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The following lines are from my textbook:

Colloidal particles being bigger aggregates, the number of particles in a colloidal solution is smaller than a true solution. Hence, the values of colligative properties (osmotic pressure, lowering in vapour pressure, depression in freezing point and elevation in boiling point) are of smaller order compared to true solutions.

What is the explanation supporting the above statement?

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Roughly (in the ideal world), colligative properties depend on the number concentration of particles, not on what those particles are. A particle can be a micelle or vesicle or other colloidal particle such as a polymer, or a nonassociated solute molecule. For instance, a protein is a polymer usually consisting of a linear chain of aminoacids. It is considered a colloidal particle. If you compute the colligative properties of a protein solution based on the total aminoacid concentration you obtain a different result than if you compute them based on the protein concentration, which may be orders of magnitude lower than the aminoacid concentration. The same applies to other polymers or to association colloids such as micelles. The concentration that determines colligative properties is that of the micelle or polymer, not that of the (associated or linked) monomers which combined make up the polymer or association colloid.

By the way, the statement "the number of particles in a colloidal solution is smaller than a true solution" is clearly potentially confusing. Presumably what is meant is that a "true" solution does not contain polymer or associated molecules (forming say micelles).

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Suppose a solution $0.01$ M of sugar or any moderately big organic compound. If $N$ is the Avogadro number, $1$ Liter of this solution contains $0.01·N$ molecules, and this is about $6·10^{21}$ molecules. The osmotic pressure will be $ p = cRT = 0.01 RT$, because $c$ is the concentration of individual particules in the solution, here molecules.

Now suppose the molecules in this solution are not individually separated, but they are grouped in micelles containing $1000$ molecules. Each micelle is an independent particule in the solution. The number of these particules (here micelles) is $6·10^{21}/1000 = 6·10^{18}$. In moles, it is $n/N = 6·10^{18}/(6·10^{23})$ = $10^{-5}$ mol. This number determines the osmotic pressure. The osmotic pressure will be : $p = cRT = 10^{-5} RT$. This is $1000$ times smaller than without micelles.

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Here is a simple example that may help from application of the Rault's Law (see this video), where colligative properties of a solution mixes apply (linearly or with deviations) to the relative molar concentration of solute compositions and not the identity of the solute.

First, take a known amount of CaCl2 and place it into a known volume of water, but do not stir (so there are large undissolved pieces of the CaCl2, or more formally, nonassociated solute molecules). Now, by Rault's law, the partial vapor pressure of the water is lowered, but only by the ratio of the moles of water present to the sum of both moles of water plus moles of dissolved CaCl2. Clearly, this is not as effective in reducing the partial vapor pressure as if you have totally dissolved the CaCl2.

Similarly, the freezing point lowering advantage of CaCl2 (employed commercially to 'melt' ice) is not achieved if pieces of the CaCl2 are too big to be effectively dissolved.

Finally, if one lets your anhydrous CaCl2 get exposed to air, there is created some insoluble masses, which do not dissolve even on boiling. However, the boiling now occurs at a low temperature (at least theoretically) than as if the particles had dissolved.

The above directly provides an explanation of the statement: "Hence, the values of colligative properties (osmotic pressure, lowering in vapour pressure, depression in freezing point and elevation in boiling point) are of smaller order compared to true solutions"

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  • $\begingroup$ "by Raoult's law, the partial vapor pressure ... sum of both mole s of water plus moles of dissolved CaCl2. " Applying the same logic to the 'freezing point situation' I believe we can conclude the same for F.P. as well But then you say,"The freezing.. is not achieved if pieces of the CaCl2 are too big to be effectively dissolved'--- here the mole fraction theoretically becomes 1. So doesn't this mean that the freezing point gets reduced even more? I'm sorry if this sounds crazy but I'm new to high school chemistry and Raoult's Law! $\endgroup$ Mar 30 '20 at 15:32

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