Download Examples of Colligative properties are

Introduction
Colligative properties ==> Properties of solutions which depend on the number of solute
particles but not on their nature.
Examples of Colligative properties are:
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Vapour Pressure lowering of a solution
Boiling Point elevation
Freezing Point depression
Osmotic Pressure
Vapour Pressure Lowering
The commonality in these properties is that the effects are entropy effects. Take, for example, the
vapour pressure of a pure liquid versus one in which a solute has been dissolved. In the former
case, the difference in entropy for the phase-change reaction is greater than that for the latter
since the process of dissolving the solute into the liquid has slightly increased the entropy of the
liquid (more random since the solute is spacing out the solvent molecules a bit).
Hence, the vapour pressure of the pure liquid is higher than that of the solution.
Psolvent = solvent P°solvent
where Psolvent = vapour pressure of the solvent in solution,
solvent = mole fraction of the solvent
= # moles of solvent molecules
Total number of moles
and P°solvent = vapour pressure of the pure solvent.
Recall that the total pressure of a solution is the sum of the partial pressures of the solvent and
solute
psolution = psolvent + psolute = solvent P°solvent + solute P°solute
If the solute is non-volatile (no vapour pressure: P°solute = 0) then the total vapour pressure of
solution is
psolution = psolvent =solvent P°solvent
To see this in graphical form, see the phase diagram for pure water, below. Three lines are
present indicating the phase transition (read equilibrium) between solid, liquid and gas. Where
all three lines meet is the triple point where all three phases are in equilibrium. We see that at an
external pressure of 1 atm. the normal melting point and normal boiling point are indicated. If the
pressure is lowered, we note that the boiling point will be lowered and the melting point raised
(very slightly; it's exaggerated here).
Now look at the following diagram indicating the phase transitions for pure water and for water
with some solute dissolved in it (not to scale).
Here we see that the curves for equilibrium between the liquid and the other phases are lowered.
Look, for example at the vertical line indicating the normal boiling point of water crosses the
solution « vapour at a lower pressure, i.e., the vapour pressure at that (and any other) temperature
is lower for the solution. One could, in principle, calculate the concentration of solute which
would be required to create a given VP lowering.
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Boiling Point Elevation
We've already seen that in a solution, the vapour pressure of the solution is lowered. Thus, it is
reasonable to conclude that one would have to raise the temperature of solution more to bring it
to the boiling point. this point is well illustrated in the Phase diagram for water and solution
figured above. There is a definite relationship between the concentration of solute and the
increase in boiling point c.f. the pure liquid.
T = kb m
m = molality of solute = # moles solute / 1000 g solvent
kb = boiling point elevation constant for the liquid. We can use boiling point elevation
measurements to determine the molar mass of an unknown solute.
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.
T = 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).
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Freezing Point Depression
At the freezing point, solid and liquid are at equilibrium. The temperature where the equilibrium
occurs at a pressure of 1.0 atm is called the normal freezing point. The same entropy effects
which cause the boiling point to be elevated in a solution cause the freezing point to be
depressed. See the figure above for a visual representation of this
Since at the freezing point, the solid and liquid have the same vapour pressure and since the
vapour pressure is lower in the solution than in the pure liquid, it requires a lower temperature to
achieve equilibrium. Actually, the solid's vapour pressure too is affected by solute particles but
to a much smaller extent than that of the liquid so we can ignore it for this hand-waving
discussion.
The expression for the freezing point depression Tf is:
Tf = kf m
where kf is the freezing point depression constant for the liquid and m is the molality of the
solute.
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 ?
Tf = 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
Tf = 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.