solvent but with some solute dissolved in it. In this case, there will be solvent passing through in both directions randomly. The rate of the two processes depends on the relative entropy of the two sides and on the relative pressure applied to the solutions.
Since the solution will have a higher entropy than the solute, expect that the spontaneous process is the one where the solvent passes through the membrane from the pure solvent side to the solution side where the entropy is higher. This will occur until the difference in the height of the two columns of liquid (h) is large enough that the pressure caused by this column of liquid exactly stops the net flow of solvent. This pressure is equal to the osmotic pressure.
Osmotic pressure is given the symbol P (Greek equivalent of P) and the equation relating the osmotic pressure to the amount of solute is:
PV = nRT
Note that n/V is concentration in mol/liter so we can also write
P = Csolute RT
P can be measured at room temperature and is therefore, more useful for measuring solutes which might decompose at higher (boiling point elevation) temperature measurements. It is also extremely sensitive to small amounts of solute and is therefore useful for measuring very large molar mass values.
Osmosis has three types:
a) Isotonic - Two solutions have same concentration of solute
b) Hypertonic- Solution which has higher concentration of solute
c) Hypotonic- Solution with lower concentration of solute
Example:
An aqueous solution containing 1.10 g of a protein in 100 mL of solution has an osmotic pressure of 3.93 × 10
-3 atm at 25°C. What is the molar mass of the protein?
note that in this case, I used a different value for the ideal gas constant R=0.08203 L-atm/molK This is simply a convenience. The alternate way is to convert to SI units.
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or in SI units
Note the very high molar mass of the protein.
4) Vapor Pressure Lowering
In Vapor Pressure Lowering, Vapor pressure of the solvent is lowered by adding solute, resulting a solution with lower vapr pressure than of solvent (liquid)
| Psolvent = csolvent P°solvent |
where Psolvent = vapour pressure of the solvent in solution,
csolvent = mole fraction of the solvent
= # moles of solvent molecules
Total number of moles
and P°solvent = vapour pressure of the pure solvent.
Total pressure of a solution is the sum of the partial pressures of the solvent and solute
psolution = psolvent + psolute = csolvent P°solvent + csolute P°solute
If the solute is non-volatile (no vapour pressure: P°solute = 0) then the total vapour pressure of solution is
psolution = psolvent = csolvent P°solvent
